Rd Sharma XII Vol 1 2020 for Class 12 Commerce Maths Chapter 18 - Indefinite Integrals

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Maths Chapter 18 Indefinite Integrals are provided here with simple step-by-step explanations. These solutions for Indefinite Integrals are extremely popular among Class 12 Commerce students for Maths Indefinite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Commerce Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 18.104:

Question 1:

$\int \frac{x}{x^{2} + 3 x + 2} d x$

Answer:

$\int \frac{x}{x^{2} + 3 x + 2} d x x = A \frac{d}{d x} (x^{2} + 3 x + 2) + B x = A (2 x + 3) + B x = (2 A x) + 3 A + B$

Comparing the Coefficients of like powers of x we get
$2 A = 1 \Rightarrow A = \frac{1}{2} 3 A + B = 0 \frac{3}{2} + B = 0 B = - \frac{3}{2} x = \frac{1}{2} (2 x + 3) - \frac{3}{2}$

$Now, \int \frac{x}{x^{2} + 3 x + 2} d x = \int [\frac{\frac{1}{2} (2 x + 3) - \frac{3}{2}}{x^{2} + 3 x + 2}] d x = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{x^{2} + 3 x + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - \frac{9}{4} + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{2} \log |x^{2} + 3 x + 2| - \frac{3}{2} \times \frac{1}{2 \times \frac{1}{2}} \log |\frac{x + \frac{3}{2} - \frac{1}{2}}{x + \frac{3}{2} + \frac{1}{2}}| + C = \frac{1}{2} \log |x^{2} + 3 x + 2| - \frac{3}{2} \log |\frac{x + 1}{x + 2}| + C$

Page No 18.104:

Question 2:

$\int \frac{x + 1}{x^{2} + x + 3} d x$

Answer:

$\int \frac{(x + 1) d x}{x^{2} + x + 3} x + 1 = \frac{A d}{d x} (x^{2} + x + 3) + B x + 1 = A (2 x + 1) + B x + 1 = 2 A x + A + B$

Comparing Coefficients of like powers of x
$2 A = 1 A = \frac{1}{2} A + B = 1 \frac{1}{2} + B = 1 B = \frac{1}{2} (x + 1) = \frac{1}{2} (2 x + 1) + \frac{1}{2}$

$Now, \int \frac{(x + 1) d x}{x^{2} + x + 3} = \int \frac{\frac{1}{2} (2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{x^{2} + x + 3} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 3} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + 3 - \frac{1}{4}} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} = \frac{1}{2} \log |x^{2} + x + 3| + \frac{1}{2} \times \frac{2}{\sqrt{11}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{11}}{2}}) + C = \frac{1}{2} \log |x^{2} + x + 3| + \frac{1}{\sqrt{11}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}}) + C$

Page No 18.104:

Question 3:

$\int \frac{x - 3}{x^{2} + 2 x - 4} d x$

Answer:

$\int (\frac{x - 3}{x^{2} + 2 x - 4}) d x x - 3 = A \frac{d}{d x} (x^{2} + 2 x - 4) + B x - 3 = A (2 x + 2) + B x - 3 = (2 A) x + 2 A + B$

Comparing Coefficients of like powers of x

$2 A = 1 A = \frac{1}{2} 2 A + B = - 3 2 \times \frac{1}{2} + B = - 3 B = - 4$

$Now, \int (\frac{x - 3}{x^{2} + 2 x - 4}) d x = \int (\frac{\frac{1}{2} (2 x + 2) - 4}{x^{2} + 2 x - 4}) d x = \frac{1}{2} \int \frac{(2 x + 2) d x}{(x^{2} + 2 x - 4)} - 4 \int \frac{d x}{x^{2} + 2 x + 1 - 1 - 4} = \frac{1}{2} \int \frac{(2 x + 2) d x}{(x^{2} + 2 x - 4)} - 4 \int \frac{d x}{{(x + 1)}^{2} - {(\sqrt{5})}^{2}} = \frac{1}{2} \log |x^{2} + 2 x - 4| - \frac{4}{2 \sqrt{5}} \log |\frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}}| + C = \frac{1}{2} \log |x^{2} + 2 x - 4| - \frac{2}{\sqrt{5}} \log |\frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}}| + C$

Page No 18.104:

Question 4:

$\int \frac{2 x - 3}{x^{2} + 6 x + 13} d x$

Answer:

$\int \frac{(2 x - 3) d x}{x^{2} + 6 x + 13} 2 x - 3 = A \frac{d}{d x} (x^{2} + 6 x + 13) + B 2 x - 3 = A (2 x + 6) + B 2 x - 3 = (2 A) x + 6 A + B$

Comparing Coefficients of like powers of x
$2 A = 2 A = 1 6 A + B = - 3 6 + B = - 3 B = - 9 ∴ 2 x - 3 = 1 (2 x + 6) - 9$

$∴ \int \frac{(2 x - 3)}{x^{2} + 6 x + 13} d x = \int (\frac{2 x + 6 - 9}{x^{2} + 6 x + 13}) d x = \int (\frac{2 x + 6}{x^{2} + 6 x + 13}) d x - \int \frac{9 d x}{x^{2} + 6 x + 13} = \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 13} - 9 \int \frac{d x}{x^{2} + 6 x + 3^{2} - 3^{2} + 13} = \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 13} - 9 \int \frac{d x}{{(x + 3)}^{2} + 2^{2}} = \log |x^{2} + 6 x + 13| - 9 \times \frac{1}{2} \tan^{- 1} (\frac{x + 3}{2}) + C = \log |x^{2} + 6 x + 13| - \frac{9}{2} \tan^{- 1} (\frac{x + 3}{2}) + C$

Page No 18.104:

Question 5:

$\int \frac{x - 1}{3 x^{2} - 4 x + 3} d x$

Answer:

$\int (\frac{x - 1}{3 x^{2} - 4 x + 3}) d x x - 1 = A \frac{d}{d x} (3 x^{2} - 4 x + 3) + B x - 1 = A (6 x - 4) + B x - 1 = (6 A) x + B - 4 A$

Comparing the Coefficients of like powers of x
$6 A = 1 A = \frac{1}{6} B - 4 A = - 1 B - 4 \times \frac{1}{6} = - 1 B = - 1 + \frac{2}{3} B = \frac{1}{3}$

$Now, \int \frac{(x - 1) d x}{3 x^{2} - 4 x + 3} = \int [\frac{\frac{1}{6} (6 x - 4) + \frac{1}{3}}{3 x^{2} - 4 x + 3}] d x = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{3} \int \frac{d x}{3 x^{2} - 4 x + 3} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + {(\frac{2}{3})}^{2} {(\frac{2}{3})}^{2} + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{{(x - \frac{2}{3})}^{2} - \frac{4}{9} + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 13} + \frac{1}{9} \int \frac{d x}{{(x - \frac{2}{3})}^{2} + {(\frac{\sqrt{5}}{3})}^{2}} = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{1}{9} \times \frac{3}{\sqrt{5}} \tan^{- 1} (\frac{x^{- \frac{2}{3}}}{\frac{\sqrt{5}}{3}}) + C = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{1}{3 \sqrt{5}} \tan^{- 1} (\frac{3 x - 2}{\sqrt{5}}) + C = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{\sqrt{5}}{15} \tan^{- 1} (\frac{3 x - 2}{\sqrt{5}}) + C$

Page No 18.104:

Question 6:

$\int \frac{2 x}{2 + x - x^{2}} d x$

Answer:

$\int \frac{2 x d x}{(2 + x - x^{2})} 2 x = A \frac{d}{d x} (2 + x - x^{2}) + B 2 x = A (0 + 1 - 2 x) + B 2 x = (- 2 A) x + A + B$

Comparing the Coefficients of like powers of x
$- 2 A = 2 A = - 1 A + B = 0 - 1 + B = 0 B = 1$

$Now, \int \frac{2 x d x}{(2 + x - x^{2})} = \int (\frac{- 1 (1 - 2 x) + 1}{- x^{2} + x + 2}) d x = - \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x + \int \frac{d x}{- x^{2} + x + 2} = - I_{1} + I_{2} . . . (1) (say) where I_{1} = \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x I_{2} = \int \frac{d x}{- x^{2} + x + 2} I_{1} = \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x let - x^{2} + x + 2 = t \Rightarrow (1 - 2 x) d x = d t I_{1} = \int \frac{d t}{t} I_{1} = \log |t| + C_{1} = \log |2 + x - x^{2}| + C_{1} . . . (2) I_{2} = \int \frac{d x}{- x^{2} + x + 2} I_{2} = \int \frac{- d x}{x^{2} - x - 2} I_{2} = \int \frac{- d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2} I_{2} = \int \frac{- d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} I_{2} = - \frac{1}{2 \times \frac{3}{2}} \log |\frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}}| + C_{2} I_{2} = - \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{2} . . . (3) from (1) (2) and (3) \int (\frac{2 x}{2 + x - x^{2}}) d x = - \log |2 + x - x^{2}| - \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{1} + C_{2} = - \log |2 + x - x^{2}| + \frac{1}{3} \log |\frac{1 + x}{x - 2}| + C where C = C_{1} + C_{2}$

Page No 18.104:

Question 7:

$\int \frac{1 - 3 x}{3 x^{2} + 4 x + 2} d x$

Answer:

$\int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} 1 - 3 x = A \frac{d}{d x} (3 x^{2} + 4 x + 2) + B 1 - 3 x = A (6 x + 4) + B 1 - 3 x = (6 A) x + 4 A + B$

Comparing the Coefficients of like powers of x

$6 A = - 3 A = \frac{- 1}{2} 4 A + B = 1 4 \times \frac{- 1}{2} + B = 1 B = 3$

$1 - 3 x = - \frac{1}{2} (6 x + 4) + 3 Now, \int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} = \int (\frac{\frac{- 1}{2} (6 x + 4) + 3}{3 x^{2} + 4 x + 2}) d x = - \frac{1}{2} \int \frac{(6 x + 4) d x}{3 x^{2} + 4 x + 2} + 3 \int \frac{d x}{3 x^{2} + 4 x + 2} = - \frac{1}{2} I_{1} + 3 I_{2} (say) . . . (1) where I_{1} = \int \frac{6 x + 4}{3 x^{2} + 4 x + 2} and I_{2} = \int \frac{d x}{3 x^{2} + 4 x + 2} I_{1} = \int (\frac{6 x + 4}{3 x^{2} + 4 x + 2}) d x let 3 x^{2} + 4 x + 2 = t \Rightarrow (6 x + 4) d x = d t I_{1} = \int \frac{d t}{t} = \log |t| + C_{1} = \log |3 x^{2} + 4 x + 2| + C_{1} . . . (2) I_{2} = \int \frac{d x}{3 x^{2} + 4 x + 2} I_{2} = \frac{1}{3} \int \frac{d x}{x^{2} + \frac{4}{3} x + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{x^{2} + \frac{4 x}{x} + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x - \frac{2}{3})}^{2} - \frac{4}{9} + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x + \frac{2}{3})}^{2} + \frac{2}{9}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x + \frac{2}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} I_{2} = \frac{1}{3} \times \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{x + \frac{2}{3}}{\frac{\sqrt{2}}{3}}) + C_{2} I_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C_{2} . . . (3) from (1), (2) and (3) \int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} = - \frac{1}{2} \log |3 x^{2} + 4 x + 2| + \frac{3 \times 1}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C_{1} + C_{2} = - \frac{1}{2} \log |3 x^{2} + 4 x + 2| + \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 8:

$\int \frac{2 x + 5}{x^{2} - x - 2} d x$

Answer:

$\int \frac{(2 x + 5) d x}{x^{2} - x - 2} 2 x + 5 = A \frac{d}{d x} (x^{2} - x - 2) + B 2 x + 5 = A (2 x - 1) + B 2 x + 5 = (2 A) x + B - A$

Comparing the Coefficients of like powers of x

$2 A = 2 A = 1 B - A = 5 B - 1 = 5 B = 6$

$∴ 2 x + 5 = 1 \cdot (2 x - 1) + 6 ∴ \int (\frac{2 x + 5}{x^{2} - x - 2}) d x \Rightarrow \int (\frac{(2 x - 1) + 6}{x^{2} - x - 2}) d x \Rightarrow \int (\frac{2 x - 1}{x^{2} - x - 2}) d x + 6 \int \frac{d x}{x^{2} - x - 2} = I_{1} + 6 I_{2} (say) . . . (1) where I_{1} = \int (\frac{2 x - 1}{x^{2} - x - 2}) d x I_{2} = \int \frac{d x}{x^{2} - x - 2} I_{1} = \int (\frac{2 x - 1}{x^{2} - x - 2}) d x let x^{2} - x - 2 = t \Rightarrow (2 x - 1) d x = d t I_{1} = \int \frac{d t}{t} I_{1} = \log (t) I_{1} = \log |x^{2} - x - 2| + C_{1} . . . (2) I_{2} = \int \frac{d x}{x^{2} - x - 2} I_{2} = \int \frac{d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2} I_{2} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} - \frac{1}{4} - 2} I_{2} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} I_{2} = \frac{1}{2 \times \frac{3}{2}} \log |\frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}}| I_{2} = \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{2} . . . (3) \int \frac{(2 x + 5) d x}{x^{2} - x - 2} = \log |x^{2} - x - 2| + \frac{6}{3} \log |\frac{x - 2}{x + 1}| + C_{1} + C_{2} = \log |x^{2} - x - 2| + 2 \log |\frac{x - 2}{x + 1}| + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 9:

$\int \frac{a x^{3} + b x}{x^{4} + c^{2}} d x$

Answer:

$\int \frac{(a x^{3} + b x)}{x^{4} + c^{2}} d x = \int \frac{a x^{3}}{x^{4} + c^{2}} d x + \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x = I_{1} + I_{2} (say) Where I_{1} = \int \frac{a x^{3}}{x^{4} + c^{2}} d x & I_{2} = \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x Now, I_{1} = \int \frac{a x^{3}}{x^{4} + c^{2}} d x let x^{4} + c^{2} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} d x = \frac{d t}{4} I_{1} = \frac{a}{4} \int \frac{d t}{t} = \frac{a}{4} \log |t| + C_{1} = \frac{a}{4} \log |x^{4} + c^{2}| + C_{1} Now, I_{2} = \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x let x^{2} = p \Rightarrow 2 x d x = d p \Rightarrow x d x = \frac{d p}{2}$

$I_{2} = \frac{b}{2} \int \frac{d p}{p^{2} + c^{2}} = \frac{b}{2} \times \frac{1}{c} \tan^{- 1} (\frac{p}{c}) + C_{2} = \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C_{2} \int \frac{(a x^{3} + b x)}{x^{4} + c^{2}} d x = \frac{a}{4} \log |x^{4} + c^{2}| + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C_{1} + C_{2} = \frac{a}{4} \log |x^{4} + c^{2}| + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 10:

$\int \frac{(3 \sin x - 2) \cos x}{5 - \cos^{2} x - 4 \sin x} d x$

Answer:

$\int \frac{(3 \sin x - 2) \cos x d x}{5 - \cos^{2} x - 4 \sin x} = \int \frac{(3 \sin x - 2) \cos x d x}{5 - (1 - \sin^{2} x) - 4 \sin x} = \int \frac{(3 \sin x - 2) \cos x d x}{\sin^{2} x - 4 \sin x + 4} Let \sin x = t \Rightarrow \cos x d x = d t \int \frac{(3 t - 2) d t}{t^{2} - 4 t + 4} 3 t - 2 = A \frac{d}{d x} (t^{2} - 4 t + 4) + B 3 t - 2 = A (2 t - 4) + B 3 t - 2 = (2 A) t + B - 4 A$

Comparing the Coefficients of like powers of t

$2 A = 3 A = \frac{3}{2} B - 4 A = - 2 B - 4 \times \frac{3}{2} = - 2 B = - 2 + 6 B = 4$

$3 t - 2 = \frac{3}{2} (2 t - 4) + 4 ∴ \int \frac{(3 t - 2) d t}{t^{2} - 4 t + 4} = \int (\frac{\frac{3}{2} (2 t - 4) + 4}{t^{2} - 4 t + 4}) d t = \frac{3}{2} \int (\frac{2 t - 4}{t^{2} - 4 t + 4}) d t + 4 \int \frac{d t}{t^{2} - 4 t + 4} = \frac{3}{2} I_{1} + 4 I_{2} . . . (1) where I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4}, I_{2} = \int \frac{d t}{t^{2} - 4 t + 4} I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4} Let t^{2} - 4 t + 4 = p \Rightarrow (2 t - 4) d t = d p I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4} = \int \frac{d p}{p} = \log |p| + C_{1} = \log |t^{2} - 4 t + 4| + C_{1} . . . (2) I_{2} = \int \frac{d t}{t^{2} - 4 t + 4} I_{2} = \int \frac{d t}{{(t - 2)}^{2}} I_{2} = \int {(t - 2)}^{- 2} d t I_{2} = \frac{{(t - 2)}^{- 2 + 1}}{- 2 + 1} + C_{2} I_{2} = \frac{- 1}{t - 2} + C_{2} . . . (3) from (1), (2) and (3) \int \frac{(3 \sin x - 2) \cos x d x}{5 - \cos^{2} x - 4 \sin x} = \frac{3}{2} \log |t^{2} - 4 t + 4| + 4 \times \frac{- 1}{t - 2} + C_{1} + C_{2} = \frac{3}{2} \log |\sin^{2} x - 4 \sin x + 4| + \frac{4}{2 - t} + C (Where C = C_{1} + C_{2}) = \frac{3}{2} \log |{(\sin x - 2)}^{2}| + \frac{4}{2 - \sin x} + C = \frac{3}{2} \times 2 \log |\sin x - 2| + \frac{4}{2 - \sin x} + C = 3 \log |2 - \sin x| + \frac{4}{2 - \sin x} + C$

Page No 18.104:

Question 11:

$\int \frac{x + 2}{2 x^{2} + 6 x + 5} d x$

Answer:

$\int (\frac{x + 2}{2 x^{2} + 6 x + 5}) d x x + 2 = A \frac{d}{d x} (2 x^{2} + 6 x + 5) + B x + 2 = A (4 x + 6) + B x + 2 = (4 A) x + 6 A + B$

Comparing the Coefficients of like powers of x

$4 A = 1 A = \frac{1}{4} 6 A + B = 2 6 \times \frac{1}{4} + B = 2 B = \frac{1}{2}$

$∴ \int (\frac{x + 2}{2 x^{2} + 6 x + 5}) d x = \int [\frac{\frac{1}{4} (4 x + 6) + \frac{1}{2}}{2 x^{2} + 6 x + 5}] d x = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{2} \int \frac{1}{2 x^{2} + 6 x + 5} d x = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + \frac{1}{4}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} = \frac{1}{4} \log |2 x^{2} + 6 x + 5| + \frac{1}{4} \times 2 \tan^{- 1} (\frac{x + \frac{3}{2}}{\frac{1}{2}}) + C = \frac{1}{4} \log |2 x^{2} + 6 x + 5| + \frac{1}{2} \tan^{- 1} (2 x + 3) + C$

Page No 18.104:

Question 12:

Evaluate the following integrals:

$\int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x$

Answer:

$Let I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \int \frac{5 x - 2}{3 x^{2} + 2 x + 1} d x We express 5 x - 2 = A (\frac{d}{d x} (3 x^{2} + 2 x + 1)) + B 5 x - 2 = A (6 x + 2) + B Equating the coefficients of x and constants, we get 5 = 6 A and - 2 = 2 A + B or A = \frac{5}{6} and B = - \frac{11}{3} ∴ I = \int \frac{\frac{5}{6} (6 x + 2) - \frac{11}{3}}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x - \frac{11}{3} \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} . . . (1) Now, I_{1} = \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x Let 3 x^{2} + 2 x + 1 = t On differentiating both sides, we get (6 x + 2) d x = d t ∴ I_{1} = \int \frac{1}{t} d t = \log |t| + c_{1} = \log |3 x^{2} + 2 x + 1| + c_{1} . . . (2) And, I_{2} = \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9} + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d x Let x + \frac{1}{3} = t On differentiating both sides, we get d x = d t ∴ I_{2} = \frac{1}{3} \int \frac{1}{{(t)}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d t = \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} \frac{3 t}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 (x + \frac{1}{3})}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{5}{6} (\log |3 x^{2} + 2 x + 1| + c_{1}) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2}) = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c Hence, \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c$

Page No 18.104:

Question 13:

$\int \frac{x + 5}{3 x^{2} + 13 x - 10} d x$

Answer:

$I = \int \frac{x + 5}{3 x^{2} + 13 x - 10} d x = \int \frac{x + 5}{3 x^{2} + 15 x - 2 x - 10} d x = \int \frac{x + 5}{3 x (x + 5) - 2 (x + 5)} d x = \int \frac{x + 5}{(3 x - 2) (x + 5)} d x$

$= \int \frac{x + 5}{(3 x - 2) (x + 5)} d x = \int \frac{1}{3 x - 2} d x ∴ I = \frac{1}{3} \ln |3 x - 2| + c$

Page No 18.104:

Question 14:

$\int \frac{(3 \sin x - 2) \cos x}{13 - \cos^{2} x - 7 \sin x} d x$

Answer:

$I = \int \frac{(3 \sin x - 2) \cos x}{13 - \cos^{2} x - 7 \sin x} d x$

$= \int \frac{(3 \sin x - 2) \cos x}{13 - (1 - \sin^{2} x) - 7 \sin x} d x (∵ \cos^{2} x = 1 - \sin^{2} x) = \int \frac{(3 \sin x - 2) \cos x}{\sin^{2} x - 7 \sin x + 12} d x = \int \frac{(3 \sin x - 2) \cos x}{\sin^{2} x - 4 \sin x - 3 \sin x + 12} d x = \int \frac{(3 \sin x - 2) \cos x}{\sin x (\sin x - 4) - 3 (\sin x - 4)} d x = \int \frac{(3 \sin x - 2) \cos x}{(\sin x - 3) (\sin x - 4)} d x$

$Let \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{(3 t - 2)}{(t - 3) (t - 4)} d t$

Using partial fraction, we get

$\frac{(3 t - 2)}{(t - 3) (t - 4)} = \frac{A}{(t - 3)} + \frac{B}{(t - 4)} = \frac{A (t - 4) + B (t - 3)}{(t - 3) (t - 4)} \Rightarrow 3 t - 2 = (A + B) t - 4 A - 3 B$

Comparing coefficients, we get

A = $-$ 7 and B = 10

So, $I = - 7 \int \frac{1}{(t - 3)} d t + 10 \int \frac{1}{(t - 4)} d t$

$\Rightarrow I = - 7 \ln |t - 3| + 10 \ln |t - 4| + c ∴ I = - 7 \ln |\sin x - 3| + 10 \ln |\sin x - 4| + c$

Page No 18.104:

Question 15:

$\int \frac{x + 7}{3 x^{2} + 25 x + 28} d x$

Answer:

$I = \int \frac{x + 7}{3 x^{2} + 25 x + 28} d x = \int \frac{x + 7}{3 x^{2} + 21 x + 4 x + 28} d x = \int \frac{x + 7}{3 x (x + 7) + 4 (x + 7)} d x = \int \frac{x + 7}{(3 x + 4) (x + 7)} d x$

$= \int \frac{1}{(3 x + 4)} d x = \frac{1}{3} \ln |3 x + 4| + c$

Page No 18.104:

Question 16:

$\int \frac{3 x + 5}{x^{2} + 3 x - 18} d x$

Answer:

