Rd Sharma XII Vol 2 2020 Solutions for Class 12 Commerce Math Chapter 1 Definite Integrals are provided here with simple step-by-step explanations. These solutions for Definite Integrals are extremely popular among Class 12 Commerce students for Math Definite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 Book of Class 12 Commerce Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 Solutions. All Rd Sharma XII Vol 2 2020 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 19.110:

Question 1:

03x+4 dx

Answer:


abfxdx=limh0 hfa+fa+h+fa+2h............+fa+n-1h, where, h=b