$I = \int \frac{3 x + 5}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = 3 \int \frac{x + \frac{5}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + \frac{10}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + 3 + \frac{1}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + 3}{x^{2} + 3 x - 18} d x + \frac{1}{2} \int \frac{1}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{x^{2} + 3 x + \frac{9}{4} - 18 - \frac{9}{4}} d x$ $(\begin{matrix} \int \frac{f' (x) d x}{f (x)} = \log |f (x)| + C \\ f (x) = x^{2} + 3 x - 18 \Rightarrow f' (x) = 2 x = 3 \end{matrix})$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{{(x + \frac{3}{2})}^{2} - \frac{81}{4}} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \times \frac{1}{2 \times \frac{9}{2}} \log |\frac{(x + \frac{3}{2}) - \frac{9}{2}}{(x + \frac{3}{2}) + \frac{9}{2}}| + C$ $(\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + C)$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{18} \log |\frac{x - 3}{x + 6}| + C$

Page No 18.104:

Question 17:

Evaluate the following integrals:
$\int \frac{x^{3}}{x^{4} + x^{2} + 1} d x$

Answer:

$I = \int \frac{x^{3}}{x^{4} + x^{2} + 1} d x = \int \frac{x^{2} \cdot x}{{(x^{2})}^{2} + x^{2} + 1} d x Let x^{2} = t or 2 x d x = d t \Rightarrow I = \frac{1}{2} \int \frac{t}{t^{2} + t + 1} d t = \frac{1}{4} \int \frac{2 t}{t^{2} + t + 1} d t = \frac{1}{4} \int \frac{2 t + 1 - 1}{t^{2} + t + 1} d t$

$= \frac{1}{4} \int [\frac{(2 t + 1)}{(t^{2} + t + 1)} - \frac{1}{(t^{2} + t + 1)}] d t = \frac{1}{4} [\log |t^{2} + t + 1| - \int \frac{1}{(t^{2} + t + \frac{1}{4} + \frac{3}{4})} d t] = \frac{1}{4} [\log |t^{2} + t + 1| - \int \frac{1}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d t] = \frac{1}{4} [\log |t^{2} + t + 1| - \frac{2}{\sqrt{3}} \tan \frac{(t + \frac{1}{2})}{(\frac{\sqrt{3}}{2})}] + c = \frac{1}{4} [\log |t^{2} + t + 1| - \frac{2}{\sqrt{3}} \tan (\frac{2 t + 1}{\sqrt{3}})] + c$

$= \frac{1}{4} [\log |x^{4} + x^{2} + 1| - \frac{2}{\sqrt{3}} \tan (\frac{2 x^{2} + 1}{\sqrt{3}})] + c$

Page No 18.104:

Question 18:

Evaluate the following integrals:
$\int \frac{x^{3} - 3 x}{x^{4} + 2 x^{2} - 4} d x$

Answer:

$I = \int \frac{x^{3} - 3 x}{x^{4} + 2 x^{2} - 4} d x$

$= \int \frac{x (x^{2} - 3)}{x^{4} + 2 x^{2} - 4} d x$

Let $x^{2} = t$ , or, $2 x d x = d t$

$\Rightarrow I = \frac{1}{2} \int \frac{(t - 3)}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int \frac{2 t - 6}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int \frac{2 t + 2 - 8}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int (\frac{2 t + 2}{t^{2} + 2 t - 4} - \frac{8}{t^{2} + 2 t - 4}) d t = \frac{1}{4} (\int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - \int \frac{8}{t^{2} + 2 t - 4} d t)$

$\Rightarrow I = \frac{1}{4} (I_{1} + I_{2}) . . . (i)$

Now,

$I_{1} = \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t$

Let $t^{2} + 2 t - 4 = u$

$or, (2 t + 2) d t = d u \Rightarrow I_{1} = \int \frac{1}{u} d u = \ln |u| + c_{1} \Rightarrow I_{1} = \ln |t^{2} + 2 t - 4| + c_{1} ∴ I_{1} = \ln |x^{4} + 2 x^{2} - 4| + c_{1}$

Now,

$I_{2} = \int \frac{- 8}{(t + 1)^{2} - 5} d t \Rightarrow I_{2} = \int \frac{8}{(\sqrt{5})^{2} - (t + 1)^{2}} d t ∴ I_{2} = \frac{8}{2 \sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}| + c_{2}$

$So, from (i), we get I = \frac{1}{4} [\ln |x^{4} + 2 x^{2} - 4| + \frac{4}{\sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}|] + C ∴ I = \frac{1}{4} \ln |x^{4} + 2 x^{2} - 4| + \frac{1}{\sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}| + C$

Page No 18.106:

Question 1:

$\int \frac{x^{2} + x + 1}{x^{2} - x} d x$

Answer:

$\int (\frac{x^{2} + x + 1}{x^{2} - x}) d x \frac{x^{2} + x + 1}{x^{2} - x} = 1 + \frac{2 x + 1}{x^{2} - x} ∴ \int (\frac{x^{2} + x + 1}{x^{2} - x}) d x = \int (1 + \frac{2 x + 1}{x^{2} - x}) d x = \int 1 + (\frac{2 x - 1 + 2}{x^{2} - x}) d x = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x} + \int \frac{2 d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x} + 2 \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = x + \log |x^{2} - x| + 2 \times \frac{1}{2 \times \frac{1}{2}} \log |\frac{x - \frac{1}{2} - \frac{1}{2}}{x - \frac{1}{2} + \frac{1}{2}}| = x + \log |x^{2} - x| + 2 \log |\frac{x - 1}{x}| + C$

Page No 18.106:

Question 2:

$\int \frac{x^{2} + x - 1}{x^{2} + x - 6} d x$

Answer:

$\int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x \frac{x^{2} + x - 1}{x^{2} + x - 6} = 1 + \frac{5}{x^{2} + x - 6} \int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x = \int d x + 5 \int \frac{d x}{x^{2} + x - 6} = \int d x + 5 \int \frac{d x}{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 6} = \int d x + 5 \int \frac{d x}{{(x + \frac{1}{2})}^{2} - \frac{1}{4} - 6} = \int d x + 5 \int \frac{d x}{{(x + \frac{1}{2})}^{2} - {(\frac{5}{2})}^{2}} = x + 5 \times \frac{1}{2 \times \frac{5}{2}} \log |\frac{x + \frac{1}{2} - \frac{5}{2}}{x + \frac{1}{2} + \frac{5}{2}}| + C = x + \log |\frac{x - 2}{x + 3}| + C$

Page No 18.106:

Question 3:

$\int \frac{(1 - x^{2})}{x (1 - 2 x)} d x$

Answer:

$We have, I = \int \frac{(1 - x^{2})}{x (1 - 2 x)} d x = \int (\frac{- x^{2} + 1}{- 2 x^{2} + x}) d x = \int \frac{1}{2} d x + \int (\frac{1 - \frac{x}{2}}{- 2 x^{2} + x}) d x = \frac{1}{2} \int d x + \frac{1}{2} \int \frac{2 - x}{- 2 x^{2} + x} d x = \frac{1}{2} [\int d x + \int \frac{2 - x}{- 2 x^{2} + x} d x] = \frac{1}{2} [I_{1} + I_{2}] (say) where I_{1} = \int d x & I_{2} = \int \frac{2 - x}{- 2 x^{2} + x} d x Now, I_{1} = \int d x = x + C_{1} I_{2} = \int \frac{2 - x}{- 2 x^{2} + x} d x Let 2 - x = A \frac{d}{d x} (- 2 x^{2} + x) + B \Rightarrow 2 - x = A (- 4 x + 1) + B \Rightarrow 2 - x = - 4 A x + A + B$

Comparing coefficients of like terms

$- 1 = - 4 A \Rightarrow A = \frac{1}{4} & A + B = 2 \Rightarrow \frac{1}{4} + B = 2 \Rightarrow B = 2 - \frac{1}{4} = \frac{8 - 1}{4} = \frac{7}{4}$

$∴ \int \frac{2 - x}{- 2 x^{2} + x} d x = \int \frac{\frac{1}{4} (- 4 x + 1) + \frac{7}{4}}{- 2 x^{2} + x} d x = \int \frac{\frac{1}{4} (- 4 x + 1)}{- 2 x^{2} + x} d x + \int \frac{\frac{7}{4}}{- 2 x^{2} + x} d x = \frac{1}{4} \log |- 2 x^{2} + x| + \frac{7}{4} \int \frac{1}{- 2 (x^{2} - \frac{x}{2} + \frac{1}{16} - \frac{1}{16})} d x = \frac{1}{4} \log |- 2 x^{2} + x| - \frac{7}{8} \int \frac{1}{\{{(x - \frac{1}{4})}^{2} - {(\frac{1}{4})}^{2}\}} d x = \frac{1}{4} \log |x (- 2 x + 1)| - \frac{7}{8} \times \frac{1}{2 \times \frac{1}{4}} \log |\frac{x - \frac{1}{4} - \frac{1}{4}}{x - \frac{1}{4} + \frac{1}{4}}| + C_{2} = \frac{1}{4} \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |\frac{x - \frac{1}{2}}{x}| + C_{2} = \frac{1}{4} \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |x - \frac{1}{2}| + \frac{7}{4} \log |x| + C_{2} = 2 \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |2 x - 1| + C_{3}, where C_{3} = C_{2} + \frac{7}{4} \log 2 = 2 \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |1 - 2 x| + C_{3} = 2 \log |x| - \frac{3}{2} \log |1 - 2 x| + C_{3} Thus, I = \frac{1}{2} [x + C_{1} + 2 \log |x| - \frac{3}{2} \log |1 - 2 x| + C_{3}] = \frac{1}{2} x + \log |x| - \frac{3}{4} \log |1 - 2 x| + C, where C = \frac{1}{2} [C_{1} + C_{3}]$

Page No 18.106:

Question 4:

$\int \frac{x^{2} + 1}{x^{2} - 5 x + 6} d x$

Answer:

$Let I \int (\frac{x^{2} + 1}{x^{2} - 5 x + 6}) d x Dividing Numerator by Denominator x^{2} - 5 x + 6 \overset{1}{x^{2} + 1} x^{2} - 5 x + 6 - + - 5 x - 5 \frac{x^{2} + 1}{x^{2} - 5 x + 6} = 1 + (\frac{5 x - 5}{x^{2} - 5 x + 6}) . . . . . (1) Also \frac{5 x - 5}{x^{2} - 5 x + 6} = \frac{5 x - 5}{(x - 2) (x - 3)} Let \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \Rightarrow \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)} \Rightarrow 5 x - 5 = A (x - 3) + B (x - 2) let x = 3 5 \times 3 - 5 = A \times 0 + B (3 - 2) 10 = B let x = 2 5 \times 2 - 5 = A (2 - 3) + B \times 0 A = - 5 \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{- 5}{x - 2} + \frac{10}{x - 3} . . . . . (2) from (1) and (2) I = \int d x - 5 \int \frac{d x}{x - 2} + 10 \int \frac{d x}{x - 3} = x - 5 \log |x - 2| + 10 \log |x - 3| + C$

Page No 18.106:

Question 5:

$\int \frac{x^{2}}{x^{2} + 7 x + 10} d x$

Answer:

$Let I = \int (\frac{x^{2}}{x^{2} + 7 x + 10}) d x Now, x^{2} + 7 x + 10 \overset{1}{x^{2}} x^{2} + 7 x + 10 - - - - 7 x - 10 ∴ \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - \frac{(7 x + 10)}{x^{2} + 7 x + 10} \Rightarrow \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - (\frac{7 x + 10}{x^{2} + 2 x + 5 x + 10}) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - (\frac{7 x + 10}{x (x + 2) + 5 (x + 2)}) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - [\frac{7 x + 10}{(x + 2) (x + 5)}] . . . . . (1) Consider, \frac{7 x + 10}{(x + 2) (x + 5)} = \frac{A}{(x + 2)} + \frac{B}{x + 5} 7 x + 10 = A (x + 5) + B (x + 2) let x + 5 = 0 x = - 5 \Rightarrow 7 (- 5) + 10 = A \times 0 + B (- 5 + 2) - 25 = B (- 3) \Rightarrow B = \frac{25}{3} let x + 2 = 0 x = - 2 7 (- 2) + 10 = A (- 2 + 5) \Rightarrow - 4 = A (3) \Rightarrow A = - \frac{4}{3} \frac{7 x + 10}{(x + 2) (x + 5)} = \frac{- 4}{3 (x + 2)} + \frac{25}{3 (x + 5)} . . . . . (2) from (1) and (2) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 + \frac{4}{3 (x + 2)} - \frac{25}{3 (x + 5)} \Rightarrow \int \frac{x^{2} d x}{x^{2} + 7 x + 10} = \int d x + \frac{4}{3} \int \frac{d x}{x + 2} - \frac{25}{3} \int \frac{d x}{x + 5} = x + \frac{4}{3} \log |x + 2| - \frac{25}{3} \log |x + 5| + C$

Page No 18.106:

Question 6:

$\int \frac{x^{2} + x + 1}{x^{2} - x + 1} d x$

Answer:

$Let I = \int (\frac{x^{2} + x + 1}{x^{2} - x + 1}) d x Now, x^{2} - x + 1 \overset{1}{x^{2} + x + 1} x^{2} - x + 1 - + - 2 x Therefore, \frac{x^{2} + x + 1}{x^{2} - x + 1} = 1 + \frac{2 x}{x^{2} - x + 1} \Rightarrow \int (\frac{x^{2} + x + 1}{x^{2} - x + 1}) d x = \int d x + \int (\frac{2 x - 1 + 1}{x^{2} - x + 1}) d x = \int d x + \int (\frac{2 x - 1}{x^{2} - x + 1}) d x + \int \frac{d x}{x^{2} - x + 1} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \int \frac{d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = x + \log |x^{2} - x + 1| + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C$

Page No 18.106:

Question 7:

$\int \frac{{(x - 1)}^{2}}{x^{2} + 2 x + 2} d x$

Answer:

$Let I = \int \frac{{(x - 1)}^{2}}{x^{2} + 2 x + 2} d x = \int (\frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2}) d x Here, x^{2} + 2 x + 2 \overset{1}{x^{2} - 2 x + 1} x^{2} + 2 x + 2 - - - - 4 x - 1 Therefore, \frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2} = 1 - \frac{(4 x + 1)}{x^{2} + 2 x + 2} . . . . . (1) Let 4 x + 1 = A \frac{d}{d x} (x^{2} + 2 x + 2) + B 4 x + 1 = A (2 x + 2) + B 4 x + 1 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 4 A = 2 2 A + B = 1 2 \times 2 + B = 1 B = - 3 \int (\frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2}) d x = \int d x - 2 \int \frac{(2 x + 2)}{x^{2} + 2 x + 2} d x + 3 \int \frac{d x}{x^{2} + 2 x + 2} = \int d x - 2 \int \frac{(2 x + 2)}{x^{2} + 2 x + 2} d x + 3 \int \frac{d x}{{(x + 1)}^{2} + 1^{2}} = x - 2 \log |x^{2} + 2 x + 2| + \frac{3}{1} \tan^{- 1} (\frac{x + 1}{1}) + C = x - 2 \log |x^{2} + 2 x + 2| + 3 \tan^{- 1} (x + 1) + C$

Page No 18.106:

Question 8:

$\int \frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1} d x$

Answer:

$Let I = \int (\frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1}) d x x^{2} - x + 1 \overset{x + 2}{x^{3} + x^{2} + 2 x + 1} x^{3} - x^{2} + x - + - 2 x^{2} + x + 1 2 x^{2} - 2 x + 2 - + - 3 x - 1 Therefore, \frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1} = x + 2 + \frac{3 x - 1}{x^{2} - x + 1} . . . . . (1) Let 3 x - 1 = A \frac{d}{d x} (x^{2} - x + 1) + B 3 x - 1 = A (2 x - 1) + B 3 x - 1 = (2 A) x + B - A Equating Coefficients of like terms 2 A = 3 A = \frac{3}{2} B - A = - 1 B - \frac{3}{2} = - 1 B = \frac{1}{2} \int (\frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1}) d x = \int (x + 2) d x + \int (\frac{\frac{3}{2} (2 x - 1) + \frac{1}{2}}{x^{2} - x + 1}) d x = \int (x + 2) d x + \frac{3}{2} \int (\frac{2 x - 1}{x^{2} - x + 1}) d x + \frac{1}{2} \int \frac{d x}{x^{2} - x + 1} = \int (x + 2) d x + \frac{3}{2} \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \frac{1}{2} \int \frac{d x}{x^{2} - x + \frac{1}{4} - \frac{1}{4} + 1} = \int (x + 2) d x + \frac{3}{2} \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{x^{2}}{2} + 2 x + \frac{3}{2} \log |x^{2} - x + 1| + \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{x^{2}}{2} + 2 x + \frac{3}{2} \log |x^{2} - x + 1| + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C$

Page No 18.106:

Question 9:

$\int \frac{x^{2} (x^{4} + 4)}{x^{2} + 4} d x$

Answer:

$Let I = \int \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} d x = \int (\frac{x^{6} + 4 x^{2}}{x^{2} + 4}) d x Now, x^{2} + 4 \overset{x^{4} - 4 x^{2} + 20}{x^{6} + 4 x^{2}} x^{6} + 4 x^{4} - - - 4 x^{4} + 4 x^{2} - 4 x^{4} - 16 x^{2} + + 20 x^{2} 20 x^{2} + 80 - - - 80 Therefore, \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} = (x^{4} - 4 x^{2} + 20) - \frac{80}{x^{2} + 4} I = \int \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} d x = \int (x^{4} - 4 x^{2} + 20) d x - 80 \int \frac{d x}{x^{2} + 2^{2}} = \int x^{4} d x - 4 \int x^{2} d x + 20 \int d x - 80 \int \frac{d x}{x^{2} + 2^{2}} = \frac{x^{4 + 1}}{4 + 1} - 4 [\frac{x^{3}}{3}] + 20 (x) - 80 \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C = \frac{x^{5}}{5} - \frac{4}{3} x^{3} + 20 x - 40 \tan^{- 1} (\frac{x}{2}) + C$

Page No 18.106:

Question 10:

$\int \frac{x^{2}}{x^{2} + 6 x + 12} d x$

Answer:

$Let I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} Now, x^{2} + 6 x + 12 \overset{1}{x^{2}} x^{2} + 6 x + 12 - - - - 6 x - 12 Therefore, \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{(6 x + 12)}{x^{2} + 6 x + 12} . . . . . (1) Let 6 x + 12 = A \frac{d}{d x} (x^{2} + 6 x + 12) + B \Rightarrow 6 x + 12 = A (2 x + 6) + B \Rightarrow 6 x + 12 = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 6 A = 3 6 A + B = 12 18 + B = 12 B = - 6 ∴ \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{3 (2 x + 6) - 6}{x^{2} + 6 x + 12} I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 9 + 3} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{{(x + 3)}^{2} + {(\sqrt{3})}^{2}} = x - 3 \log |x^{2} + 6 x + 12| + \frac{6}{\sqrt{3}} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C = x - 3 \log |x^{2} + 6 x + 12| + 2 \sqrt{3} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C$

Page No 18.110:

Question 1:

$\int \frac{x}{\sqrt{x^{2} + 6 x + 10}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} x = A \frac{d}{d x} (x^{2} + 6 x + 10) + B x = A (2 x + 6) + B x = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 1 A = \frac{1}{2} 6 A + B = 0 6 \times \frac{1}{2} + B = 0 B = - 3 I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} = \int (\frac{\frac{1}{2} (2 x + 6) - 3}{\sqrt{x^{2} + 6 x + 10}}) d x = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 3^{2} - 3^{2} + 10}} = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1^{2}}} let x^{2} + 6 x + 10 = t \Rightarrow (2 x + 6) d x = d t I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1}} = \frac{1}{2} \times 2 \sqrt{t} - 3 \log |x + 3 + \sqrt{{(x + 3)}^{2} + 1}| + C = \sqrt{t} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C = \sqrt{x^{2} + 6 x + 10} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C$

Page No 18.110:

Question 2:

$\int \frac{2 x + 1}{\sqrt{x^{2} + 2 x - 1}} d x$

Answer:

$Let I = \int \frac{(2 x + 1) d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2 - 1) d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 - 1 - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} let x^{2} + 2 x - 1 = t \Rightarrow (2 x + 2) d x = d t I = \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} = 2 \sqrt{t} - \log |x + 1 + \sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}| + C = 2 \sqrt{x^{2} + 2 x - 1} - \log |x + 1 + \sqrt{x^{2} + 2 x - 1}| + C$

Page No 18.110:

Question 3:

$\int \frac{x + 1}{\sqrt{4 + 5 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(x + 1) d x}{\sqrt{4 + 5 x - x^{2}}} Also, x + 1 = A \frac{d}{d x} (4 + 5 x - x^{2}) + B x + 1 = A (5 - 2 x) + B x + 1 = (- 2 A) x + 5 A + B Equating Coefficients of like terms - 2 A = 1 \Rightarrow A = - \frac{1}{2} And 5 A + B = 1 \Rightarrow - \frac{5}{2} + B = 1 B = \frac{7}{2} I = \int \frac{(x + 1) d x}{\sqrt{4 + 5 x - x^{2}}} = \int (\frac{- \frac{1}{2} (5 - 2 x) + \frac{7}{2}}{\sqrt{4 + 5 x - x^{2}}}) d x = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - (x^{2} - 5 x)}} = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - [x^{2} - 5 x + {(\frac{5}{2})}^{2} - {(\frac{5}{2})}^{2}]}} = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - {(x - \frac{5}{2})}^{2} + \frac{25}{4}}} = - \frac{1}{2} \int (\frac{5 - 2 x}{\sqrt{4 + 5 x - x^{2}}}) d x + \frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}} = - \frac{1}{2} \int (\frac{5 - 2 x}{\sqrt{4 + 5 x - x^{2}}}) d x + \frac{7}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{5}{2})}^{2}}} let 4 + 5 x - x^{2} = t \Rightarrow (5 - 2 x) d x = d t Then, I = - \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \frac{7}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{5}{2})}^{2}}} = - \frac{1}{2} \times 2 \sqrt{t} + \frac{7}{2} \times \sin^{- 1} (\frac{x - \frac{5}{2}}{\frac{\sqrt{41}}{2}}) + C = - \sqrt{t} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + C = - \sqrt{4 + 5 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + C$

Page No 18.110:

Question 4:

$\int \frac{6 x - 5}{\sqrt{3 x^{2} - 5 x + 1}} d x$

Answer:

$Let I = \int \frac{6 x - 5}{\sqrt{3 x^{2} - 5 x + 1}} d x Putting 3 x^{2} - 5 x + 1 = t \Rightarrow (6 x - 5) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} = 2 \sqrt{t} + C = 2 \sqrt{3 x^{2} - 5 x + 1} + C$

Page No 18.110:

Question 5:

$\int \frac{3 x + 1}{\sqrt{5 - 2 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(3 x + 1) d x}{\sqrt{5 - 2 x - x^{2}}} Consider, 3 x + 1 = A \frac{d}{d x} (5 - 2 x - x^{2}) + B \Rightarrow 3 x + 1 = A (- 2 - 2 x) + B \Rightarrow 3 x + 1 = (- 2 A) x - 2 A + B Equating Coefficients of like terms - 2 A = 3 \Rightarrow A = - \frac{3}{2} And - 2 A + B = 1 \Rightarrow - 2 \times - \frac{3}{2} + B = 1 \Rightarrow B = 1 - 3 \Rightarrow B = - 2 ∴ I = \int [\frac{- \frac{3}{2} (- 2 - 2 x) - 2}{\sqrt{5 - 2 x - x^{2}}}] d x = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - 2 x - x^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x + 1 - 1)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{6 - {(x + 1)}^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} let 5 - 2 x - x^{2} = t \Rightarrow (- 2 - 2 x) d x = d t ∴ I = - \frac{3}{2} \int \frac{d t}{\sqrt{t}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} = - \frac{3}{2} \times 2 \sqrt{t} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C = - 3 \sqrt{5 - 2 x - x^{2}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C$

Page No 18.110:

Question 6:

$\int \frac{x}{\sqrt{8 + x - x^{2}}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{8 + x - x^{2}}} Consider, x = A \frac{d}{d x} (8 + x - x^{2}) + B x = A (1 - 2 x) + B x = (- 2 A) x + A + B Equating Coefficients of like terms - 2 A = 1 \Rightarrow A = - \frac{1}{2} And A + B = 0 \Rightarrow - \frac{1}{2} + B = 0 \Rightarrow B = \frac{1}{2} ∴ x = - \frac{1}{2} (1 - 2 x) + \frac{1}{2} Then, I = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 + x - x^{2}}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 - (x^{2} - x)}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 - (x^{2} - x + \frac{1}{4} - \frac{1}{4})}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 + \frac{1}{4} - {(x - \frac{1}{2})}^{2}}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{33}}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} let 8 + x - x^{2} = t \Rightarrow (1 - 2 x) d x = d t ∴ I = - \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \frac{1}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{33}}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} = - \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{33}}{2}}) + C = - \sqrt{t} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + C = - \sqrt{8 + x - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + C$

Page No 18.110:

Question 7:

$\int \frac{x + 2}{\sqrt{x^{2} + 2 x - 1}} d x$

Answer:

$Let I = \int \frac{(x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} Consider, x + 2 = A \frac{d}{d x} (x^{2} + 2 x - 1) + B \Rightarrow x + 2 = A (2 x + 2) + B \Rightarrow x + 2 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} And 2 A + B = 2 \Rightarrow 2 \times \frac{1}{2} + B = 2 \Rightarrow B = 1 T h e n, I = \int \frac{[\frac{1}{2} (2 x + 2) + 1]}{\sqrt{x^{2} + 2 x - 1}} d x = \frac{1}{2} \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} + \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} let x^{2} + 2 x - 1 = t \Rightarrow (2 x + 2) d x = d t ∴ I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 - 2}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} = \sqrt{t} + \log |x + 1 + \sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}| + C = \sqrt{x^{2} + 2 x - 1} + \log |x + 1 + \sqrt{x^{2} + 2 x - 1}| + C$

Page No 18.110:

Question 8:

$\int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$

Answer:

$Let I = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x = \int \frac{x}{\sqrt{x^{2} - 1}} d x + 2 \int \frac{d x}{\sqrt{x^{2} - 1}} let x^{2} - 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + 2 \int \frac{d x}{\sqrt{x^{2} - 1^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + 2 \int \frac{d x}{\sqrt{x^{2} - 1^{2}}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + 2 \log |x + \sqrt{x^{2} - 1}| + C = \sqrt{t} + 2 \log |x + \sqrt{x^{2} - 1}| + C = \sqrt{x^{2} - 1} + 2 \log |x + \sqrt{x^{2} - 1}| + C$

Page No 18.110:

Question 9:

$\int \frac{x - 1}{\sqrt{x^{2} + 1}} d x$

Answer:

$Let I = \int (\frac{x - 1}{\sqrt{x^{2} + 1}}) d x = \int \frac{x d x}{\sqrt{x^{2} + 1}} - \int \frac{d x}{\sqrt{x^{2} + 1}} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \sqrt{t} - \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{x^{2} + 1} - \log |x + \sqrt{x^{2} + 1}| + C$

Page No 18.110:

Question 10:

$\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{x^{2} + x + 1}} Consider, x = A \frac{d}{d x} (x^{2} + x + 1) + B \Rightarrow x = A (2 x + 1) + B \Rightarrow x = (2 A) x + A + B Equating Coefficient of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} And A + B = 0 \Rightarrow \frac{1}{2} + B = 0 \Rightarrow B = - \frac{1}{2} ∴ I = \int \frac{(\frac{1}{2} (2 x + 1) - \frac{1}{2})}{\sqrt{x^{2} + x + 1}} d x = \frac{1}{2} \int (\frac{2 x + 1}{\sqrt{x^{2} + x + 1}}) d x - \frac{1}{2} \int \frac{d x}{\sqrt{x^{2} + x + \frac{1}{4} - \frac{1}{4} + 1}} Putting x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - \frac{1}{2} \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}| + C = \frac{1}{2} |\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}| - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C = \sqrt{t} - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C = \sqrt{x^{2} + x + 1} - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

Page No 18.110:

Question 11:

$\int \frac{x + 1}{\sqrt{x^{2} + 1}} d x$

Answer:

$Let I = \int (\frac{x + 1}{\sqrt{x^{2} + 1}}) d x = \int \frac{x d x}{\sqrt{x^{2} + 1}} + \int \frac{d x}{\sqrt{x^{2} + 1}} Putting, x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \int \frac{d x}{\sqrt{x^{2} + 1}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \int \frac{d x}{\sqrt{x^{2} + 1}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{t} + \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{x^{2} + 1} + \log |x + \sqrt{x^{2} + 1}| + C$

Page No 18.110:

Question 12:

$\int \frac{2 x + 5}{\sqrt{x^{2} + 2 x + 5}} d x$

Answer:

$Let I = \int \frac{(2 x + 5) d x}{\sqrt{x^{2} + 2 x + 5}} Consider, 2 x + 5 = A \frac{d}{d x} (x^{2} + 2 x + 5) + B \Rightarrow 2 x + 5 = A (2 x + 2) + B \Rightarrow 2 x + 5 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 2 \Rightarrow A = 1 And 2 A + B = 5 \Rightarrow B = 3 ∴ I = \int (\frac{2 x + 2 + 3}{\sqrt{x^{2} + 2 x + 5}}) d x = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x + 5}} + 3 \int \frac{d x}{\sqrt{x^{2} + 2 x + 5}} let x^{2} + 2 x + 5 = t \Rightarrow (2 x + 2) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} + 3 \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 + 4}} = \int t^{- \frac{1}{2}} d t + 3 \int \frac{d x}{\sqrt{{(x + 1)}^{2} + 2^{2}}} = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + 3 \log |x + 1 + \sqrt{{(x + 1)}^{2} + 4}| + C = 2 \sqrt{t} + 3 \log |x + 1 + \sqrt{x^{2} + 2 x + 5}| + C = 2 \sqrt{x^{2} + 2 x + 5} + 3 \log |x + 1 + \sqrt{x^{2} + 2 x + 5}| + C$

Page No 18.110:

Question 13:

$\int \frac{3 x + 1}{\sqrt{5 - 2 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(3 x + 1) d x}{\sqrt{5 - 2 x - x^{2}}} Consider, 3 x + 1 = A \frac{d}{d x} (5 - 2 x - x^{2}) + B \Rightarrow 3 x + 1 = A (- 2 - 2 x) + B \Rightarrow 3 x + 1 = (- 2 A) x + (- 2 A + B) Equating Coefficients of like terms - 2 A = 3 \Rightarrow A = - \frac{3}{2} And - 2 A + B = 1 \Rightarrow - 2 \times - \frac{3}{2} + B = 1 \Rightarrow B = - 2 ∴ I = \int [\frac{- \frac{3}{2} (- 2 - 2 x) - 2}{\sqrt{5 - 2 x - x^{2}}}] d x = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - 2 x - x^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x + 1 - 1)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{6 - {(x + 1)}^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} Putting, 5 - 2 x - x^{2} = t \Rightarrow (- 2 - 2 x) d x = d t Then, I = - \frac{3}{2} \int \frac{d t}{\sqrt{t}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C_{1} = - \frac{3}{2} \times 2 \sqrt{t} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C = - 3 \sqrt{5 - 2 x - x^{2}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C$

Page No 18.110:

Question 14:

$\int \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$Let I = \int \sqrt{\frac{1 - x}{1 + x}} d x = \int \sqrt{\frac{(1 - x) (1 - x)}{(1 + x) (1 - x)}} d x = \int (\frac{1 - x}{\sqrt{1 - x^{2}}}) d x = \int \frac{d x}{\sqrt{1 - x^{2}}} - \int \frac{x d x}{\sqrt{1 - x^{2}}} Putting 1 - x^{2} = t \Rightarrow - 2 x d x = d t \Rightarrow x d x = - \frac{d t}{2} Then, I = \int \frac{d x}{\sqrt{1 - x^{2}}} + \frac{1}{2} \int \frac{d t}{\sqrt{t}} = \sin^{- 1} (x) + \frac{1}{2} \times 2 \sqrt{t} + C = \sin^{- 1} (x) + \sqrt{1 - x^{2}} + C$

Page No 18.111:

Question 15:

$\int \frac{2 x + 1}{\sqrt{x^{2} + 4 x + 3}} d x$

Answer:

$Let I = \int \frac{(2 x + 1) d x}{\sqrt{x^{2} + 4 x + 3}} Consider, 2 x + 1 = A \frac{d}{d x} (x^{2} + 4 x + 3) + B \Rightarrow 2 x + 1 = A (2 x + 4) + B \Rightarrow 2 x + 1 = (2 A) x + 4 A + B Equating Coefficients of like terms 2 A = 2 \Rightarrow A = 1 And 4 A + B = 1 \Rightarrow 4 + B = 1 \Rightarrow B = - 3 ∴ I = \int (\frac{2 x + 4 - 3}{\sqrt{x^{2} + 4 x + 3}}) d x = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 3}} - 3 \int \frac{d x}{\sqrt{x^{2} + 4 x + 4 - 4 + 3}} = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 3}} - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} Let x^{2} + 4 x + 3 = t \Rightarrow (2 x + 4) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} = \int t^{- \frac{1}{2}} d t - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] - 3 \log |x + 2 + \sqrt{{(x + 2)}^{2} - 1}| + C = 2 \sqrt{t} - 3 \log |x + 2 + \sqrt{x^{2} + 4 x + 3}| + C = 2 \sqrt{x^{2} + 4 x + 3} - 3 \log |x + 2 + \sqrt{x^{2} + 4 x + 3}| + C$

Page No 18.111:

Question 16:

$\int \frac{2 x + 3}{\sqrt{x^{2} + 4 x + 5}} d x$

Answer:

$Let I = \int \frac{(2 x + 3) d x}{\sqrt{x^{2} + 4 x + 5}} = \int \frac{(2 x + 4 - 1)}{\sqrt{x^{2} + 4 x + 5}} d x = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 5}} - \int \frac{d x}{\sqrt{x^{2} + 4 x + 5}} = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 5}} - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1}} Consider, x^{2} + 4 x + 5 = t \Rightarrow (2 x + 4) d x = d t ∴ I = \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1^{2}}} = \int t^{- \frac{1}{2}} d t - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1^{2}}} = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} - \log |x + 2 + \sqrt{{(x + 2)}^{2} + 1}| + C = 2 \sqrt{x^{2} + 4 x + 5} - \log |x + 2 + \sqrt{x^{2} + 4 x + 5}| + C$

Page No 18.111:

Question 17:

$\int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x$

Answer:

$Let I = \int \frac{(5 x + 3) d x}{\sqrt{x^{2} + 4 x + 10}} Consider, 5 x + 3 = A \frac{d}{d x} (x^{2} + 4 x + 10) + B \Rightarrow 5 x + 3 = A (2 x + 4) + B \Rightarrow 5 x + 3 = (2 A) x + 4 A + B Equating Coefficients of like terms 2 A = 5 \Rightarrow A = \frac{5}{2} And 4 A + B = 3 \Rightarrow 4 \times \frac{5}{2} + B = 3 \Rightarrow B = - 7 ∴ I = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{x^{2} + 4 x + 10}} = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{x^{2} + 4 x + 4 - 4 + 10}} = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} Putting, x^{2} + 4 x + 10 = t \Rightarrow (2 x + 4) d x = d t Then, I = \frac{5}{2} \int \frac{d t}{\sqrt{t}} - 7 \log |x + 2 + \sqrt{{(x + 2)}^{2} + 6}| + C = \frac{5}{2} \int t^{- \frac{1}{2}} d t - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C = \frac{5}{2} \times 2 \sqrt{t} - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C = 5 \sqrt{x^{2} + 4 x + 10} - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C$

Page No 18.111:

Question 18:

Evaluate the following integrals:

$\int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x$

Answer:

$Let I = \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x We express x + 2 = A (\frac{d}{d x} (x^{2} + 2 x + 3)) + B x + 2 = A (2 x + 2) + B Equating the coefficients of x and constants, we get 1 = 2 A and 2 = 2 A + B or A = \frac{1}{2} and B = 1 ∴ I = \int \frac{\frac{1}{2} (2 x + 2) + 1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} I_{1} + I_{2} . . . (1) Now, I_{1} = \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x Let x^{2} + 2 x + 3 = u On differentiating both sides, we get (2 x + 2) d x = d u ∴ I_{1} = \int \frac{1}{\sqrt{u}} d u = 2 \sqrt{u} + c_{1} = 2 \sqrt{x^{2} + 2 x + 3} + c_{1} . . . (2) And, I_{2} = \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \int \frac{1}{\sqrt{x^{2} + 2 x + 1 - 1 + 3}} d x = \int \frac{1}{\sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}}} d x Let (x + 1) = u On differentiating both sides, we get d x = d u ∴ I_{2} = \int \frac{1}{\sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}} d u = \log |u + \sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}| + c_{2} = \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{1}{2} (2 \sqrt{x^{2} + 2 x + 3} + c_{1}) + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c Hence, \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c$

Page No 18.114:

Question 1:

$\int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\frac{1}{\cos^{2} x}}{4 + 9 \tan^{2} x} d x = \int \frac{\sec^{2} x}{4 + 9 \tan^{2} x} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{4 + 9 t^{2}} = \frac{1}{9} \int \frac{d t}{\frac{4}{9} + t^{2}} = \frac{1}{9} \int \frac{d t}{{(\frac{2}{3})}^{2} + t^{2}} = \frac{1}{9} \times \frac{3}{2} \tan^{- 1} (\frac{t}{\frac{2}{3}}) + C = \frac{1}{6} \tan^{- 1} (\frac{3 t}{2}) + C = \frac{1}{6} \tan^{- 1} (\frac{3 \tan x}{2}) + C$

Page No 18.114:

Question 2:

$\int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x Dividing numerator & denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{4 \tan^{2} x + 5} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{4 t^{2} + 5} = \frac{1}{4} \int \frac{d t}{t^{2} + \frac{5}{4}} = \frac{1}{4} \int \frac{d t}{t^{2} + {(\frac{\sqrt{5}}{2})}^{2}} = \frac{1}{4} \times \frac{2}{\sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{5}} \times 2) + C = \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{2 \tan x}{\sqrt{5}}) + C$

Page No 18.114:

Question 3:

$\int \frac{2}{2 + \sin 2 x} d x$

Answer:

$Let I = \int \frac{2}{2 + \sin (2 x)} d x = \int \frac{2}{2 + 2 \sin x \cos x} d x = \int \frac{1}{1 + \sin x \cos x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x d x}{\sec^{2} x + \tan x} = \int \frac{\sec^{2} x d x}{1 + \tan^{2} x + \tan x} Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t^{2} + t + 1} = \int \frac{d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4} + 1} = \int \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) + C = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + C$

Page No 18.114:

Question 4:

$\int \frac{\cos x}{\cos 3 x} d x$

Answer:

$Let I = \int \frac{\cos x}{\cos 3 x} d x = \int \frac{\cos x}{(4 \cos^{3} x - 3 \cos x)} d x [\cos 3 A = 4 \cos^{3} A - 3 \cos A] = \int \frac{1}{4 \cos^{2} x - 3} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{4 - 3 \sec^{2} x} d x = \int \frac{\sec^{2} x}{4 - 3 (1 + \tan^{2} x)} d x = \int \frac{\sec^{2} x}{1 - 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 - {(\sqrt{3} \tan x)}^{2}} d x Let \sqrt{3} \tan x = t \Rightarrow \sqrt{3} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{3}} ∴ I = \frac{1}{\sqrt{3}} \int \frac{d t}{1^{2} - t^{2}} = \frac{1}{\sqrt{3}} \times \frac{1}{2} \ln |\frac{1 + t}{1 - t}| + C = \frac{1}{2 \sqrt{3}} \ln |\frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x}| + C$

Page No 18.114:

Question 5:

$\int \frac{1}{1 + 3 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{1 + 3 \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + \tan^{2} x + 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + 4 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + {(2 \tan x)}^{2}} d x Let 2 \tan x = t \Rightarrow 2 \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{1 + t^{2}} = \frac{1}{2} \tan^{- 1} (t) + C = \frac{1}{2} \tan^{- 1} (2 \tan x) + C$

Page No 18.114:

Question 6:

$\int \frac{1}{3 + 2 \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1}{3 + 2 \cos^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{3 \sec^{2} x + 2} d x = \int \frac{\sec^{2} x}{3 (1 + \tan^{2} x) + 2} d x = \int \frac{\sec^{2} x}{3 \tan^{2} x + 5} d x = \int \frac{\sec^{2} x}{{(\sqrt{5})}^{2} + {(\sqrt{3} \tan x)}^{2}} d x Let \sqrt{3} \tan x = t \Rightarrow \sqrt{3} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{3}} ∴ I = \frac{1}{\sqrt{3}} \int \frac{d t}{{(\sqrt{5})}^{2} + t^{2}} = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{5}}) + C = \frac{1}{\sqrt{15}} \tan^{- 1} (\frac{\sqrt{3} \tan x}{\sqrt{5}}) + C$

Page No 18.114:

Question 7:

$\int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$

Answer:

$Let I = \int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{(\frac{\sin x - 2 \cos x}{\cos x}) \times (\frac{2 \sin x + \cos x}{\cos x})} d x = \int \frac{\sec^{2} x}{(\tan x - 2) (2 \tan x + 1)} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{(t - 2) (2 t + 1)} = \int \frac{d t}{2 t^{2} + t - 4 t - 2} = \int \frac{d t}{2 t^{2} - 3 t - 2} = \frac{1}{2} \int \frac{d t}{t^{2} - \frac{3}{2} t - 1} = \frac{1}{2} \int \frac{d t}{t^{2} - \frac{3}{2} t + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2} - 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{3}{4})}^{2} - \frac{9}{16} - 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{3}{4})}^{2} - {(\frac{5}{4})}^{2}} = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log |\frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}}| + C = \frac{1}{5} \ln |\frac{t - 2}{t + \frac{1}{2}}| + C = \frac{1}{5} \ln |\frac{{(t - 2)}^{2}}{2 t + 1}| + C = \frac{1}{5} \ln |\frac{t - 2}{2 t + 1}| + \frac{1}{5} \ln (2) + C = \frac{1}{5} \ln |\frac{t - 2}{2 t + 1}| + C' where C' = C + \frac{1}{5} \ln (2) = \frac{1}{5} \ln |\frac{\tan x - 2}{2 \tan x + 1}| + C$

Page No 18.114:

Question 8:

$\int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let I = \int \frac{\sin (2 x)}{\sin^{4} x + \cos^{4} x} d x = \int \frac{2 \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x Dividing numerator & denominator by \cos^{4} x \Rightarrow I = \int \frac{(\frac{2 \sin x \cos x}{\cos^{4} x})}{\tan^{4} x + 1} d x = \int \frac{2 \tan x . \sec^{2} x}{{(\tan^{2} x)}^{2} + 1} d x Let \tan^{2} x = t \Rightarrow 2 \tan x \sec^{2} x d x = d t = \int \frac{d t}{t^{2} + 1} ∴ I = \tan^{- 1} (t) + C = \tan^{- 1} (\tan^{2} x) + C$

Page No 18.114:

Question 9:

$\int \frac{1}{\cos x (\sin x + 2 \cos x)} d x$

Answer:

$Let I = \int \frac{1}{\cos x (\sin x + 2 \cos x)} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\frac{\cos x}{\cos x} \times (\frac{\sin x + 2 \cos x}{\cos x})} d x = \int \frac{\sec^{2} x}{(\tan x + 2)} d x Let \tan x + 2 = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t} = \ln |t| + C = \ln |\tan x + 2| + C$

Page No 18.114:

Question 10:

$\int \frac{1}{\sin^{2} x + \sin 2 x} d x$

Answer:

$Let I = \int \frac{1}{\sin^{2} x + \sin (2 x)} d x = \int \frac{1}{\sin^{2} x + 2 \sin x \cos x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\tan^{2} x + 2 \tan x} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t^{2} + 2 t} = \int \frac{d t}{t^{2} + 2 t + 1 - 1} = \int \frac{d t}{{(t + 1)}^{2} - {(- 1)}^{2}} = \frac{1}{2} \ln |\frac{t + 1 - 1}{t + 1 + 1}| + C = \frac{1}{2} \ln |\frac{t}{t + 2}| + C = \frac{1}{2} \ln |\frac{\tan x}{\tan x + 2}| + C$

Page No 18.114:

Question 11:

$\int \frac{1}{\cos 2 x + 3 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{\cos (2 x) + 3 \sin^{2} x} d x = \int \frac{1}{(1 - 2 \sin^{2} x) + 3 \sin^{2} x} d x = \int \frac{1}{1 + \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\sec^{2} x + \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + \tan^{2} x + \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + 2 \tan^{2}} d x = \int \frac{\sec^{2} x}{1 + {(\sqrt{2} \tan x)}^{2}} d x Let \sqrt{2} \tan x = t \Rightarrow \sqrt{2} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{2}} ∴ I = \frac{1}{\sqrt{2}} \int \frac{d t}{1 + t^{2}} = \frac{1}{\sqrt{2}} \tan^{- 1} (t) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{2} \tan x) + C$

Page No 18.117:

Question 1:

$\int \frac{1}{5 + 4 \cos x} d x$

Answer:

$Let I = \int \frac{1}{5 + 4 \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} I = \int \frac{1}{5 + 4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) + 4 (1 - \tan^{2} \frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2} d x}{5 + 5 \tan^{2} \frac{x}{2} + 4 - 4 \tan^{2} \frac{x}{2}} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{\tan^{2} (\frac{x}{2}) + 9} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} + 3^{2}} = \frac{2}{3} \tan^{- 1} (\frac{t}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{\tan (\frac{x}{2})}{3}) + C$

Page No 18.117:

Question 2:

$\int \frac{1}{5 - 4 \sin x} d x$

Answer:

$Let I = \int \frac{1}{5 - 4 \sin x} d x Putting \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} \Rightarrow I = \int \frac{1}{5 - 4 \times \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) - 8 \tan \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{5 \tan^{2} (\frac{x}{2}) - 8 \tan (\frac{x}{2}) + 5} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{5 t^{2} - 8 t + 5} = \frac{2}{5} \int \frac{d t}{t^{2} - \frac{8}{5} t + 1} = \frac{2}{5} \int \frac{d t}{t^{2} - \frac{8}{5} t + {(\frac{4}{5})}^{2} - {(\frac{4}{5})}^{2} + 1} = \frac{2}{5} \int \frac{d t}{{(t - \frac{4}{5})}^{2} - \frac{16}{25} + 1} = \frac{2}{5} \int \frac{d t}{{(t - \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} = \frac{2}{5} \times \frac{5}{3} \tan^{- 1} (\frac{t - \frac{4}{5}}{\frac{3}{5}}) + C = \frac{2}{3} \tan^{- 1} (\frac{5 t - 4}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{5 \tan (\frac{x}{2}) - 4}{3}) + C$

Page No 18.117:

Question 3:

$\int \frac{1}{1 - 2 \sin x} d x$

Answer:

$Let I = \int \frac{1}{1 - 2 \sin x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{1 - 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2}) - 4 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) - 4 \tan (\frac{x}{2}) + 1} d x Let \tan (\frac{x}{2}) = t \Rightarrow \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} - 4 t + 1} = 2 \int \frac{d t}{t^{2} - 4 t + 4 - 4 + 1} = 2 \int \frac{d t}{{(t - 2)}^{2} - 3} = 2 \int \frac{d t}{{(t - 2)}^{2} - {(\sqrt{3})}^{2}} = 2 \times \frac{1}{2 \sqrt{3}} \ln |\frac{t - 2 - \sqrt{3}}{t - 2 + \sqrt{3}}| + C = \frac{1}{\sqrt{3}} \ln |\frac{\tan (\frac{x}{2}) - 2 - \sqrt{3}}{\tan (\frac{x}{2}) - 2 + \sqrt{3}}| + C$

Page No 18.117:

Question 4:

$\int \frac{1}{4 \cos x - 1} d x$

Answer:

$Let I = \int \frac{1}{4 \cos x - 1} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - 1} d x = \int \frac{1}{\frac{4 (1 - \tan^{2} \frac{x}{2}) - (1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2})}} = \int \frac{(1 + \tan^{2} \frac{x}{2}) d x}{4 - 4 \tan^{2} (\frac{x}{2}) - 1 - \tan^{2} (\frac{x}{2})} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{3 - 5 \tan^{2} (\frac{x}{2})} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{3 - 5 t^{2}} = \frac{2}{5} \int \frac{d t}{\frac{3}{5} - t^{2}} = \frac{2}{5} \int \frac{d t}{{(\frac{\sqrt{3}}{\sqrt{5}})}^{2} - t^{2}} = \frac{2}{5} \times \frac{\sqrt{5}}{2 \sqrt{3}} \ln |\frac{\frac{\sqrt{3}}{\sqrt{5}} + t}{\frac{\sqrt{3}}{\sqrt{5}} - t}| + C = \frac{1}{\sqrt{15}} \ln |\frac{\sqrt{3} + \sqrt{5} t}{\sqrt{3} - \sqrt{5} t}| + C = \frac{1}{\sqrt{15}} \ln |\frac{\sqrt{3} + \sqrt{5} \tan (\frac{x}{2})}{\sqrt{3} - \sqrt{5} \tan (\frac{x}{2})}| + C$

Page No 18.117:

Question 5:

$\int \frac{1}{1 - \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{1 - \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} = \int \frac{1}{1 - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2}) - 2 \tan x (2 + 1 - \tan^{2} \frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{2 - 2 \tan (\frac{x}{2})} d x = \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2})}{1 - \tan (\frac{x}{2})} d x Let [1 - \tan (\frac{x}{2})] = t \Rightarrow - \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = - 2 d t ∴ I = \frac{1}{2} \int \frac{- 2 d t}{t} = - \int \frac{d t}{t} = - \ln |t| + C = - \ln |1 - \tan \frac{x}{2}| + C$

Page No 18.117:

Question 6:

$\int \frac{1}{3 + 2 \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{3 + 2 \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{3 + 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{3 (1 + \tan^{2} \frac{x}{2}) + 4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{3 + 3 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{2 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 4} d x = \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 2} d x Let \tan (\frac{x}{2}) = t \Rightarrow \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \frac{1}{2} \int \frac{2 d t}{t^{2} + 2 t + 2} = \int \frac{d t}{t^{2} + 2 t + 1 + 1} = \int \frac{d t}{{(t + 1)}^{2} + 1^{2}} = \tan^{- 1} (\frac{t + 1}{1}) + C = \tan^{- 1} (1 + \tan \frac{x}{2}) + C$

Page No 18.117:

Question 7:

$\int \frac{1}{13 + 3 \cos x + 4 \sin x} d x$

Answer:

$L e t I = \int \frac{1}{13 + 3 \cos x + 4 \sin x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} ∴ I = \int \frac{1}{13 + 3 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + 4 \times 2 \frac{\tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{13 (1 + \tan^{2} \frac{x}{2}) + 3 - 3 \tan^{2} \frac{x}{2} + 8 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{13 \tan^{2} \frac{x}{2} - 3 \tan^{2} \frac{x}{2} + 16 + 8 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{10 \tan^{2} (\frac{x}{2}) + 8 \tan (\frac{x}{2}) + 16} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{10 t^{2} + 8 t + 16} = \int \frac{d t}{5 t^{2} + 4 t + 8} = \frac{1}{5} \int \frac{d t}{t^{2} + \frac{4}{5} t + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{t^{2} + \frac{4}{5} t + {(\frac{2}{5})}^{2} - {(\frac{2}{5})}^{2} + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} - \frac{4}{25} + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} + \frac{- 4 + 40}{25}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} + {(\frac{6}{5})}^{2}} = \frac{1}{5} \times \frac{5}{6} \tan^{- 1} (\frac{t + \frac{2}{5}}{\frac{6}{5}}) + C = \frac{1}{6} \tan^{- 1} (\frac{5 t + 2}{6}) + C = \frac{1}{6} \tan^{- 1} (\frac{5 \tan \frac{x}{2} + 2}{6}) + C$

Page No 18.117:

Question 8:

$\int \frac{1}{\cos x - \sin x} d x$

Answer:

$Let I = \int \frac{d x}{\cos x - \sin x} Putting \cos x = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} and \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{d x}{\frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{1 - \tan^{2} (\frac{x}{2}) - 2 \tan (\frac{x}{2})} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{1 - t^{2} - 2 t} = \int \frac{- 2 d t}{t^{2} + 2 t - 1} = \int \frac{- 2 d t}{t^{2} + 2 t + 1 - 2} = - \int \frac{2 d t}{{(t + 1)}^{2} - {(\sqrt{2})}^{2}} = \int \frac{2 d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = \frac{2}{2 \sqrt{2}} \ln |\frac{\sqrt{2} + t + 1}{\sqrt{2} - t - 1}| + C = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2} + \tan \frac{x}{2} + 1}{\sqrt{2} - \tan \frac{x}{2} - 1}| + C$

Page No 18.117:

Question 9:

$\int \frac{1}{\sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{\sin x + \cos x} d x Putting \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} and \cos x = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} = \int \frac{1}{\frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} + \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{1 - \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2})} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{1 - t^{2} + 2 t} = - 2 \int \frac{d t}{t^{2} - 2 t - 1} = - 2 \int \frac{d t}{t^{2} - 2 t + 1 - 2} = 2 \int \frac{d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = 2 \times \frac{1}{2 \sqrt{2}} \ln |\frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1}| + C = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2} + \tan \frac{x}{2} - 1}{\sqrt{2} - \tan \frac{x}{2} + 1}| + C$

Page No 18.117:

Question 10:

$\int \frac{1}{5 - 4 \cos x} d x$

Answer:

$Let I = \int \frac{1}{5 - 4 \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{5 - 4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) - 4 + 4 \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{9 \tan^{2} \frac{x}{2} + 1} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{9 t^{2} + 1} = \frac{2}{9} \int \frac{d t}{t^{2} + \frac{1}{9}} = \frac{2}{9} \int \frac{d t}{t^{2} + {(\frac{1}{3})}^{2}} = \frac{2}{9} \times 3 \tan^{- 1} (\frac{t}{\frac{1}{3}}) + C = \frac{2}{3} \tan^{- 1} (3 t) + C = \frac{2}{3} \tan^{- 1} (3 \tan \frac{x}{2}) + C$

Page No 18.117:

Question 11:

$\int \frac{1}{2 + \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{2 + \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{2 + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 (1 + \tan^{2} \frac{x}{2}) + 2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{2 + 2 \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + 3} d x Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} + 2 t + 3} = 2 \int \frac{d t}{t^{2} + 2 t + 1 + 2} = 2 \int \frac{d t}{{(t + 1)}^{2} + {(\sqrt{2})}^{2}} = 2 \times \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t + 1}{\sqrt{2}}) + C = \sqrt{2} \tan^{- 1} (\frac{\tan \frac{x}{2} + 1}{\sqrt{2}}) + C$

Page No 18.117:

Question 12:

$\int \frac{1}{\sin x + \sqrt{3} \cos x} d x$

Answer:

$Let I = \int \frac{1}{\sin x + \sqrt{3} \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \sqrt{3} \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + \sqrt{3} - \sqrt{3} \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{- \sqrt{3} \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + \sqrt{3}} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{- \sqrt{3} t^{2} + 2 t + \sqrt{3}} = - \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} - \frac{2}{\sqrt{3}} t - 1} = - \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} - \frac{2}{\sqrt{3}} t + {(\frac{1}{\sqrt{3}})}^{2} - {(\frac{1}{\sqrt{3}})}^{2} - 1} = - \frac{2}{\sqrt{3}} \int \frac{d t}{{(t - \frac{1}{\sqrt{3}})}^{2} - {(\frac{2}{\sqrt{3}})}^{2}} = - \frac{2}{\sqrt{3}} \times \frac{1}{2 \frac{2}{\sqrt{3}}} \log |\frac{t - \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t - \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}| + C$

$= - \frac{1}{2} \log |\frac{t - \frac{3}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}}}| + C = - \frac{1}{2} \log |\frac{\sqrt{3} t - 3}{\sqrt{3} t + 1}| + C = \frac{1}{2} \log |\frac{\sqrt{3} t + 1}{\sqrt{3} t - 3}| + C = \frac{1}{2} \log |\frac{\sqrt{3} \tan \frac{x}{2} + 1}{\sqrt{3} \tan \frac{x}{2} - 3}| + C or, \frac{1}{2} \log |\frac{1 + \sqrt{3} \tan \frac{x}{2}}{3 - \sqrt{3} \tan \frac{x}{2}}| + C$

Page No 18.117:

Question 13:

$\int \frac{1}{\sqrt{3} \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{d x}{\sqrt{3} \sin x + \cos x} Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\sqrt{3} \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \sqrt{3} \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{- \tan^{2} \frac{x}{2} + 2 \sqrt{3} \tan \frac{x}{2} + 1} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{- t^{2} + 2 \sqrt{3} t + 1} = - 2 \int \frac{d t}{t^{2} - 2 \sqrt{3} t - 1} = - 2 \int \frac{d t}{t^{2} - 2 \sqrt{3} t + {(\sqrt{3})}^{2} - {(\sqrt{3})}^{2} - 1} = - 2 \int \frac{d t}{{(t - \sqrt{3})}^{2} - {(2)}^{2}} = - \frac{2}{2 \times 2} \log |\frac{t - \sqrt{3} - 2}{t - \sqrt{3} + 2}| + C$

$= - \frac{1}{2} \log |\frac{\tan \frac{x}{2} - 2 - \sqrt{3}}{\tan \frac{x}{2} + 2 - \sqrt{3}}| + C = \frac{1}{2} \log |\frac{\tan \frac{x}{2} + 2 - \sqrt{3}}{\tan \frac{x}{2} + 2 - \sqrt{3}}| + C$

Page No 18.117:

Question 14:

$\int \frac{1}{\sin x - \sqrt{3} \cos x} d x$

Answer:

$Let I = \int \frac{d x}{\sin x - \sqrt{3} \cos x} Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} - \sqrt{3} \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} - \sqrt{3} + \sqrt{3} \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{\sqrt{3} \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} - \sqrt{3}} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{\sqrt{3} t^{2} + 2 t - \sqrt{3}} = \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} + \frac{2}{\sqrt{3}} t - 1} = 2 \int \frac{d t}{t^{2} + \frac{2}{\sqrt{3}} t + {(\frac{1}{\sqrt{3}})}^{2} - 1 - {(\frac{1}{\sqrt{3}})}^{2}} = 2 \int \frac{d t}{{(t + \frac{1}{\sqrt{3}})}^{2} - {(\frac{2}{\sqrt{3}})}^{2}} = \frac{2}{2 \times \frac{2}{\sqrt{3}}} \log |\frac{t + \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}| + C$
$= \frac{\sqrt{3}}{2} \log |\frac{\tan \frac{x}{2} - \frac{1}{\sqrt{3}}}{\tan \frac{x}{2} + \frac{3}{\sqrt{3}}}| + C = \frac{\sqrt{3}}{2} \log |\frac{\sqrt{3} \tan \frac{x}{2} - 1}{\sqrt{3} \tan \frac{x}{2} + 3}| + C$

Page No 18.117:

Question 15:

$\int \frac{1}{5 + 7 \cos x + \sin x} d x$

Answer:

$Let I = \int \frac{1}{5 + 7 \cos x + \sin x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{5 + 7 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) + 7 - 7 \tan^{2} \frac{x}{2} + 2 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{- 2 \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 12} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{- 2 t^{2} + 2 t + 12} = \int \frac{d t}{- t^{2} + t + 6} = \int \frac{- d t}{t^{2} - t - 6} = \int \frac{- d t}{t^{2} - t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 6} = \int \frac{- d t}{{(t - \frac{1}{2})}^{2} - \frac{1}{4} - 6} = \int \frac{- d t}{{(t - \frac{1}{2})}^{2} - {(\frac{5}{2})}^{2}} = \int \frac{d t}{{(\frac{5}{2})}^{2} - {(t - \frac{1}{2})}^{2}} = \frac{1}{2 \times \frac{5}{2}} \log |\frac{\frac{5}{2} + t - \frac{1}{2}}{\frac{5}{2} - t + \frac{1}{2}}| + C = \frac{1}{5} \log |\frac{2 + t}{3 - t}| + C = \frac{1}{5} \log |\frac{2 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}}| + C$

Page No 18.122:

Question 1:

$\int \frac{1}{1 - \cot x} d x$

Answer:

$Let I = \int \frac{1}{1 - \cot x} d x = \int \frac{1}{1 - \frac{\cos x}{\sin x}} d x = \int \frac{\sin x}{\sin x - \cos x} d x = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} d x = \frac{1}{2} \int [\frac{\sin x + \cos x + \sin x - \cos x}{\sin x - \cos x}] d x = \frac{1}{2} \int (\frac{\sin x + \cos x}{\sin x - \cos x}) d x + \frac{1}{2} \int d x Putting \sin x - \cos x = t \Rightarrow (\cos x + \sin x) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{t} d t + \frac{1}{2} \int d x = \frac{1}{2} \ln |t| + \frac{x}{2} + C = \frac{x}{2} + \frac{1}{2} \ln |\sin x - \cos x| + C$

Page No 18.122:

Question 2:

$\int \frac{1}{1 - \tan x} d x$

Answer:

$Let I = \int \frac{1}{1 - \tan x} d x = \int \frac{1}{1 - \frac{\sin x}{\cos x}} d x = \int \frac{\cos x}{\cos x - \sin x} d x = \frac{1}{2} \int \frac{2 \cos x}{\cos x - \sin x} d x = \frac{1}{2} \int (\frac{\cos x + \sin x + \cos x - \sin x}{\cos x - \sin x}) d x = \frac{1}{2} \int (\frac{\cos x + \sin x}{\cos x - \sin x}) d x + \frac{1}{2} \int d x Putting \cos x - \sin x = t \Rightarrow (- \sin x - \cos x) d x = d t \Rightarrow (\sin x + \cos x) d x = - d t ∴ I = - \frac{1}{2} \int \frac{d t}{t} + \frac{x}{2} + C = - \frac{1}{2} \ln |\cos x - \sin x| + \frac{x}{2} + C = \frac{x}{2} - \frac{1}{2} \ln |\cos x - \sin x| + C$

Page No 18.122:

Question 3:

$\int \frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} d x$

Answer:

$Let I = \int (\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3}) d x Let 3 + 2 \cos x + 4 \sin x = A (2 \sin x + \cos x + 3) + B (2 \cos x - \sin x) + C \Rightarrow 3 + 2 \cos x + 4 \sin x = (2 A - B) \sin x + (A + 2 B) \cos x + 3 A + C$

Comparing the coefficients of like terms
$2 A - B = 4 . . . (1) A + 2 B = 2 . . . (2) 3 A + C = 3 . . . (3)$

Multiplying eq (1) by 2 and adding it to eq (2) we get ,

$\Rightarrow 4 A - 2 B + A + 2 B = 8 + 2 \Rightarrow 5 A = 10 \Rightarrow A = 2$

Putting value of A = 2 in eq (1)

$\Rightarrow 2 \times 2 - B = 4 \Rightarrow B = 0 Putting value of A in eq (3) \Rightarrow 3 \times 2 + C = 3 \Rightarrow C = - 3$

$∴ I = \int [\frac{2 (2 \sin x + \cos x + 3) - 3}{2 \sin x + \cos x + 3}] d x = 2 \int d x - 3 \int \frac{1}{2 \sin x + \cos x + 3} d x Substituting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = 2 \int d x - 3 \int \frac{1}{2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + 3} d x = 2 \int d x - 3 \int \frac{(1 + \tan^{2} \frac{x}{2})}{4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2}) + 3 (1 + \tan^{2} \frac{x}{2})} d x = 2 \int d x - 3 \int \frac{\sec^{2} (\frac{x}{2})}{2 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 4} d x = 2 \int d x - \frac{3}{2} \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 2} d x Putting \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int d x - \frac{3}{2} \int \frac{2}{t^{2} + 2 t + 2} d t = 2 \int d x - 3 \int \frac{1}{t^{2} + 2 t + 1 + 1} d t = 2 \int d x - 3 \int \frac{1}{{(t + 1)}^{2} + {(1)}^{2}} d t = 2 x - \frac{3}{1} \tan^{- 1} (\frac{t + 1}{1}) + C = 2 x - 3 \tan^{- 1} (\tan \frac{x}{2} + 1) + C [∵ t = \tan \frac{x}{2}]$

Page No 18.122:

Question 4:

$\int \frac{1}{p + q \tan x} d x$

Answer:

$Let I = \int \frac{d x}{p + q \tan x} = \int \frac{1}{p + \frac{q \sin x}{\cos x}} d x = \int \frac{\cos x}{q \sin x + p \cos x} d x Let \cos x = A (q \sin x + p \cos x) + B (q \cos x - p \sin x) \Rightarrow \cos x = (A p + B q) \cos x + (A q - B p) \sin x$

Comparing coefficients of like terms

$A p + B q = 1 . . . (1) A q - B p = 0 . . . (2)$

Multiplying eq (1) by p and eq (2) by q and then adding

$\Rightarrow A p^{2} + B p q = p \Rightarrow A q^{2} - B p q = 0 \Rightarrow A = \frac{p}{p^{2} q^{2}}$

Putting value of A in eq (1)

\frac{p^{2}}{p^{2} + q^{2}} + B q = 1 \Rightarrow B q = 1 - \frac{p^{2}}{p^{2} + q^{2}} \Rightarrow B q = \frac{p^{2} + q^{2} - p^{2}}{p^{2} + q^{2}} \Rightarrow B = \frac{q}{p^{2} + q^{2}} ∴ I = \int [\frac{p}{p^{2} + q^{2}} \times \frac{(q \sin x + p \cos x)}{(q \sin x + p \cos x)} + \frac{q}{p^{2} + q^{2}} \times \frac{(q \cos x - p \sin x)}{(q \sin x + p \cos x)}] d x = \frac{p}{p^{2} + q^{2}} \int d x + \frac{q}{p^{2} + q^{2}} \int (\frac{q \cos x - p \sin x}{q \sin x + p \cos x}) d x Putting q \sin x + p \cos x = t \Rightarrow (q \cos x - p \sin x) d x = d t ∴ I = \frac{p}{p^{2} + q^{2}} \int d x + \frac{q}{p^{2} + q^{2}} \int \frac{1}{t} d t = \frac{p}{p^{2} + q^{2}} x + \frac{q}{p^{2} + q^{2}} \ln |q \sin x + p \cos x| + C

Page No 18.122:

Question 5:

$\int \frac{5 \cos x + 6}{2 \cos x + \sin x + 3} d x$

Answer:

$Let I = \int (\frac{5 \cos x + 6}{2 \cos x + \sin x + 3}) d x & let 5 \cos x + 6 = A (2 \cos x + \sin x + 3) + B (- 2 \sin x + \cos x) + C . . . . (1) \Rightarrow 5 \cos x + 6 = (A - 2 B) \sin x + (2 A + B) \cos x + 3 A + C$

Comparing coefficients of like terms

$A - 2 B = 0 . . . (2) 2 A + B = 5 . . . (3) 3 A + C = 6 . . . (4)$

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10
$\Rightarrow$ A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

$By putting the values of A, B and C in eq (1) we get, ∴ I = \int [\frac{2 (2 \cos x + \sin x + 3) + (- 2 \sin x + \cos x)}{(2 \cos x + \sin x + 3)}] d x = 2 \int d x + \int (\frac{- 2 \sin x + \cos x}{2 \cos x + \sin x + 3}) d x Putting 2 \cos x + \sin x + 3 = t \Rightarrow (- 2 \sin x + \cos x) d x = d t ∴ I = 2 \int d x + \int \frac{1}{t} d t = 2 x + \ln |2 \cos x + \sin x + 3| + C$

Page No 18.122:

Question 6:

$\int \frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} d x$

Answer:

$Let I = \int (\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x}) d x & let 2 \sin x + 3 \cos x = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . (1) \Rightarrow 2 \sin x + 3 \cos x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x$

By comparing the coefficients of like terms we get,

$3 A - 4 B = 2 . . . (2) 4 A - 3 B = 3 . . . (3)$

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

$9 A - 12 B + 16 A + 12 B = 6 + 12 \Rightarrow 25 A = 18 \Rightarrow A = \frac{18}{25} Putting value of A = \frac{18}{25} in eq (2) we get, 3 \times \frac{18}{25} - 4 B = 2 \Rightarrow \frac{54}{25} - 2 = 4 B \Rightarrow \frac{4}{25 \times 4} = B \Rightarrow B = \frac{1}{25}$
Thus, substituting the values of A,B and C in eq (1) we get ,

$I = \int [\frac{\frac{18}{25} (3 \sin x + 4 \cos x) + \frac{1}{25} (3 \cos x - 4 \sin x)}{(3 \sin x + 4 \cos x)}] d x = \frac{18}{25} \int d x + \frac{1}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{18}{25} \int d x + \frac{1}{25} \int \frac{1}{t} d t = \frac{18 x}{25} + \frac{1}{25} \ln |t| + C = \frac{18 x}{25} + \frac{1}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 7:

$\int \frac{1}{3 + 4 \cot x} d x$

Answer:

$Let I = \int \frac{1}{3 + 4 \cot x} d x = \int \frac{1}{3 + \frac{4 \cos x}{\sin x}} d x = \int \frac{\sin x}{3 \sin x + 4 \cos x} d x Let \sin x = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . (1) \Rightarrow \sin x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x By comparing the coefficients of both sides we get, 3 A - 4 B = 1 . . . (2) 4 A + 3 B = 0 . . . (3)$

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

$9 A - 12 B + 16 A + 12 B = 3 + 0 \Rightarrow 25 A = 3 \Rightarrow A = \frac{3}{25} Putting value of A in eq (3) we get, 4 \times \frac{3}{25} + 3 B = 0 \Rightarrow 3 B = - \frac{12}{25} \Rightarrow B = - \frac{4}{25}$
$Thus, by substituting the value of A and B in eq (1) we get I = \int [\frac{\frac{3}{25} (3 \sin x + 4 \cos x) - \frac{4}{25} (3 \cos x - 4 \sin x)}{3 \sin x + 4 \cos x}] d x = \frac{3}{25} \int d x - \frac{4}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{3}{25} \int d x - \frac{4}{25} \int \frac{d t}{t} = \frac{3}{25} x - \frac{4}{25} \ln |t| + C = \frac{3 x}{25} - \frac{4}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 8:

$\int \frac{2 \tan x + 3}{3 \tan x + 4} d x$

Answer:

$Let I = \int (\frac{2 \tan x + 3}{3 \tan x + 4}) d x = \int (\frac{\frac{2 \sin x}{\cos x} + 3}{\frac{3 \sin x}{\cos x} + 4}) d x = \int (\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x}) d x Let (2 \sin x + 3 \cos x) = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . . (1) \Rightarrow 2 \sin x + 3 \cos x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x Equating the coefficients of like terms 3 A - 4 B = 2 . . . (2) 4 A + 3 B = 3 . . . (3)$

Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get

$\begin{matrix} 9 A & - & 12 B & = & 6 \\ 16 A & + & 12 B & = & 12 \end{matrix} \begin{matrix} 25 A & = & 18 \end{matrix} \Rightarrow A = \frac{18}{25} Putting value of A in eq (2) we get, \Rightarrow B = \frac{1}{25}$
$Thus, by substituting the values of A and B in eq (1) we get, I = \int \{\frac{\frac{18}{25} (3 \sin x + 4 \cos x) + \frac{1}{25} (3 \cos x - 4 \sin x)}{3 \sin x + 4 \cos x}\} d x = \frac{18}{25} \int d x + \frac{1}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{18}{25} x + \frac{1}{25} \int \frac{1}{t} d t = \frac{18 x}{25} + \frac{1}{25} \ln |t| + C = \frac{18 x}{25} + \frac{1}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 9:

$\int \frac{1}{4 + 3 \tan x} d x$

Answer:

$Let I = \int \frac{d x}{4 + 3 \tan x} = \int \frac{d x}{4 + \frac{3 \sin x}{\cos x}} = \int \frac{\cos x d x}{4 \cos x + 3 \sin x} Consider, \cos x = A (4 \cos x + 3 \sin x) + B \frac{d}{dx} (4 \cos x + 3 \sin x) \Rightarrow \cos x = A (4 \cos x + 3 \sin x) + B (- 4 \sin x + 3 \cos x) \Rightarrow \cos x = (4 A + 3 B) \cos x + (3 A - 4 B) \sin x Equating the coefficients of like terms 4 A + 3 B = 1 . . . . . (1) 3 A - 4 B = 0 . . . . . (2)$
Solving (1) and (2), we get
$A = \frac{4}{25} and B = \frac{3}{25}$
$\int [\frac{\frac{4}{25} (4 \cos x + 3 \sin x) + (- 4 \sin x + 3 \cos x) \frac{3}{25}}{4 \cos x + 3 \sin x}] d x = \frac{4}{25} \int d x + \frac{3}{25} \int (\frac{- 4 \sin x + 3 \cos x}{4 \cos x + 3 \sin x}) d x let 4 \cos x + 3 \sin x = t \Rightarrow (- 4 \sin x + 3 \cos x) d x = d t Then, I = \frac{4}{25} \int d x + \frac{3}{25} \int \frac{d t}{t} = \frac{4 x}{25} + \frac{3}{25} \log |t| + C = \frac{4 x}{25} + \frac{3}{25} \log |4 \cos x + 3 \sin x| + C$

Page No 18.122:

Question 10:

$\int \frac{8 \cot x + 1}{3 \cot x + 2} d x$

Answer:

$L e t I = \int (\frac{8 \cot x + 1}{3 \cot x + 2}) d x = \int (\frac{8 \frac{\cos x}{\sin x} + 1}{\frac{3 \cos x}{\sin x} + 2}) d x = \int (\frac{8 \cos x + \sin x}{3 \cos x + 2 \sin x}) d x Now, let 8 \cos x + \sin x = A (3 \cos x + 2 \sin x) + B (- 3 \sin x + 2 \cos x) . . . (1) \Rightarrow 8 \cos x + \sin x = 3 A \cos x + 2 A \sin x - 3 B \sin x + 2 B \cos x \Rightarrow 8 \cos x + \sin x = (3 A + 2 B) \cos x + (2 A - 3 B) \sin x Equating the coefficients of like terms we get, 2 A - 3 B = 1 . . . (2) 3 A + 2 B = 8 . . . (3)$

Solving eq (2) and eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,

$I = \int [\frac{2 (3 \cos x + 2 \sin x) + 1 (- 3 \sin x + 2 \cos x)}{(3 \cos x + 2 \sin x)}] d x = 2 \int (\frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x}) d x + \int (\frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x}) d x = 2 \int d x + \int (\frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x}) d x Putting 3 \cos x + 2 \sin x = t \Rightarrow (- 3 \sin x + 2 \cos x) d x = d t ∴ I = 2 \int d x + \int \frac{1}{t} d t = 2 x + \ln |t| + C = 2 x + \ln |3 \cos x + 2 \sin x| + C$

Page No 18.122:

Question 11:

$\int \frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} d x$

Answer:

$Let I = \int (\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x}) d x & let 4 \sin x + 5 \cos x = A (5 \sin x + 4 \cos x) + B (5 \cos x - 4 \sin x) . . . (1) \Rightarrow 4 \sin x + 5 \cos x = (5 A - 4 B) \sin x + (4 A + 5 B) \cos x By equating the coefficients of like terms we get, 5 A - 4 B = 4 . . . (2) 4 A + 5 B = 5 . . . (3)$

By solving eq (2) and eq (3) we get,
$A = \frac{40}{41}, B = \frac{9}{41} Thus, by substituting the values of A and B in eq (1), we get I = \int [\frac{\frac{40}{41} (5 \sin x + 4 \cos x) + \frac{9}{41} (5 \cos x - 4 \sin x)}{(5 \sin x + 4 \cos x)}] d x = \frac{40}{41} \int d x + \frac{9}{41} \int (\frac{5 \cos x - 4 \sin x}{5 \sin x + 4 \cos x}) d x Putting 5 \sin x + 4 \cos x = t \Rightarrow (5 \cos x - 4 \sin x) d x = d t ∴ I = \frac{40}{41} x + \frac{9}{41} \int \frac{1}{t} d t = \frac{40}{41} x + \frac{9}{41} \ln |t| + C = \frac{40}{41} x + \frac{9}{41} \ln |5 \sin x + 4 \cos x| + C$

Page No 18.133:

Question 1:

$\int x \cos x d x$

Answer:

$\int x \cos x d x Taking x as the first function and \cos x as the second function . = x \int \cos x d x - \int \{\frac{d}{d x} (x) \int \cos x d x\} d x = x \sin x - \int \sin x d x = x \sin x + \cos x + C$

Page No 18.133:

Question 2:

$\int \log (x + 1) d x$

Answer:

$\int \log (x + 1) d x = \int 1 . \log (x + 1) d x Taking \log (x + 1) as the first function and 1 as the second function . = \log (x + 1) \int 1 d x - \int [\frac{d}{d x} \{\log (x + 1)\} \int 1 d x] d x = x \log (x + 1) - \int \frac{x}{x + 1} d x = x \log (x + 1) - \int \frac{x + 1}{x + 1} - \frac{1}{x + 1} d x = x \log (x + 1) - x + \log |x + 1| + C$

Page No 18.133:

Question 3:

$\int x^{3} \log x d x$

Answer:

$\int x^{3} \log x d x Taking \log x as the first function and x^{3} as the second function . = \log x \int x^{3} d x - \int (\frac{d}{d x} \log x \int x^{3} d x) d x = (\log x) \frac{x^{4}}{4} - \int \frac{1}{x} (\frac{x^{4}}{4}) d x = \frac{x^{4}}{x} \log x - \frac{x^{4}}{16} + C$

Page No 18.133:

Question 4:

$\int x e^{x} d x$

Answer:

$\int x e^{x} d x Taking x as the first function and e^{x} as the second function . = x \int e^{x} d x - \int \{\frac{d}{d x} (x) \int e^{x} d x\} d x = x e^{x} - \int 1 (e^{x}) d x = x e^{x} - e^{x} + C = (x - 1) e^{x} + C$

Page No 18.133:

Question 5:

$\int x e^{2 x} d x$

Answer:

$\int x e^{2 x} d x Taking x as the first function and e^{2 x} as the second function . = x \int e^{2 x} d x - \int \{\frac{d}{d x} (x) \int e^{2 x} d x\} d x = \frac{x e^{2 x}}{2} - \int (\frac{e^{2 x}}{2}) d x = \frac{x}{2} e^{2 x} - \frac{e^{2 x}}{4} + C = e^{2 x} (\frac{x}{2} - \frac{1}{4}) + C$

Page No 18.133:

Question 6:

$\int x^{2} e^{- x} d x$

Answer:

$\int x^{2} e^{- x} d x Taking x^{2} as the first function and e^{- x} as the second function . = x^{2} \int e^{- x} d x - \int (\frac{d}{d x} x^{2} \int e^{- x} d x) d x = - x^{2} e^{- x} - \int 2 x (e^{- x}) (- 1) d x = - x^{2} e^{- x} + 2 \int x e^{- x} d x = - x^{2} e^{- x} + 2 [- x e^{- x} + \int e^{- x} d x] = - x^{2} e^{- x} + 2 [- x e^{- x} - e^{- x}] + C = - e^{- x} [x^{2} + 2 x + 2] + C$

Page No 18.133:

Question 7:

$\int x^{2} \cos x d x$

Answer:

$\int x^{2} \cos x d x Taking x^{2} as the first function and \cos x as the second function . = x^{2} \int \cos x d x - \int (\frac{d}{d x} x^{2} \int \cos x d x) d x = x^{2} \sin x - \int 2 x \sin x d x = x^{2} \sin x - 2 [x \int \sin x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = x^{2} \sin x - 2 [- x \cos x + \int \cos x d x] = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 18.133:

Question 8:

$\int x^{2} \cos 2 x d x$

Answer:

$\int x^{2} \cos 2 x d x Taking x^{2} as the first function and \cos 2 x as the second function . = x^{2} \int \cos 2 x d x - \int (2 x \int \cos 2 x d x) d x = \frac{x^{2} \sin 2 x}{2} - \int \frac{2 x \sin 2 x}{2} d x = \frac{x^{2}}{2} \sin 2 x - \int x \sin 2 x d x = \frac{x^{2}}{2} \sin 2 x - [x \int \sin 2 x - \int (\int \sin 2 x d x) d x] = \frac{x^{2}}{2} \sin 2 x - [\frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x] = \frac{x^{2}}{2} \sin 2 x + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + C$

Page No 18.133:

Question 9:

$\int x \sin 2 x d x$

Answer:

$\int x \sin 2 x d x Taking x as the first function and \sin 2 x as the second function . = x \int \sin 2 x d x - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x = \frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x = \frac{- x \cos 2 x}{2} + \frac{\sin 2 x}{4} + C$

Page No 18.133:

Question 10:

$\int \frac{\log (\log x)}{x} d x$

Answer:

$\int \frac{\log (\log x)}{x} d x Taking \log \log x as the first function and \frac{1}{x} as the second function . = \log \log x \int \frac{1}{x} d x - \int \{\frac{d}{d x} \log (\log x) \int \frac{1}{x} d x\} d x = \log x . \log (\log x) - \int \frac{1}{x \log x} (\log x) d x = \log x . \log (\log x) - \int \frac{1}{x} d x = \log x . \log (\log x) - \log x + C$

$= \log x [\log (\log x) - 1] + C$

Page No 18.133:

Question 11:

$\int x^{2} \cos x d x$

Answer:

$\int x^{2} \cos x d x Taking x^{2} as the first function and \cos x as the second function . = x^{2} \int \cos x d x - \int \{\frac{d}{d x} (x^{2}) \int \cos x d x\} d x = x^{2} \sin x - \int 2 x \sin x d x = x^{2} \sin x - 2 [x \int \sin x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = x^{2} \sin x + 2 x \cos x - 2 \int \cos x d x = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 18.133:

Question 12:

$\int x {cosec}^{2} x d x$

Answer:

$\int x {cosec}^{2} x d x Taking x as the first function and {cosec}^{2} x as the second function . = x \int {cosec}^{2} x d x - \int \{\frac{d}{d x} (x) \int {cosec}^{2} x d x\} d x = - x \cot x + \int \cot x d x = - x \cot x + \log |\sin x| + c$

Page No 18.133:

Question 13:

$\int x \cos^{2} x d x$

Answer:

$\int x \cos^{2} x d x Taking x as the first function and \cos^{2} x as the second function . = x \int \frac{1 + \cos 2 x}{2} d x - \int \{\frac{d}{d x} (x) \int \frac{1 + \cos 2 x}{2} d x\} d x = \frac{x}{2} [x + \frac{\sin 2 x}{2}] - \int \frac{1}{2} (x + \frac{\sin 2 x}{2}) d x = \frac{x}{2} [x + \frac{\sin 2 x}{2}] - [\frac{x^{2}}{4} - \frac{\cos 2 x}{8}] + C = \frac{x^{2}}{2} + \frac{x \sin 2 x}{2} - \frac{x^{2}}{4} + \frac{\cos 2 x}{8} + C = \frac{x^{2}}{4} + \frac{x \sin 2 x}{2} + \frac{\cos 2 x}{8} + C$

Page No 18.133:

Question 14:

$\int x^{n} \cdot \log x d x$

Answer:

$\int x^{n} \log x d x Taking \log x as the first function and x^{n} as the second function . = \log x \int x^{n} d x - \int (\frac{d}{d x} \log x \int x^{n} d x) d x = \log x (\frac{x^{n + 1}}{n + 1}) - \int \frac{1}{x} (\frac{x^{n + 1}}{n + 1}) d x = \log x (\frac{x^{n + 1}}{n + 1}) - \int \frac{x^{n}}{n + 1} d x = \log x (\frac{x^{n + 1}}{n + 1}) - \frac{x^{n + 1}}{{(n + 1)}^{2}} + C$

Page No 18.133:

Question 15:

$\int \frac{\log x}{x^{n}} d x$

Answer:

$\int \frac{1}{x^{n}} \log x d x Taking \log x as the first function and \frac{1}{x^{n}} as the second function . = \log x \int \frac{1}{x^{n}} d x - \int (\frac{d}{d x} \log x \int \frac{1}{x^{n}} d x) d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \int \frac{1}{x} (\frac{x^{- n + 1}}{- n + 1}) d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \int \frac{x^{- n}}{- n + 1} d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \frac{x^{- n + 1}}{{(- n + 1)}^{2}} + C = \log x (\frac{x^{1 - n}}{1 - n}) - \frac{x^{1 - n}}{{(1 - n)}^{2}} + C$

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Question 16:

$\int x^{2} \sin^{2} x d x$

Answer:

$\int x^{2} \sin^{2} x d x Taking x^{2} as the first function and \sin^{2} x as the second function . = x^{2} \int \frac{1 - \cos 2 x}{2} - \int \{\frac{d}{d x} (x^{2}) \int \frac{1 - \cos 2 x}{2} d x\} d x = \frac{x^{2}}{2} [x - \frac{\sin 2 x}{2}] - \int 2 x (\frac{x - \frac{\sin 2 x}{2}}{2}) d x = \frac{x^{2}}{2} (x - \frac{\sin 2 x}{2}) - \int x^{2} d x + \int \frac{x \sin 2 x}{2} d x [Here, taking x as the first function and \sin 2 x as the second function .] = \frac{x^{3}}{2} - \frac{x^{2} \sin 2 x}{4} - \frac{x^{3}}{3} + \frac{1}{2} [x \int \sin 2 x - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x] = \frac{x^{3}}{2} - \frac{x^{2} \sin 2 x}{4} - \frac{x^{3}}{3} + \frac{1}{2} [\frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x d x}{4}] = \frac{x^{3}}{6} - \frac{x^{2} \sin 2 x}{4} - \frac{x \cos 2 x}{4} + \frac{\sin 2 x}{8} + C$

Page No 18.133:

Question 17:

$\int 2 x^{3} e^{x^{2}} d x$

Answer:

$\int 2 x^{3} \cdot e^{x^{2}} d x = \int x^{2} \cdot (e^{x^{2}}) \cdot 2 x d x L et x^{2} = t \Rightarrow 2 x d x = d t = \int \underset{I}{t} \cdot {\underset{II}{e}}^{t} d t = t \cdot e^{t} - \int 1 \cdot e^{t} d t = t e^{t} - e^{t} + C = x^{2} e^{x^{2}} - e^{x^{2}} + C = e^{x^{2}} (x^{2} - 1) + C$

Page No 18.133:

Question 18:

$\int x^{3} \cos x^{2} d x$

Answer:

$\int x^{3} \cos x^{2} d x Let x^{2} = t \Rightarrow 2 x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{2 x} = \frac{1}{2} [\int t \cos t d t] Taking t as the first function and \cos t as the second function . = \frac{1}{2} [t \sin t - \int \sin t d t] = \frac{1}{2} [t \sin t + \cos t] . . . (1) Substituting the value of t in eq (1) = \frac{x^{2} \sin x^{2}}{2} + \frac{\cos x^{2}}{2} + c$

Page No 18.133:

Question 19:

$\int x \sin x \cos x d x$

Answer:

$\int x \sin x \cdot \cos x d x = \frac{1}{2} \int x (2 \sin x \cos x) d x = \frac{1}{2} \int \underset{}{x} \cdot \underset{}{\sin (2 x)} d x Taking x as the first function and \sin 2 x as the second function . = \frac{1}{2} [x \int \sin 2 x dx - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x] = \frac{1}{2} [x \times \frac{- \cos (2 x)}{2} - \int 1 \cdot (\frac{- \cos 2 x}{2}) d x] = \frac{1}{2} [\frac{- x \cos (2 x)}{2} + \frac{\sin (2 x)}{4}] + C = \frac{- x \cos (2 x)}{4} + \frac{\sin (2 x)}{8} + C$

Page No 18.133:

Question 20:

$\int \sin x \log (\cos x) d x$

Answer:

$Let I = \int \sin x \cdot \log (\cos x) d x Let \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int \log t d t = - \int \underset{}{1} \cdot \underset{}{\log t} d t Taking \log t as the first function and 1 as the second function . = \log t \int 1 d t - \int \{\frac{d}{d t} (\log t) \int 1 d t\} d t = - [\log t \cdot t - \int \frac{1}{t} \times t d t] = - [\log t \cdot t - t] + C = - t (\log t - 1) + C . . . . (1) Substituting the value of t in eq (1) = - \cos x \{\log (\cos x) - 1\} + C = \cos x \{1 - \log (\cos x)\} + C$

Page No 18.133:

Question 21:

$\int {(\log x)}^{2} \cdot x d x$

Answer:

$\int {\underset{}{(\log x)}}^{2} \underset{}{x \cdot} d x Taking {(\log x)}^{2} as the first function and x as the second function . = {(\log x)}^{2} \int x d x - \int \{\frac{d}{d x} {(\log x)}^{2} \int x d x\} d x = {(\log x)}^{2} \cdot \frac{x^{2}}{2} - \int \frac{(2 \log x)}{x} \times \frac{x^{2}}{2} d x = {(\log x)}^{2} \times \frac{x^{2}}{2} - \int \underset{II}{x} \underset{I}{\log x} d x = {(\log x)}^{2} \times \frac{x^{2}}{2} - [\log x \int x d x - \int \{\frac{d}{d x} (\log x) \int x d x\} d x] = {(\log x)}^{2} \times \frac{x^{2}}{2} - [\log x \cdot \frac{x^{2}}{2} - \int \frac{1}{x} \times \frac{x^{2}}{2} d x] = {(\log x)}^{2} \times \frac{x^{2}}{2} - \log x \cdot \frac{x^{2}}{2} + \frac{x^{2}}{4} + C = \frac{x^{2}}{2} [{(\log x)}^{2} - \log x + \frac{1}{2}] + C$

Page No 18.133:

Question 22:

$\int e^{\sqrt{x}} d x$

Answer:

$Let I = \int e^{\sqrt{x}} d x = \int \sqrt{x} \cdot \frac{e^{\sqrt{x}}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{}{t} \cdot \underset{}{e^{t}} d t Taking t as the first function and e^{t} as the second function . = 2 [t \int e^{t} d t - \int \{\frac{d}{d t} (t) \int e^{t} d t\} d t] = 2 [t \cdot e^{t} - \int 1 \cdot e^{t} d t] + C . . . (1) Substituting the value of t in eq (1) = 2 [\sqrt{x} e^{\sqrt{x}} - e^{\sqrt{x}}] + C = 2 e^{\sqrt{x}} (\sqrt{x} - 1) + C$

Page No 18.133:

Question 23:

$\int \frac{\log (x + 2)}{{(x + 2)}^{2}} d x$

Answer:

$Let I = \int \frac{\log (x + 2) d x}{{(x + 2)}^{2}} Let \log (x + 2) = t \Rightarrow x + 2 = e^{t} \Rightarrow \frac{1}{(x + 2)} d x = d t ∴ I = \int \frac{t}{e^{t}} d t = \int t e^{- t} d t Taking t as the first function and e^{- t} as the second function . = t \int e^{- t} - \int \{\frac{d}{d t} (t) \int e^{- 2 t} d t\} d t = t \times \frac{e^{- t}}{- 1} - \int 1 \cdot e^{- t} d t = - t e^{- t} + \frac{e^{- t}}{- 1} + C = - e^{- t} (t + 1) + C = - \frac{(t + 1)}{e^{t}} + C . . . (1) Substituting the value of t in eq (1) = \frac{- [\log (x + 2) + 1]}{x + 2} + C = - \frac{\log (x + 2)}{x + 2} - \frac{1}{(x + 2)} + C$

Page No 18.133:

Question 24:

$\int \frac{x + \sin x}{1 + \cos x} d x$

Answer:

$\int (\frac{x + \sin x}{1 + \cos x}) d x = \int [\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x}] d x = \int [\frac{x}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}] d x = \frac{1}{2} \int \underset{I}{x} \cdot \underset{II}{\sec^{2}} \frac{x}{2} d x + \int \tan \frac{x}{2} d x = \frac{1}{2} [x \cdot \frac{\tan (\frac{x}{2})}{\frac{1}{2}} - \int 1 \times 2 \tan (\frac{x}{2}) d x] + \frac{\log |s e c \frac{x}{2}|}{\frac{1}{2}} + C = x \tan (\frac{x}{2}) - \frac{\log |\sec \frac{x}{2}|}{\frac{1}{2}} + \log \frac{|\sec \frac{x}{2}|}{\frac{1}{2}} + C = x \tan (\frac{x}{2}) + C$

Page No 18.133:

Question 25:

$\int \log_{10} x d x$

Answer:

$\int \log_{10} x d x = \int \frac{\log x}{\log 10} d x = \frac{1}{\log 10} \int \underset{}{1} \cdot \underset{}{\log x} d x Taking \log x as the first function and 1 as the second function = \frac{1}{\log 10} [\log x \int 1 d x - \int \{\frac{d}{d x} (\log x) \int 1 d x\} d x] = \frac{1}{\log 10} [\log x \cdot x - \int \frac{1}{x} \cdot x d x] = \frac{1}{\log 10} [x \log x - x] + C = \frac{1}{\log 10} [x (\log x - 1)] + C$

Page No 18.133:

Question 26:

$\int \cos \sqrt{x} d x$

Answer:

$Let I = \int \cos \sqrt{x} d x = \int \frac{\sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{}{t} \cdot \underset{}{\cos (t)} \cdot d t Taking t as the first function and \cos t as the second function . = 2 [t \cdot \sin t - \int 1 \cdot \sin t d t] = 2 [t \cdot \sin t + \cos t] + C . . . . (1) Substituting the value of t in eq (1) = 2 [\sqrt{x} \cdot \sin \sqrt{x} + \cos \sqrt{x}] + C$

Page No 18.133:

Question 27:

Evaluate the following integrals:

$\int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x Let the first function be \cos^{- 1} x and second function be \frac{x}{\sqrt{1 - x^{2}}} . First we find the integral of the second function, i . e ., \int \frac{x}{\sqrt{1 - x^{2}}} d x . Put t = 1 - x^{2} . Then d t = - 2 x d x Therefore, \int \frac{x}{\sqrt{1 - x^{2}}} d x = - \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = - \sqrt{t} = - \sqrt{1 - x^{2}} Hence, using integration by parts, we get \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x = (\cos^{- 1} x) \int \frac{x}{\sqrt{1 - x^{2}}} d x - \int [(\frac{d (\cos^{- 1} x)}{d x}) \int (\frac{x}{\sqrt{1 - x^{2}}} d x)] d x = (\cos^{- 1} x) (- \sqrt{1 - x^{2}}) - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) (- \sqrt{1 - x^{2}}) d x = - \sqrt{1 - x^{2}} \cos^{- 1} x - x + c Hence, \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x = - \sqrt{1 - x^{2}} \cos^{- 1} x - x + c$

Page No 18.133:

Question 28:

Evaluate the following integrals:

$\int \frac{\log x}{{(x + 1)}^{2}} d x$

Answer:

$Let I = \int \frac{\log x}{{(x + 1)}^{2}} d x Let the first function be (\log x) and second function be \frac{1}{{(x + 1)}^{2}} . First we find the integral of the second function, i . e ., \int \frac{1}{{(x + 1)}^{2}} d x . Put t = (x + 1) . Then d t = d x Therefore, \int \frac{1}{{(x + 1)}^{2}} d x = \int t^{- 2} d t = - \frac{1}{t} = - \frac{1}{1 + x} Hence, using integration by parts, we get \int \frac{\log x}{{(x + 1)}^{2}} d x = (\log x) \int \frac{1}{{(x + 1)}^{2}} d x - \int [(\frac{d (\log x)}{d x}) \int \frac{1}{{(x + 1)}^{2}} d x] d x = (\log x) (- \frac{1}{1 + x}) - \int (\frac{1}{x}) (- \frac{1}{1 + x}) d x = - \frac{\log x}{1 + x} + \int (\frac{1}{x^{2} + x}) d x = - \frac{\log x}{1 + x} + \int \frac{1}{x^{2} + x + \frac{1}{4} - \frac{1}{4}} d x = - \frac{\log x}{1 + x} + \int \frac{1}{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x = - \frac{\log x}{1 + x} + \frac{1}{2 \times \frac{1}{2}} \log |\frac{x + \frac{1}{2} - \frac{1}{2}}{x + \frac{1}{2} + \frac{1}{2}}| + c = - \frac{\log x}{1 + x} + \log |\frac{x}{x + 1}| + c Hence, \int \frac{\log x}{{(x + 1)}^{2}} d x = - \frac{\log x}{1 + x} + \log |\frac{x}{x + 1}| + c$

Page No 18.133:

Question 29:

$\int {cosec}^{3} x d x$

Answer:

$Let I = \int {cosec}^{3} x d x = \int {cosec}^{2} x \cdot cosec x d x = \int {cosec}^{2} x \cdot \sqrt{1 + \cot^{2} x} d x Let \cot x = t \Rightarrow - {cosec}^{2} x d x = d t ∴ I = - \int \sqrt{1 + t^{2}} d t = - \frac{t}{2} \sqrt{1 + t^{2}} - \frac{1^{2}}{2} \log |t + \sqrt{1 + t^{2}}| + C . . . (1) Substituting the value of t in eq (1) = - \frac{\cot x}{2} \cdot cosec x - \frac{1}{2} \log |\cot x + cosec x| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\frac{\cos x}{\sin x} + \frac{1}{\sin x}| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\cot \frac{x}{2}| + C = - \frac{1}{2} cosec x \cot x + \frac{1}{2} \log |\tan \frac{x}{2}| + C (∵ \log |\cot \frac{x}{2}| = \log |\frac{1}{\tan \frac{x}{2}}| \Rightarrow - \log |\tan \frac{x}{2}|)$

Page No 18.133:

Question 30:

$\int \sec^{- 1} \sqrt{x} d x$

Answer:

$\int \underset{II}{1} . \underset{I}{\sec^{- 1} \sqrt{x}} d x = \underset{}{\sec^{- 1} \sqrt{x}} \int 1 d x - \int \{\frac{d}{d x} (\sec^{- 1} \sqrt{x}) \int 1 d x\} d x = \sec^{- 1} \sqrt{x} . x - \int \frac{1}{\sqrt{x} \sqrt{1 - x}} \times \frac{1}{2 \sqrt{x}} \times x d x = x \sec^{- 1} \sqrt{x} - \frac{1}{2} \int {(1 - x)}^{- \frac{1}{2}} d x = x \sec^{- 1} x - \frac{1}{2} [\frac{{(1 - x)}^{- \frac{1}{2} + 1}}{(- \frac{1}{2} + 1) (- 1)}] + C = x \sec^{- 1} x + {(1 - x)}^{\frac{1}{2}} + C$

Page No 18.134:

Question 31:

$\int \sin^{- 1} \sqrt{x} d x$

Answer:

$Let I = \int \sin^{- 1} \sqrt{x} d x = \int \frac{\sqrt{x} . \sin^{- 1} \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = \int \underset{II}{t} . \underset{I}{\sin^{- 1} t} d t = \sin^{- 1} t \int t d t - \int \{\frac{d}{d t} (\sin^{- 1} t) \int t d t\} d t = 2 [\sin^{- 1} t . \frac{t^{2}}{2} - \int \frac{1}{\sqrt{1 - t^{2}}} \times \frac{t^{2}}{2} d t] = \sin^{- 1} t . t^{2} - \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t = \sin^{- 1} t . t^{2} + \int (\frac{1 - t^{2} - 1}{\sqrt{1 - t^{2}}}) d t = \sin^{- 1} t . t^{2} + \int \sqrt{1 - t^{2}} d t - \int \frac{d t}{\sqrt{1 - t^{2}}} = \sin^{- 1} t . t^{2} + \frac{t}{2} \sqrt{1 - t^{2}} + \frac{1}{2} \sin^{- 1} t - \sin^{- 1} t + C = \sin^{- 1} t . t^{2} + \frac{t}{2} \sqrt{1 - t^{2}} - \frac{1}{2} \sin^{- 1} t + C = x . \sin^{- 1} \sqrt{x} + \frac{\sqrt{x}}{2} \sqrt{1 - x} - \frac{1}{2} \sin^{- 1} (\sqrt{x}) + C (∵ \sqrt{x} = t) = \frac{(2 x - 1) \sin^{- 1} \sqrt{x}}{2} + \frac{\sqrt{x - x^{2}}}{2} + C$

Page No 18.134:

Question 32:

$\int x \tan^{2} x d x$

Answer:

$\int$ x tan² x dx
= ∫ x (sec² x – 1) dx
$= \int \underset{I}{x} . \underset{II}{\sec^{2} x} d x - \int x d x = x \int \sec^{2} x - \int \{\frac{d}{d x} (x) \int \sec^{2} x d x\} d x - \frac{x^{2}}{2} + C_{1} = x . \tan x - \int 1 . \tan x d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C_{1} + C_{2} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C (where C = C_{1} + C_{2})$

Page No 18.134:

Question 33:

$\int x (\frac{\sec 2 x - 1}{\sec 2 x + 1}) d x$

Answer:

$\int x (\frac{\sec 2 x - 1}{\sec 2 x + 1}) d x = \int x (\frac{\frac{1}{\cos 2 x} - 1}{\frac{1}{\cos 2 x} + 1}) d x = \int x (\frac{1 - \cos 2 x}{1 + \cos 2 x}) d x = \int x (\frac{2 \sin^{2} x}{2 \cos^{2} x}) d x [∵ (1 - \cos 2 x) = 2 \sin^{2} x and (1 + \cos 2 x) = 2 \cos^{2} x] = \int x . \tan^{2} x d x = \int x . (\sec^{2} x - 1) d x = \int \underset{I}{x} . \underset{II}{\sec^{2} x} d x - \int x d x = x \int \sec^{2} x d x - \int \{\frac{d}{d x} (x) \int \sec^{2} x d x\} d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \int 1 . \tan x d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C_{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C (where C = C_{1} + C_{2})$

Page No 18.134:

Question 34:

$\int (x + 1) e^{x} \log (x e^{x}) d x$

Answer:

$\int (x + 1) e^{x} . \log (x e^{x}) d x Let x e^{x} = t \Rightarrow (x . e^{x} + 1 . e^{x}) d x = d t ∴ \int (x + 1) e^{x} . \log (x e^{x}) d x = \int \underset{II}{1} . \underset{I}{\log (t)} d t = \log t \int 1 d t - \int \{\frac{d}{d t} (\log t) - \int 1 d t\} d t = \log (t) \times t - \int \frac{1}{t} \times t d t = t \log (t) - t + C . . . (1) Substituting the value of t in eq (1) \Rightarrow \int (x + 1) e^{x} . \log (x e^{x}) d x = (x e^{x}) . \log (x e^{x}) - x e^{x} + C = x e^{x} \{\log (x e^{x}) - 1\} + C$

Page No 18.134:

Question 35:

$\int \sin^{- 1} (3 x - 4 x^{3}) d x$

Answer:

$\int$ sin^–1 (3x – 4x³)dx
Let x = sin θ
⇒ dx = cos θ.dθ
& θ = sin^–1 x
$\int$ sin^–1 (3x – 4x³)dx = $\int$ sin^–1 (3 sin θ – 4 sin³ θ) . cos θ dθ
= ∫ sin^–1 (sin 3θ) . cos θ dθ
$= 3 \int \underset{I}{θ} . \underset{II}{\cos θ} d θ = 3 [θ \int \cos θ d θ - \int \{\frac{d}{d θ} (θ) - \int \cos θ d θ\} d θ] = 3 [θ . \sin θ - \int 1 . \sin θ d θ] = 3 [θ . \sin θ + \cos θ] + C = 3 [θ . \sin θ + \sqrt{1 - \sin^{2} θ}] + C = 3 [(\sin^{- 1} x) . x + \sqrt{1 - x^{2}}] + C (∵ θ = \sin^{- 1} x)$

Page No 18.134:

Question 36:

$\int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x$

Answer:

$\int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x Let x = \tan θ d x = \sec^{2} θ d θ ∴ \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x = \int \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) . \sec^{2} θ d θ = \int \sin^{- 1} (\sin 2 θ) . \sec^{2} θ d θ = \int (2 θ) \sec^{2} θ d θ = 2 \int \underset{I}{θ} \underset{II}{\sec^{2} θ} d θ = 2 [θ \int \sec^{2} θ d θ - \int \{\frac{d}{d θ} (θ) \int s {ec}^{2} θ d θ\} d θ] = 2 [θ . \tan θ - \int 1 . \tan θ d θ] = 2 [θ \tan θ - \log |\sec θ|] + C = 2 [θ \tan θ - \log {|1 + \tan^{2} θ|}^{\frac{1}{2}}] + C = 2 [(\tan^{- 1} x) \times x - \log {(1 + x^{2})}^{\frac{1}{2}}] + C = 2 x \tan^{- 1} x - 2 \times \frac{1}{2} \log |1 + x^{2}| + C = 2 x \tan^{- 1} x - \log |1 + x^{2}| + C$

Page No 18.134:

Question 37:

$\int \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x$

Answer:

$Let I = \int \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x = \int 3 \tan^{- 1} (x) d x = 3 \int [\tan^{- 1} (x) \times 1] d x = 3 [\tan^{- 1} x \times x - \int \frac{1}{1 + x^{2}} \times x d x] = 3 x \tan^{- 1} x - 3 \int \frac{x}{1 + x^{2}} d x let 1 + x^{2} = t \Rightarrow 2 x d x = d t Then, I = 3 x \tan^{- 1} x - \frac{3}{2} \int \frac{d t}{t} = 3 x \tan^{- 1} x - \frac{3}{2} \log |t| + C = 3 x \tan^{- 1} x - \frac{3}{2} \log |1 + x^{2}| + C$

Page No 18.134:

Question 38:

$\int x^{2} \sin^{- 1} x d x$

Answer:

$\int \underset{II}{x^{2}} . \underset{I}{\sin^{- 1}} x d x = \underset{}{\sin^{- 1}} x \int x^{2} d x - \int \{\frac{d}{d x} (\underset{}{\sin^{- 1}} x) \int x^{2} d x\} d x = \sin^{- 1} x . \frac{x^{3}}{3} - \int \frac{1}{\sqrt{1 - x^{2}}} \frac{x^{3}}{3} d x Let 1 - x^{2} = t \Rightarrow x^{2} = 1 - t \Rightarrow - 2 x d x = d t \Rightarrow x d x = - \frac{d t}{2} ∴ \int \underset{}{x^{2}} . \underset{}{\sin^{- 1}} x d x = \sin^{- 1} x . \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{2} . x}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int \frac{(1 - t)}{\sqrt{t}} d t = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{1}{6} \int t^{- \frac{1}{2}} d t - \frac{1}{6} \int t^{\frac{1}{2}} d t = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{1}{6} \times 2 \sqrt{t} - \frac{1}{6} \times \frac{2}{3} t^{\frac{3}{2}} + C = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{\sqrt{1 - x^{2}}}{3} - \frac{1}{9} {(1 - x^{2})}^{\frac{3}{2}} + C (∵ 1 - x^{2} = t)$

Page No 18.134:

Question 39:

$\int \frac{\sin^{- 1} x}{x^{2}} d x$

Answer:

$Let I = \int \frac{\sin^{- 1} x}{x^{2}} d x Putting x = \sin θ \Rightarrow θ = \sin^{- 1} x & d x = \cos θ d θ ∴ I = \int \frac{θ . \cos θ}{\sin^{2} θ} d θ = \int θ . (\frac{\cos θ}{\sin θ}) \times \frac{1}{\sin θ} d θ = \int \underset{I}{θ} . \underset{II}{cosec θ} \cot θ d θ = θ \int cosec θ \cot θ d θ - \int \{\frac{d}{d θ} (θ) \int cosec θ \cot θ d θ\} d θ = θ (- cosec θ) - \int 1 . (- cosec θ) d θ = - θ cosec θ + \int cosec θ d θ = - θ cosec θ + \ln |cosec θ - \cot θ| + C = \frac{- θ}{\sin θ} + \ln |\frac{1 - \cos θ}{\sin θ}| + C = \frac{- θ}{\sin θ} + \ln |\frac{1 - \sqrt{1 - \sin^{2} θ}}{\sin θ}| + C = \frac{- \sin^{- 1} x}{x} + \ln |\frac{1 - \sqrt{1 - x^{2}}}{x}| + C [∵ θ = \sin^{- 1} x]$

Page No 18.134:

Question 40:

$\int \frac{x^{2} \tan^{- 1} x}{1 + x^{2}} d x$

Answer:

$Let I = \int (\frac{x^{2} \tan^{- 1} x}{1 + x^{2}}) d x = \int (\frac{x^{2} + 1 - 1}{x^{2} + 1}) \tan^{- 1} x d x = \int (1 - \frac{1}{x^{2} + 1}) \tan^{- 1} x d x = \int \underset{II}{1} . \underset{I}{\tan^{- 1}} x d x - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x = [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int 1 d x\} d x] - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x = t [{an}^{- 1} x \times x - \int \frac{x}{1 + x^{2}} d x] - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x Putting x^{2} + 1 = t in the first integral and \tan^{- 1} x = p in the second integral \Rightarrow 2 x d x = d t and \frac{1}{1 + x^{2}} d x = d p \Rightarrow x d x = \frac{d t}{2} and \frac{1}{1 + x^{2}} d x = d p ∴ I = \tan^{- 1} x . x - \frac{1}{2} \int \frac{d t}{t} - \int p . d p = x \tan^{- 1} x - \frac{1}{2} \ln |t| - \frac{p^{2}}{2} + C = x \tan^{- 1} x - \frac{1}{2} \ln |1 + x^{2}| - \frac{{(\tan^{- 1} x)}^{2}}{2} + C [∵ t = x^{2} + 1 and p = \tan^{- 1} x]$

Page No 18.134:

Question 41:

$\int \cos^{- 1} (4 x^{3} - 3 x) d x$

Answer:

$\int \cos^{- 1} (4 x^{3} - 3 x) d x Let x = \cos θ \Rightarrow θ = \cos^{- 1} x & d x = - \sin θ d θ ∴ \int \cos^{- 1} (4 x^{3} - 3 x) d x = \int \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) . (- \sin θ) d θ = \int \cos^{- 1} (\cos 3 θ) . (- \sin θ) d θ (∵ \cos 3 θ = 4 \cos^{3} θ - 3 \cos θ) = - 3 \int \underset{I}{θ} \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin θ d θ\} d θ = 3 [θ (- \cos θ) - \int 1 . (- \cos θ) d θ] = 3 θ \cos θ - 3 \sin θ + C = 3 \cos^{- 1} x . x - 3 \sqrt{1 - x^{2}} + C (∵ x = \cos θ)$

Page No 18.134:

Question 42:

$\int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$

Answer:

$Let I = \int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x = 2 \int \underset{II}{1} . \underset{I}{\tan^{- 1}} x d x [∵ \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 \tan^{- 1} x] = 2 [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int 1 d x\} d x] = 2 [\tan^{- 1} x . x - \int \frac{1}{1 + x^{2}} \times x d x] = 2 x \tan^{- 1} x - \int \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t ∴ I = 2 x \tan^{- 1} x - \int \frac{d t}{t} = 2 x \tan^{- 1} x - \ln |t| + C = 2 x \tan^{- 1} x - \ln |1 + x^{2}| + C [∵ t = 1 + x^{2}]$

Page No 18.134:

Question 43:

$\int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$

Answer:

$Let I = \int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x = 2 \int \underset{II}{1} . \underset{I}{\tan^{- 1} x} d x = 2 [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} \{\tan^{- 1} x\} \int 1 d x\} d x] = 2 [\tan^{- 1} x . x - \int \frac{1}{1 + x^{2}} \times x d x] = 2 \tan^{- 1} x . x - \int \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t ∴ I = 2 x \tan^{- 1} x - \int \frac{d t}{t} = 2 x \tan^{- 1} x - \ln |t| + C = 2 x \tan^{- 1} x - \ln |1 + x^{2}| + C [∵ t = 1 + x^{2}]$

Page No 18.134:

Question 44:

$\int (x + 1) \log x d x$

Answer:

$\int \underset{II}{(x + 1)} . \underset{I}{\log x} d x = \log x \int (x + 1) d x - \int \{\frac{d}{d x} (\log x) \int (x + 1) d x\} d x = \log x [\frac{x^{2}}{2} + x] - \int \frac{1}{x} (\frac{x^{2}}{2} + x) d x = \log x (\frac{x^{2}}{2} + x) - \int (\frac{x}{2} + 1) d x = \log x (\frac{x^{2}}{2} + x) - (\frac{x^{2}}{4} + x) + C$

Page No 18.134:

Question 45:

$\int x^{2} \tan^{- 1} x d x$

Answer:

$Let I = \int \underset{II}{x^{2}} . \underset{I}{\tan^{- 1} x} d x = \tan^{- 1} x \int x^{2} d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int x^{2} d x\} d x = \tan^{- 1} x \times \frac{x^{3}}{3} - \int (\frac{1}{1 + x^{2}}) \times \frac{x^{3}}{3} d x = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{2} . x}{1 + x^{2}} d x Let 1 + x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int \frac{(t - 1)}{t} . d t = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int d t + \frac{1}{6} \int \frac{d t}{t} = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{t}{6} + \frac{1}{6} \log |t| + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{(1 + x^{2})}{6} + \frac{1}{6} \log (1 + x^{2}) + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{x^{2}}{6} + \frac{1}{6} \log (1 + x^{2}) - \frac{1}{6} + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{x^{2}}{6} + \frac{1}{6} \log |1 + x^{2}| + C' where C' = C - \frac{1}{6}$

Page No 18.134:

Question 46:

$\int (e^{\log x} + \sin x) \cos x d x$

Answer:

$\int (e^{\log x} + \sin x) \cos x d x = \int (x + \sin x) \cos x d x (∵ e^{\log x} = x) = \int (x \cos x + \sin x \cos x) d x = \int x \cos x d x + \frac{1}{2} \int 2 \sin x \cos x d x = \int \underset{I}{x} \underset{II}{\cos x} d x + \frac{1}{2} \int \sin 2 x d x = [x \int \cos x d x - \int \{\frac{d}{d x} (x) \int \cos x d x\} d x] + \frac{1}{2} \int \sin 2 x d x = x \sin x - \int 1 . \sin x d x + \frac{1}{2} [\frac{- \cos 2 x}{2}] + C = x \sin x - (- \cos x) - \frac{1}{4} \cos 2 x + C = x \sin x + \cos x - \frac{1}{4} (1 - 2 \sin^{2} x) + C = x \sin x + \cos x + \frac{\sin^{2} x}{2} - \frac{1}{4} + C = x \sin x + \cos x + \frac{\sin^{2} x}{2} + C' where C' = C - \frac{1}{4}$

Page No 18.134:

Question 47:

$\int \frac{(x \tan^{- 1} x)}{{(1 + x^{2})}^{3 / 2}} d x$

Answer:

$Let I = \int \frac{x \tan^{- 1} x}{{(1 + x^{2})}^{\frac{3}{2}}} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ & θ = \tan^{- 1} x ∴ I = \int \frac{(\tan θ) . θ . \sec^{2} θ d θ}{{(1 + \tan^{2} θ)}^{\frac{3}{2}}} = \int \frac{θ . \tan θ \sec^{2} θ d θ}{{(\sec^{2} θ)}^{\frac{3}{2}}} = \int \frac{θ \tan θ . \sec^{2} θ d θ}{\sec^{3} θ} = \int \frac{θ . \tan θ}{\sec θ} d θ = \int \underset{I}{θ} . \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin d θ\} d θ = θ (- \cos θ) - \int 1 . (- \cos θ) d θ = - θ \cos θ + \sin θ + C = \frac{- θ}{\sec θ} + \frac{1}{cose c θ} + C = \frac{- θ}{\sqrt{1 + \tan^{2} θ}} + \frac{1}{\sqrt{1 + \cot^{2} θ}} + C = \frac{- θ}{\sqrt{1 + \tan^{2} θ}} + \frac{\tan θ}{\sqrt{\tan^{2} θ + 1}} + C = \frac{- \tan^{- 1} x}{\sqrt{1 + x^{2}}} + \frac{x}{\sqrt{x^{2} + 1}} + C$

Page No 18.134:

Question 48:

$\int \tan^{- 1} (\sqrt{x}) d x$

Answer:

$Let I = \int \tan^{- 1} \sqrt{x} d x = \int \frac{\sqrt{x} . \tan^{- 1} \sqrt{x} d x}{\sqrt{x}} Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t = \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{II}{t} . \underset{I}{\tan^{- 1} (t)} d t = 2 [\tan^{- 1} t \int t d t - \int \{\frac{d}{d t} (\tan^{- 1} t) \int t d t\} d t] = 2 [\tan^{- 1} (t) . \frac{t^{2}}{2} - \int \frac{1}{1 + t^{2}} . \frac{t^{2}}{2} d t] = \tan^{- 1} (t) . t^{2} - \int \frac{t^{2}}{1 + t^{2}} d t = \tan^{- 1} (t) . t^{2} - \int (\frac{1 + t^{2} - 1}{1 + t^{2}}) d t = \tan^{- 1} (t) . t^{2} - \int d t + \int \frac{d t}{1 + t^{2}} = \tan^{- 1} (t) . t^{2} - t + \tan^{- 1} (t) + C (∵ \sqrt{x} = t) = \tan^{- 1} (\sqrt{x}) . x - \sqrt{x} + \tan^{- 1} \sqrt{x} + C = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Page No 18.134:

Question 49:

$\int x^{3} \tan^{- 1} x d x$

Answer:

$\int \underset{II}{x^{3}} . \underset{I}{\tan^{- 1} x} d x = \tan^{- 1} x \int x^{3} d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int x^{3} d x\} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \int \frac{1}{1 + x^{2}} \times \frac{x^{4}}{4} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int \frac{x^{4} d x}{x^{2} + 1} = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (\frac{x^{4} - 1 + 1}{x^{2} + 1}) d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (\frac{x^{4} - 1}{x^{2} + 1}) d x - \frac{1}{4} \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} d x - \frac{1}{4} \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (x^{2} - 1) d x - \frac{1}{4} \tan^{- 1} x + C = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} (\frac{x^{3}}{3} - x) - \frac{1}{4} \tan^{- 1} x + C = (\frac{x^{4} - 1}{4}) \tan^{- 1} x - \frac{1}{12} (x^{3} - 3 x) + C$

Page No 18.134:

Question 50:

$\int x \sin x \cos 2 x d x$

Answer:

$\int x . \cos 2 x \sin x d x = \frac{1}{2} \int x (2 \cos 2 x \sin x) d x [∵ 2 \cos A \sin B = \sin (A + B) - \sin (A - B)] = \frac{1}{2} \int x (\sin 3 x - \sin x) d x = \frac{1}{2} \int x . \sin 3 x d x - \frac{1}{2} \int x \sin x d x = \frac{1}{2} [x \int \sin 3 x d x - \int \{\frac{d}{d x} (x) \int \sin 3 x d x\} d x] - \frac{1}{2} [x \int \sin x d x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = \frac{1}{2} [x . (\frac{- \cos 3 x}{3}) - \int 1 . (\frac{- \cos 3 x}{3}) d x] - \frac{1}{2} [x . (- \cos x) - \int 1 . (- \cos x) d x] = \frac{1}{2} [x . (\frac{- \cos 3 x}{3}) + \frac{1}{9} \sin 3 x] - \frac{1}{2} [x . (- \cos x) + \sin x] = - \frac{x \cos 3 x}{6} + \frac{\sin 3 x}{18} + \frac{x \cos x}{2} - \frac{\sin x}{2} + C$

Page No 18.134:

Question 51:

$\int (\tan^{- 1} x^{2}) x d x$

Answer:

Let I = $\int$ (tan^–1 x²) x dx
Putting x² = t
⇒ 2x dx = dt
$\Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \underset{II}{1} . \underset{I}{\tan^{- 1} t} . d t = \frac{1}{2} \tan^{- 1} t \int 1 d t - \int \{\frac{d}{d t} (\tan^{- 1} t) \int 1 d t\} d t = \frac{1}{2} [\tan^{- 1} t . t - \int \frac{t}{1 + t^{2}} d t] Now putting 1 + t^{2} = p \Rightarrow 2 t d t = d p \Rightarrow t d t = \frac{d p}{2} ∴ I = \frac{1}{2} t . \tan^{- 1} t - \frac{1}{2} \int \frac{t d t}{1 + t^{2}} = \frac{t . \tan^{- 1} t}{2} - \frac{1}{2 x^{2}} \int \frac{d p}{p} = \frac{t . \tan^{- 1} t}{2} - \frac{1}{4} \ln p + C = \frac{x^{2} . \tan^{- 1} x^{2}}{2} - \frac{1}{4} \ln |1 + x^{4}| + C [∵ p = 1 + t^{2}]$

Page No 18.134:

Question 52:

$\int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Answer:

$\int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x Let \sin^{- 1} x = θ x = \sin θ d x = \cos θ d θ ∴ \int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x = \int \frac{(\sin θ) . θ}{\sqrt{1 - \sin^{2} θ}} . \cos θ d θ = \int \frac{(\sin θ) . θ}{\cos θ} . \cos θ d θ = \int \underset{I}{θ} . \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin θ d θ\} d θ = θ (- \cos θ) - \int 1 . (- \cos θ) d θ = - θ \cos θ + \sin θ + C = - θ \sqrt{1 - \sin^{2} θ} + \sin θ + C = - \sin^{- 1} x \sqrt{1 - x^{2}} + x + C (∵ \sin^{- 1} x = θ)$

Page No 18.134:

Question 53:

$\int \sin^{- 1} 2 x d x$

Answer:

$\int \sin^{- 1} 2 x d x$

$= \int \sin^{- 1} 2 x \times 1 d x$

$= \sin^{- 1} 2 x \int d x - \int (\frac{d}{d x} \sin^{- 1} 2 x \int d x) d x$

$= x \sin^{- 1} 2 x - \int \frac{2}{\sqrt{1 - 4 x^{2}}} \times x d x$

$= x \sin^{- 1} 2 x - \int \frac{2 x}{\sqrt{1 - 4 x^{2}}} d x$

$= x \sin^{- 1} 2 x + \frac{1}{4} \int {(1 - 4 x^{2})}^{- \frac{1}{2}} (- 8 x) d x$

$= x \sin^{- 1} 2 x + \frac{1}{4} \times \frac{{(1 - 4 x^{2})}^{\frac{1}{2}}}{\frac{1}{2}} + C [\begin{matrix} \int {[f (x)]}^{n} f' (x) d x = \frac{{[f (x)]}^{n + 1}}{n + 1} + C \\ f (x) = 1 - 4 x^{2} \Rightarrow f' (x) = - 8 x \end{matrix}]$

$= x \sin^{- 1} 2 x + \frac{1}{2} \sqrt{1 - 4 x^{2}} + C$

Page No 18.134:

Question 54:

$\int \sin^{3} \sqrt{x} d x$

Answer:

$Let, I = \int \sin^{3} \sqrt{x} d x . . . . . (1) Consider, \sqrt{x} = t . . . . . (2) Differentiating both sides we get, \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \Rightarrow d x = 2 t d t Therefore, (1) becomes, I = \int \sin^{3} t 2 t d t = 2 \int t \sin^{3} t d t = 2 \int t (\frac{3 \sin t - \sin 3 t}{4}) d t (Since, \sin 3 A = 3 \sin A - 4 \sin^{3} A) = \frac{3}{2} \int t si n t d t - \frac{1}{2} \int t \sin 3 t d t = \frac{3}{2} [t \int \sin t d t - \int (\frac{d t}{d t} \int \sin t d t) d t] - \frac{1}{2} [t \int \sin 3 t d t - \int (\frac{d t}{d t} \int \sin 3 t d t) d t] = \frac{3}{2} [- t co s t + \int \cos t d t] - \frac{1}{2} [- \frac{t \cos 3 t}{3} + \frac{1}{3} \int \cos 3 t d t] = \frac{3}{2} [- t \cos t + \sin t] - \frac{1}{2} [- \frac{t \cos 3 t}{3} + \frac{1}{9} \sin 3 t] + C = - \frac{3}{2} t \cos t + \frac{3}{2} \sin t + \frac{1}{6} t \cos 3 t - \frac{1}{18} \sin 3 t + C = - \frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{1}{6} \sqrt{x} \cos (3 \sqrt{x}) - \frac{1}{18} \sin (3 \sqrt{x}) + C$

Note: The answer in indefinite integration may vary depending on the integral constant.

Page No 18.134:

Question 55:

$\int x \sin^{3} x d x$

Answer:

Let I = $\int x \sin^{3} x d x$
sin (3A) = 3 sin A – 4 sin³ A
$\sin^{3} A = \frac{1}{4} [3 \sin A - \sin 3 A] ∴ I = \frac{1}{4} \int x . (3 \sin x - \sin 3 x) d x = \frac{3}{4} \int \underset{I}{x} . \underset{II}{\sin x} d x - \frac{1}{4} \int \underset{I}{x} \underset{II}{. \sin (3 x)} d x = \frac{3}{4} [x (- \cos x) - \int 1 . (- \cos x) d x] - \frac{1}{4} [x (- \frac{\cos 3 x}{3}) - \int 1 . (- \frac{\cos 3 x}{3}) d x] = - \frac{3 x \cos x}{4} + \frac{3}{4} \sin x + \frac{x \cos 3 x}{12} - \frac{1}{36} \sin 3 x + C$

Page No 18.134:

Question 56:

$\int \cos^{3} \sqrt{x} d x$

Answer:

$Let, I = \int \cos^{3} \sqrt{x} d x . . . . . (1) Consider, \sqrt{x} = t . . . . . (2) Differentiating both sides we get, \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \Rightarrow d x = 2 t d t Therefore, (1) becomes, I = \int \cos^{3} t 2 t d t = 2 \int t \cos^{3} t d t = 2 \int t (\frac{3 \cos t + \cos 3 t}{4}) d t (Since, \cos 3 A = 4 \cos^{3} A - 3 \cos A) = \frac{3}{2} \int t \cos t d t + \frac{1}{2} \int t \cos 3 t d t = \frac{3}{2} [t \int \cos t d t - \int (\frac{d t}{d t} \int \cos t d t) d t] + \frac{1}{2} [t \int \cos 3 t d t - \int (\frac{d t}{d t} \int \cos 3 t d t) d t] = \frac{3}{2} [t \sin t - \int \sin t d t] + \frac{1}{2} [\frac{t \sin 3 t}{3} - \frac{1}{3} \int \sin 3 t d t] = \frac{3}{2} [t \sin t + \cos t] + \frac{1}{2} [\frac{t \sin 3 t}{3} + \frac{1}{9} \cos 3 t] + C = \frac{3}{2} t \sin t + \frac{3}{2} \cos t + \frac{1}{6} t \sin 3 t + \frac{1}{18} \cos 3 t + C = \frac{3}{2} \sqrt{x} \sin \sqrt{x} + \frac{3}{2} \cos \sqrt{x} + \frac{1}{6} \sqrt{x} \sin (3 \sqrt{x}) + \frac{1}{18} \cos (3 \sqrt{x}) + C$

Note: The final answer in indefinite integration may vary based on the integration constant.

Page No 18.134:

Question 57:

$\int x \cos^{3} x d x$

Answer:

Let I= $\int x \cos^{3} x d x$

$As we know, \cos 3 x = 4 \cos^{3} x - 3 \cos x \Rightarrow \cos^{3} x = \frac{1}{4} (\cos 3 x + 3 \cos x)$

$∴ I = \frac{1}{4} \int x . (\cos 3 x + 3 \cos x) d x = \frac{1}{4} \int \underset{I}{x} . \underset{II}{\cos} (3 x) d x + \frac{3}{4} \int \underset{I}{x} . \underset{II}{\cos x} d x = \frac{1}{4} [x . \int \cos 3 x d x - \int \{\frac{d}{d x} (x) . \int \cos 3 x d x\} d x] + \frac{3}{4} [x \int \cos x - \int \{\frac{d}{d x} (x) . \int \cos x d x\} d x] = \frac{1}{4} [x . \frac{\sin 3 x}{3} - \int 1 . \frac{\sin 3 x}{3} d x] + \frac{3}{4} [x (\sin x) - \int 1 . \sin x d x] = \frac{x \sin 3 x}{12} + \frac{\cos 3 x}{36} + \frac{3}{4} x \sin x + \frac{3}{4} \cos x + C$

Page No 18.134:

Question 58:

$\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$Let I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x Putting x = \cos θ \Rightarrow d x = - \sin θ d θ & θ = \cos^{- 1} x ∴ I = \int \tan^{- 1} \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} (- \sin θ) d θ = \int \tan^{- 1} \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}} (- \sin θ) d θ = \int \tan^{- 1} (\tan \frac{θ}{2}) (- \sin θ) d θ = - \frac{1}{2} \int \underset{I}{θ} \underset{II}{\sin θ} d θ = - \frac{1}{2} [θ \int \sin θ d θ - \int \{(\frac{d}{d θ} θ) \int \sin θ d θ\} d θ] = - \frac{1}{2} [θ (- \cos θ) - \int 1 . (- \cos θ) d θ] = - \frac{1}{2} [- θ \cos θ + \sin θ] + C = - \frac{1}{2} [- θ . \cos θ + \sqrt{1 - \cos^{2} θ}] + C = - \frac{1}{2} [- \cos^{- 1} x . x + \sqrt{1 - x^{2}}] + C [∵ θ = \cos^{- 1} x] = \frac{x \cos^{- 1} x}{2} - \frac{\sqrt{1 - x^{2}}}{2} + C$

Page No 18.134:

Question 59:

$\int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$

Answer:

$Let I = \int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x Putting x = a \tan^{2} θ \Rightarrow \sqrt{\frac{x}{a}} = \tan θ \Rightarrow d x = a (2 \tan θ) \sec^{2} θ d θ ∴ I = \int \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}} (2 a \tan θ) \sec^{2} θ d θ = \int \sin^{- 1} \sqrt{\frac{\tan^{2} θ}{\sec^{2} θ}} (2 a \tan θ \sec^{2} θ) d θ = 2 a \int [\sin^{- 1} (\sin θ) \tan θ \sec^{2} θ] d θ$

$= 2 a \int \underset{I}{θ} \underset{II}{\tan θ \sec^{2} θ} d θ = 2 a [θ \frac{\tan^{2} θ}{2} - \int 1 \frac{\tan^{2} θ}{2} d θ] = 2 a [\frac{θ . \tan^{2} θ}{2} - \frac{1}{2} \int (s e c^{2} θ - 1) d θ] = a θ \tan^{2} θ - a \tan θ + a θ + C = a (\frac{x}{a}) \tan^{- 1} (\frac{\sqrt{x}}{\sqrt{a}}) - a \sqrt{\frac{x}{a}} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C = x \tan^{- 1} \sqrt{\frac{x}{a}} - \sqrt{a x} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C$

Page No 18.134:

Question 60:

$\int \frac{x^{3} \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x$

Answer:

$Let I = \int \frac{x^{3} \times \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x Putting \sin^{- 1} x^{2} = t \Rightarrow x^{2} = \sin t \Rightarrow \frac{1 \times 2 x d x}{\sqrt{1 - {(x^{2})}^{2}}} = d t \Rightarrow \frac{x d x}{\sqrt{1 - x^{4}}} = \frac{d t}{2} ∴ I = \int x^{2} . \frac{\sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} . x d x = \int (\sin t) . t . \frac{d t}{2} = \frac{1}{2} \int \underset{I}{t} . \underset{II}{\sin t} d t = \frac{1}{2} [t \int \sin t d t - \int \{\frac{d}{d t} (t) \int \sin t d t\} d t] = \frac{1}{2} [t . (- \cos t) - \int 1 . (- \cos t) d t] = \frac{1}{2} [- t \cos t + \sin t] + C = \frac{1}{2} [- t \sqrt{1 - \sin^{2} t} + \sin t] + C = \frac{1}{2} [- \sin^{- 1} (x^{2}) \sqrt{1 - x^{4}} + x^{2}] + C$

Page No 18.134:

Question 61:

$\int \frac{x^{2} \sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} d x$

Answer:

$Let I = \int \frac{x^{2} . \sin^{- 1} x d x}{{(1 - x^{2})}^{\frac{3}{2}}} Putting x = \sin θ \Rightarrow d x = \cos θ d θ & θ = \sin^{- 1} x ∴ I = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{{(1 - \sin^{2} θ)}^{\frac{3}{2}}} = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{{(\cos^{2} θ)}^{\frac{3}{2}}} = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{\cos^{3} θ} = \int \tan^{2} θ . θ . d θ = \int (\sec^{2} θ - 1) θ . d θ = \int \underset{I}{θ} . \underset{II}{\sec^{2} θ} d θ - \int θ . d θ = θ \int \sec^{2} θ d θ - \int \{\frac{d}{d θ} (θ) \int \sec^{2} θ d θ\} d θ - \int θ . d θ = θ \tan θ - \int 1 . \tan θ d θ - \frac{θ^{2}}{2} = θ . \tan θ - \ln |\sec θ| - \frac{θ^{2}}{2} + C = θ . \frac{\sin θ}{\cos θ} + \ln |\cos θ| - \frac{θ^{2}}{2} + C = θ . \frac{\sin θ}{\cos θ} + \ln |\sqrt{1 - \sin^{2} θ}| - \frac{θ^{2}}{2} + C = \frac{θ . \sin θ}{\sqrt{1 - \sin^{2} θ}} + \frac{1}{2} \ln |1 - \sin^{2} θ| - \frac{θ^{2}}{2} + C = \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{1}{2} \ln (1 - x^{2}) - \frac{1}{2} {(\sin^{- 1} x)}^{2} + C [∵ θ = \sin^{- 1} x]$

Page No 18.14:

Question 1:

$\int (3 x \sqrt{x} + 4 \sqrt{x} + 5) d x$

Answer:

$\int (3 x \sqrt{x} + 4 \sqrt{x} + 5) d x = \int (3 x^{1} \cdot x^{\frac{1}{2}} + 4 x^{\frac{1}{2}} + 5) d x = 3 \int x^{\frac{3}{2}} d x + 4 \int x^{\frac{1}{2}} d x + 5 \int d x = 3 [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + 4 [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 5 x + C = 3 \times \frac{2}{5} x^{\frac{5}{2}} + 4 \times \frac{2}{3} x^{\frac{3}{2}} + 5 x + C = \frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + 5 x + C$

Page No 18.14:

Question 2:

$\int (2^{x} + \frac{5}{x} - \frac{1}{x^{1 / 3}}) d x$

Answer:

$\int (2^{x} + \frac{5}{x} - \frac{1}{x^{\frac{1}{3}}}) d x = \int 2^{x} d x + 5 \int \frac{d x}{x} - \int \frac{d x}{x^{\frac{1}{3}}} = \int 2^{x} d x + 5 \int \frac{d x}{x} - \int x^{- \frac{1}{3}} d x = \frac{2^{x}}{\ln 2} + 5 \ln x - [\frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}] + C = \frac{2^{x}}{\ln 2} + 5 \ln x - \frac{3}{2} x^{\frac{2}{3}} + C$

Page No 18.14:

Question 3:

$\int \{\sqrt{x} (a x^{2} + b x + c)\} d x$

Answer:

$\int \sqrt{x} (a x^{2} + b x + c) d x = \int x^{\frac{1}{2}} (a x^{2} + b x + c) d x = \int (a x^{2 + \frac{1}{2}} + b x^{\frac{1}{2} + 1} + c x^{\frac{1}{2}}) d x = a \int x^{\frac{5}{2}} d x + b \int x^{\frac{3}{2}} d x + c \int x^{\frac{1}{2}} d x = a [\frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1}] + b [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + c [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2 a}{7} x^{\frac{7}{2}} + \frac{2 b}{5} x^{\frac{3}{2}} + \frac{2 c}{3} x^{\frac{3}{2}} + C$

Page No 18.14:

Question 4:

$\int (2 - 3 x) (3 + 2 x) (1 - 2 x) d x$

Answer:

$\int (2 - 3 x) (3 + 2 x) (1 - 2 x) d x = \int (2 - 3 x) (3 - 6 x + 2 x - 4 x^{2}) d x = \int (2 - 3 x) (- 4 x^{2} - 4 x + 3) d x = \int (- 8 x^{2} - 8 x + 6 + 12 x^{3} + 12 x^{2} - 9 x) d x = \int (12 x^{3} + 4 x^{2} - 17 x + 6) d x = \frac{12 x^{4}}{4} + \frac{4 x^{3}}{3} - \frac{17 x^{2}}{2} + 6 x + C = 3 x^{4} + \frac{4}{3} x^{3} - \frac{17}{2} x^{2} + 6 x + C$

Page No 18.14:

Question 5:

$\int (\frac{m}{x} + \frac{x}{m} + m^{x} + x^{m} + m x) d x$

Answer:

$\int (\frac{m}{x} + \frac{x}{m} + m^{x} + x^{m} + m x) d x = m \int \frac{1}{x} d x + \frac{1}{m} \int x d x + \int m^{x} d x + \int x^{m} d x + m \int x d x = m \ln |x| + \frac{1}{m} [\frac{x^{1 + 1}}{1 + 1}] + [\frac{m^{x}}{\ln m}] + [\frac{x^{m + 1}}{m + 1}] + m [\frac{x^{1 + 1}}{1 + 1}] = m \ln |x| + \frac{x^{2}}{2 m} + \frac{m^{x}}{\ln m} + \frac{x^{m + 1}}{m + 1} + \frac{m x^{2}}{2} + C$

Page No 18.14:

Question 6:

$\int {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} d x$

Answer:

$\int {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} d x = \int (x + \frac{1}{x} - 2) d x = \int x d x + \int \frac{d x}{x} - 2 \int d x = \frac{x^{2}}{2} + \ln |x| - 2 x + C$

Page No 18.14:

Question 7:

$\int \frac{{(1 + x)}^{3}}{\sqrt{x}} d x$

Answer:

$\int \frac{{(1 + x)}^{3}}{\sqrt{x}} d x = \int (\frac{1 + x^{3} + 3 {(1)}^{2} x + 3 (1) x^{2}}{\sqrt{x}}) d x = \int (\frac{1 + x^{3} + 3 x + 3 x^{2}}{\sqrt{x}}) d x = \int (\frac{1}{\sqrt{x}} + \frac{x^{3}}{\sqrt{x}} + \frac{3 x}{\sqrt{x}} + \frac{3 x^{2}}{\sqrt{x}}) d x = \int (x^{- \frac{1}{2}} + x^{\frac{5}{2}} + 3 x^{\frac{1}{2}} + 3 x^{\frac{3}{2}}) d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + \frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + 3 \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + 3 \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = 2 \sqrt{x} + \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + \frac{6}{5} x^{\frac{5}{2}} + C$

Page No 18.14:

Question 8:

$\int \{x^{2} + e^{\log x} + {(\frac{e}{2})}^{x}\} d x$

Answer:

$\int (x^{2} + e^{\log x} + {(\frac{e}{2})}^{x}) d x = \int x^{2} d x + \int x d x + \int {(\frac{e}{2})}^{x} d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} + \frac{{(\frac{e}{2})}^{x}}{\ln (\frac{e}{2})} + C$

Page No 18.14:

Question 9:

$\int (x^{e} + e^{x} + e^{e}) d x$

Answer:

$\int (x^{e} + e^{x} + e^{e}) d x = \int x^{e} d x + \int e^{x} d x + e^{e} \int 1 d x = \frac{x^{e + 1}}{e + 1} + e^{x} + x \cdot e^{e} + C$

Page No 18.14:

Question 10:

$\int \sqrt{x} (x^{3} - \frac{2}{x}) d x$

Answer:

$\int \sqrt{x} (x^{3} - \frac{2}{x}) d x = \int (x^{\frac{7}{2}} - \frac{2}{\sqrt{x}}) d x = \int (x^{\frac{7}{2}} - 2 x^{- \frac{1}{2}}) d x = \frac{x^{\frac{7}{2} + 1}}{\frac{7}{2} + 1} - 2 \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = \frac{2}{9} x^{\frac{9}{2}} - 4 x^{\frac{1}{2}} + C = \frac{2}{9} x^{\frac{9}{2}} - 4 \sqrt{x} + C$

Page No 18.14:

Question 11:

$\int \frac{1}{\sqrt{x}} (1 + \frac{1}{x}) d x$

Answer:

$\int \frac{1}{\sqrt{x}} (1 + \frac{1}{x}) d x = \int x^{- \frac{1}{2}} (1 + \frac{1}{x}) d x = \int (x^{- \frac{1}{2}} + \frac{1}{x^{\frac{3}{2}}}) d x = \int x^{- \frac{1}{2}} d x + \int x^{- \frac{3}{2}} d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + [\frac{x^{- \frac{3}{2} + 1}}{- \frac{3}{2} + 1}] = 2 \sqrt{x} - \frac{2}{\sqrt{x}} + C$

Page No 18.14:

Question 12:

$\int \frac{x^{6} + 1}{x^{2} + 1} d x$

Answer:

$\int (\frac{x^{6} + 1}{x^{2} + 1}) d x = \int [\frac{{(x^{2})}^{3} + 1^{3}}{x^{2} + 1}] d x A^{3} + B^{3} = (A + B) (A^{2} - AB + B^{2}) = \int \frac{(x^{2} + 1) (x^{4} - x^{2} + 1)}{(x^{2} + 1)} d x = \int (x^{4} - x^{2} + 1) d x = \int x^{4} d x + \int x^{2} d x + \int 1 d x = \frac{x^{4 + 1}}{4 + 1} - \frac{x^{2 + 1}}{2 + 1} + x + C = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x + C$

Page No 18.14:

Question 13:

$\int \frac{x^{- 1 / 3} + \sqrt{x} + 2}{\sqrt[3]{x}} d x$

Answer:

$\int (\frac{x^{- \frac{1}{3}} + \sqrt{x} + 2}{x^{\frac{1}{3}}}) d x = \int (\frac{x^{- \frac{1}{3}}}{x^{\frac{1}{3}}} + \frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} + \frac{2}{x^{\frac{1}{3}}}) d x = \int (x^{- \frac{2}{3}} + x^{\frac{1}{6}} + 2 x^{- \frac{1}{3}}) d x = [\frac{x^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1} + \frac{x^{\frac{1}{6} + 1}}{\frac{1}{6} + 1} + 2 \frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}] = [\frac{x^{\frac{1}{3}}}{\frac{1}{3}} + \frac{x^{\frac{7}{6}}}{\frac{7}{6}} + 3 x^{\frac{2}{3}}] + C = 3 x^{\frac{1}{3}} + \frac{6}{7} x^{\frac{7}{6}} + 3 x^{\frac{2}{3}} + C$

Page No 18.14:

Question 14:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x$

Answer:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x = \int (\frac{1}{\sqrt{x}} + \frac{x}{\sqrt{x}} + 2 \frac{\sqrt{x}}{\sqrt{x}}) d x = \int (x^{- \frac{1}{2}} + x^{\frac{1}{2}} + 2) d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 2 x + C = 2 \sqrt{x} + \frac{2}{3} x^{\frac{3}{2}} + 2 x + C$

Page No 18.143:

Question 1:

$\int e^{x} (\cos x - \sin x) d x$

Answer:

$Let I = \int e^{x} (\cos x - \sin x) d x let e^{x} \cos x = t Diff both sides w . r . t x e^{x} \cdot \cos x + e^{x} (- \sin x) = \frac{d t}{d x} Put e^{x} f (x) = t \Rightarrow e^{x} (\cos x - \sin x) d x = d t ∴ \int e^{x} (\cos x - \sin x) d x = \int d t \Rightarrow I = t + C = e^{x} \cos x + C$

Page No 18.143:

Question 2:

$\int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x$

Answer:

$Let I = \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x Also let e^{x} \times \frac{1}{x^{2}} = t Diff both sides w . r . t x e^{x} \times \frac{1}{x^{2}} + e^{x} (\frac{- 2}{x^{3}}) = \frac{d t}{d x} \Rightarrow e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x = d t ∴ \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x = \int d t = t + C = \frac{e^{x}}{x^{2}} + C$

Page No 18.143:

Question 3:

$\int e^{x} (\frac{1 + \sin x}{1 + \cos x}) d x$

Answer:

$Let I = \int e^{x} (\frac{1 + \sin x}{1 + \cos x}) d x = \int e^{x} (\frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x}) d x = \int e^{x} (\frac{1}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) d x = \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) d x Putting e^{x} \tan \frac{x}{2} = t Diff both sides w . r . t . x e^{x} \cdot \tan (\frac{x}{2}) + e^{x} \times \frac{1}{2} \sec^{2} \frac{x}{2} = \frac{d t}{d x} \Rightarrow e^{x} [\tan \frac{x}{2} + \frac{1}{2} \sec^{2} (\frac{x}{2})] d x = d t ∴ \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) d x = \int d t = t + C = e^{x} \tan (\frac{x}{2}) + C$

Page No 18.143:

Question 4:

$\int e^{x} (\cot x - {cosec}^{2} x) d x$

Answer:

$Let I = \int e^{x} (\cot x - {cosec}^{2} x) d x here f (x) = \cot x put e^{x} f (x) = t f' (x) = - {cosec}^{2} x let e^{x} \cot x = t Diff both sides w . r . t x e^{x} \cot x + e^{x} (- {cosec}^{2} x) = \frac{d t}{d x} \Rightarrow e^{x} (\cot x - {cosec}^{2} x) d x = d t ∴ \int e^{x} (\cot x - {cosec}^{2} x) d x = \int d t = t + C = e^{x} \cot x + C$

Page No 18.143:

Question 5:

$\int e^{x} (\frac{x - 1}{2 x^{2}}) d x$

Answer:

$Let I = \int e^{x} (\frac{x - 1}{2 x^{2}}) d x = \frac{1}{2} \int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x here \frac{1}{x} = f (x) Put e^{x} f (x) = t \Rightarrow - \frac{1}{x^{2}} = f' (x) let e^{x} \frac{1}{x} = t Diff both sides w . r . t x (e^{x} \frac{1}{x} + e^{x} \frac{- 1}{x^{2}}) = \frac{d t}{d x} \Rightarrow e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x = d t ∴ I = \frac{1}{2} \int d t = \frac{t}{2} + C = \frac{e^{x}}{2 x} + C$

Page No 18.143:

Question 6:

$\int e^{x} \sec x (1 + \tan x) d x$

Answer:

$Let I = \int e^{x} \sec x (1 + \tan x) d x = \int e^{x} (\sec x + \sec x \tan x) d x Here, f (x) = \sec x Put e^{x} f (x) = t \Rightarrow f' (x) = \sec x \tan x let e^{x} \sec x = t Diff both sides w . r . t x e^{x} \sec x + e^{x} \sec x \tan x = \frac{d t}{d x} \Rightarrow e^{x} (\sec x + \tan x) d x = d t ∴ \int e^{x} (\sec x + \sec x \tan x) d x = \int d t = t + C = e^{x} \sec x + C$

Page No 18.143:

Question 7:

$\int e^{x} (\tan x - \log \cos x) d x$

Answer:

$Let I = \int e^{x} (\tan x - \log \cos x) d x here f (x) = - \log \cos x Put e^{x} f (x) = t \Rightarrow f' (x) = \tan x let - e^{x} \log \cos x = t Diff both sides w . r . t x - [e^{x} \log (\cos x) + e^{x} \frac{1}{\cos x} \times (- \sin x)] = \frac{d t}{d x} \Rightarrow [- e^{x} \log (\cos x) + e^{x} \tan x] d x = d t ∴ \int e^{x} (\tan x - \log \cos x) d x = \int d t = t + C = - e^{x} \log (\cos x) + C = e^{x} \log (\sec x) + C$

Page No 18.143:

Question 8:

$\int e^{x} [\sec x + \log (\sec x + \tan x)] d x$

Answer:

$Let I = \int e^{x} [\sec x + \log (\sec x + \tan x)] d x Here, f (x) = \log (\sec x + \tan x) Put e^{x} f (x) = t \Rightarrow f' (x) = \sec x let e^{x} \log (\sec x + \tan x) = t Diff both sides w . r . t x e^{x} \log (\sec x + \tan x) + e^{x} \frac{1}{\sec x + \tan x} (\sec x + \tan x + \sec^{2} x) = \frac{d t}{d x} \Rightarrow [e^{x} \log (\sec x + \tan x) + e^{x} (\sec x)] d x = d t \Rightarrow e^{x} [\sec x + \log (\sec x + \tan x)] d x = d t ∴ \int e^{x} [\sec x + \log (\sec x + \tan x)] d x = \int d t = t + C = e^{x} \log (\sec x + \tan x) + C$

Page No 18.143:

Question 9:

$\int e^{x} (\cot x + \log \sin x) d x$

Answer:

$Let I = \int e^{x} (\cot x + \log \sin x) d x Here, f (x) = \log \sin x Put e^{x} f (x) = t \Rightarrow f' (x) = \cot x let e^{x} \log \sin x = t Diff both sides w . r . t x e^{x} \log (\sin x) + e^{x} \times \frac{1}{\sin x} \times \cos x = \frac{d t}{d x} \Rightarrow [e^{x} \log (\sin x) + e^{x} \cot x] d x = d t \Rightarrow e^{x} (\cot x + \log \sin x) d x = d t ∴ \int e^{x} (\cot x + \log \sin x) d x = \int d t = t + C = e^{x} \log \sin x + C$

Page No 18.143:

Question 10:

$\int e^{x} \frac{x - 1}{{(x + 1)}^{3}} d x$

Answer:

$Let I = \int e^{x} [\frac{x - 1}{{(x - 1)}^{3}}] d x = \int e^{x} [\frac{x + 1 - 2}{{(x + 1)}^{3}}] d x = \int e^{x} [\frac{1}{{(x - 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x Here, f (x) = \frac{1}{{(x + 1)}^{2}} \Rightarrow f' (x) = \frac{- 2}{{(x + 1)}^{2}} Put e^{x} f (x) = t let e^{x} \frac{1}{{(x + 1)}^{2}} = t Diff both sides e^{x} \frac{1}{{(x + 1)}^{2}} + e^{x} \frac{(- 2)}{{(x + 1)}^{3}} = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{{(x + 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x = d t ∴ \int e^{x} [\frac{1}{{(x + 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x = \int d t = t + C = \frac{e^{x}}{{(x + 1)}^{2}} + C$

Page No 18.143:

Question 11:

$\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) d x$

Answer:

$Let I = \int e^{x} [\frac{\sin 4 x - 4}{1 - \cos 4 x}] d x = \int e^{x} [\frac{2 \sin 2 x \cos 2 x}{2 \sin^{2} (2 x)} - \frac{4}{2 \sin^{2} 2 x}] d x = \int e^{x} [\cot (2 x) - 2 {cosec}^{2} (2 x)] d x Here, f (x) = \cot (2 x) \Rightarrow f' (x) = - 2 {cosec}^{2} (2 x) Put e^{x} f (x) = t let e^{x} \cot (2 x) = t Diff both sides w . r . t x e^{x} \cot (2 x) + e^{x} \times [- 2 {cosec}^{2} (2 x)] = \frac{d t}{d x} \Rightarrow e^{x} [\cot (2 x) - 2 {cosec}^{2} (2 x)] d x = d t ∴ \int e^{x} [\cot 2 x - 2 {cosec}^{2} 2 x] d x = \int d t \Rightarrow I = t + C = e^{x} \cot (2 x) + C$

Page No 18.143:

Question 12:

$\int e^{x} \frac{{(1 - x)}^{2}}{{(1 + x^{2})}^{2}} d x$

Answer:

$Let I = \int e^{x} [\frac{{(1 - x)}^{2}}{{(1 + x^{2})}^{2}}] d x = \int e^{x} [\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}] d x = \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x Here, f (x) = \frac{1}{1 + x^{2}} \Rightarrow f' (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{1 + x^{2}} = t Diff both sides w . r . t x e^{x} \frac{1}{1 + x^{2}} + e^{x} \frac{- 1}{{(1 + x^{2})}^{2}} 2 x = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x = d t ∴ \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x = \int d t \Rightarrow I = t + C = \frac{e^{x}}{1 + x^{2}} + C$

Page No 18.143:

Question 13:

$\int e^{x} \frac{1 + x}{{(2 + x)}^{2}} d x$

Answer:

$Let I = \int e^{x} [\frac{1 + x}{{(2 + x)}^{2}}] d x = \int e^{x} (\frac{2 + x - 1}{{(2 + x)}^{2}}) d x = \int e^{x} [\frac{1}{(2 + x)} - \frac{1}{{(2 + x)}^{2}}] d x Here, f (x) = \frac{1}{2 + x} \Rightarrow f' (x) = \frac{- 1}{{(2 + x)}^{2}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{x + 2} = t Diff both sides w . r . t x e^{x} \frac{1}{x + 2} + e^{x} \frac{- 1}{{(x + 2)}^{2}} = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{x + 2} - \frac{1}{{(x + 2)}^{2}}] d x = d t ∴ \int e^{x} [\frac{1}{(2 + x)} - \frac{1}{{(2 + x)}^{2}}] d x = \int d t \Rightarrow I = t + C = \frac{e^{x}}{2 + x} + C$

Page No 18.143:

Question 14:

$\int \frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x / 2} d x$

Answer:

$Let I = \int (\frac{\sqrt{1 - \sin x}}{1 + \cos x}) e^{\frac{- x}{2}} d x = \int (\frac{\sqrt{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}}{2 \cos^{2} \frac{x}{2}}) e^{\frac{- x}{2}} d x = \int \frac{\sqrt{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}}{2 \cos^{2} \frac{x}{2}} e^{\frac{- x}{2}} d x = \int (\frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) e^{\frac{- x}{2}} d x = \int [\frac{1}{2} \sec \frac{x}{2} \tan \frac{x}{2} - \frac{1}{2} \sec (\frac{x}{2})] e^{\frac{- x}{2}} d x = \frac{1}{2} \int (\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}) e^{\frac{- x}{2}} d x let e^{\frac{- x}{2}} \sec (\frac{x}{2}) = t Diff both sides w . r . t x e^{\frac{- x}{2}} \frac{\sec (\frac{x}{2}) \tan (\frac{x}{2})}{2} + \sec (\frac{x}{2}) \times e^{\frac{- x}{2}} \times \frac{- 1}{2} = \frac{d t}{d x} \Rightarrow {\frac{e}{2}}^{\frac{- x}{2}} [\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}] d x = d t ∴ \frac{1}{2} \int (\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}) e^{\frac{- x}{2}} d x = \int d t \Rightarrow I = \int t + C = e^{\frac{- x}{2}} \sec (\frac{x}{2}) + C$

Page No 18.143:

Question 15:

$\int e^{x} (\log x + \frac{1}{x}) d x$

Answer:

$Let I = \int e^{x} (\log x + \frac{1}{x}) d x Here, f (x) = \log x \Rightarrow f' (x) = \frac{1}{x} put e^{x} f (x) = t \Rightarrow e^{x} \log x = t Diff both sides w . r . t x e^{x} \log x + e^{x} \frac{1}{x} = \frac{d t}{d x} \Rightarrow e^{x} (\log x + \frac{1}{x}) d x = d t ∴ \int e^{x} [\log x + \frac{1}{x}] d x = \int d t \Rightarrow I = t + C = e^{x} \log x + C$

Page No 18.143:

Question 16:

$\int e^{x} (\log x + \frac{1}{x^{2}}) d x$

Answer:

$Let I = \int (\log x + \frac{1}{x^{2}}) e^{x} d x = \int e^{x} (\log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^{2}}) d x = \int e^{x} (\log x + \frac{1}{x}) d x + \int e^{x} (- \frac{1}{x} + \frac{1}{x^{2}}) d x \begin{matrix} let e^{x} \log x = t \\ Diff both sides \\ (e^{x} \log x + e^{x} \frac{1}{x}) d x = d t \end{matrix} \begin{matrix} let e^{x} (- \frac{1}{x}) = p \\ Diff both sides \\ (e^{x} \frac{- 1}{x} + e^{x} \frac{1}{x^{2}}) d x = d p \end{matrix} ∴ I = \int d t + \int d p = t + p + C = e^{x} \log x + e^{x} \frac{- 1}{x} + C = e^{x} (\log x - \frac{1}{x}) + C$

Page No 18.143:

Question 17:

$\int \frac{e^{x}}{x} \{x {(\log x)}^{2} + 2 \log x\} d x$

Answer:

$Let I = \int \frac{e^{x}}{x} [x {(\log x)}^{2} + 2 \log x] d x = \int e^{x} [{(\log x)}^{2} + \frac{2 \log x}{x}] d x Here, f (x) = {(\log x)}^{2} \Rightarrow f' (x) = \frac{2 \log x}{x} put e^{x} f (x) = t \Rightarrow e^{x} {(\log x)}^{2} = t Diff both sides w . r . t x [e^{x} {(\log x)}^{2} + e^{x} \frac{2 \log x}{x}] d x = d t ∴ I = \int d t = t + C = e^{x} {(\log x)}^{2} + C$

Page No 18.143:

Question 18:

$\int e^{x} \cdot \frac{\sqrt{1 - x^{2}} \sin^{- 1} x + 1}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int e^{x} [\frac{\sqrt{1 - x^{2}} \sin^{- 1} x + 1}{\sqrt{1 - x^{2}}}] d x = \int e^{x} [\sin^{- 1} x + \frac{1}{\sqrt{1 - x^{2}}}] d x Here, f (x) = \sin^{- 1} x \Rightarrow f' (x) = \frac{1}{\sqrt{1 - x^{2}}} Put e^{x} f (x) = t \Rightarrow e^{x} \sin^{- 1} x = t D i f f b o t h s i d e s w . r . t x (e^{x} \sin^{- 1} x + e^{x} \times \frac{1}{\sqrt{1 - x^{2}}}) d x = d t ∴ I = \int d t = t + C = e^{x} \sin^{- 1} x + C$

Page No 18.143:

Question 19:

$\int e^{2 x} (- \sin x + 2 \cos x) d x$

Answer:

$Let I = \int e^{2 x} (- \sin x + 2 \cos x) d x Put e^{2 x} \cos x = t Diff both sides w . r . t x [2 e^{2 x} \cos x + e^{2 x} \times (- \sin x)] d x = d t ∴ I = \int d t = t + C = e^{2 x} \cos x + C$

Page No 18.143:

Question 20:

$\int (\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}) d x$

Answer:

$Let I = \int (\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}) d x Put \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t ∴ I = \int e^{t} (\frac{1}{t} - \frac{1}{t^{2}}) d t Here, f (t) = \frac{1}{t} \Rightarrow f' (t) = \frac{- 1}{t^{2}} let e^{t} \times \frac{1}{t} = p Diff both sides w . r . t t (e^{t} \times \frac{1}{t} + e^{t} \times \frac{- 1}{t^{2}}) d t = d p ∴ I = \int d p = p + C = \frac{e^{t}}{t} + C = \frac{x}{\log x} + C$

Page No 18.143:

Question 21:

$\int e^{x} (\frac{\sin x \cos x - 1}{\sin^{2} x}) d x$

Answer:

$Let I = \int e^{x} (\frac{\sin x \cos x - 1}{\sin^{2} x}) d x = \int e^{x} [\cot x - {cosec}^{2} x] d x Here, f (x) = \cot x \Rightarrow f' (x) = - {cosec}^{2} x Put e^{x} f (x) = t \Rightarrow e^{x} \cot x = t Diff both sides w . r . t x e^{x} (\cot x - {cosec}^{2} x) d x = d t ∴ I = \int d t = t + C = e^{x} \cot x + C$

Page No 18.143:

Question 22:

$\int \{\tan (\log x) + \sec^{2} (\log x)\} d x$

Answer:

$Let I = \int [\tan (\log x) + \sec^{2} (\log x)] d x Put \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t ∴ I = \int (\tan t + \sec^{2} t) e^{t} d t Here, f (t) = \tan t \Rightarrow f' (t) = \sec^{2} t let e^{t} \tan (t) = p Diff both sides w . r . t t e^{t} [\tan t + \sec^{2} t] = \frac{d p}{d t} \Rightarrow e^{t} [\tan t + \sec^{2} t] d t = d p ∴ I = \int d p = p + C = e^{t} \tan t + C = x \tan (\log x) + C$

Page No 18.143:

Question 23:

$\int \frac{e^{x} (x - 4)}{{(x - 2)}^{3}} d x$

Answer:

$Let I = \int e^{x} (\frac{x - 4}{{(x - 2)}^{3}}) d x = \int e^{x} [\frac{x - 2 - 2}{{(x - 2)}^{3}}] d x = \int e^{x} [\frac{1}{{(x - 2)}^{2}} - \frac{2}{{(x - 2)}^{3}}] d x Here, f (x) = \frac{1}{{(x - 2)}^{2}} \Rightarrow f' (x) = \frac{- 2}{{(x - 2)}^{3}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{{(x - 2)}^{2}} = t Diff both sides w . r . t x [e^{x} \frac{1}{{(x - 2)}^{2}} + e^{x} \frac{- 2}{{(x - 2)}^{3}}] d x = d t ∴ I = \int d t = t + C = \frac{e^{x}}{{(x - 2)}^{2}} + C$

Page No 18.143:

Question 24:

Evaluate the following integrals:

$\int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x$

Answer:

$We have, I = \int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x = \int e^{2 x} (\frac{1 - 2 sinx \cos x}{2 \sin^{2} x}) d x Put t = 2 x . Then d t = 2 d x Therefore, I = \frac{1}{2} \int e^{t} (\frac{1 - 2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} (\frac{1 - 2 \sin \frac{t}{2} \cos \frac{t}{2}}{\sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} (\frac{1}{\sin^{2} \frac{t}{2}} - \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{\sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} ({cosec}^{2} \frac{t}{2} - 2 \cot \frac{t}{2}) d t = - \frac{1}{4} \int e^{t} (2 \cot \frac{t}{2} - {cosec}^{2} \frac{t}{2}) d t Consider, f (x) = 2 \cot \frac{t}{2}, then f^{'} (x) = - {cosec}^{2} \frac{t}{2} Thus, the given integrand is of the form e^{x} [f (x) + f^{'} (x)] . Therefore, I = - \frac{1}{4} (2 \cot \frac{t}{2}) e^{t} + c = - \frac{1}{4} (2 \cot \frac{2 x}{2}) e^{2 x} + c Hence, \int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x = - \frac{1}{2} (\cot x) e^{2 x} + c$

Page No 18.149:

Question 1:

$\int e^{a x} \cos b x d x$

Answer:

$Let I = \int e^{a x} \cos (b x) d x Considering \cos (b x) as first function and e^{a x} as second function I = \cos (b x) \frac{e^{a x}}{a} - \int - \sin (b x) \times b \times \frac{e^{a x}}{a} d x \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b}{a} \int \sin (b x) e^{a x} d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} \int e^{a x} \times \sin (b x) d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} I_{1} . . . . . (1) where I_{1} = \int e^{a x} \sin (b x) d x Now, I_{1} = \int e^{a x} \sin (b x) d x Considering \sin (b x) as first function e^{a x} as second function I_{1} = \sin (b x) \frac{e^{a x}}{a} - \int \cos (b x) b \frac{e^{a x}}{a} d x \Rightarrow I_{1} = \frac{\sin (b x) e^{a x}}{a} - \frac{b}{a} \int e^{a x} \cos (b x) d x \Rightarrow I_{1} = \frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I . . . . . (2) From (1) and (2) I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} [\frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I] \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b e^{a x} \sin (b x)}{a^{2}} - \frac{b^{2}}{a^{2}} I \Rightarrow I (1 + \frac{b^{2}}{a^{2}}) = e^{a x} [\frac{a \cos (b x) + b \sin (b x)}{a^{2}}] ∴ I = \frac{e^{a x} [a \cos b x + b \sin (b x)]}{(a^{2} + b^{2})} + C$

Page No 18.149:

Question 2:

$\int e^{a x} \sin (b x + C) d x$

Answer:

$Let I = <$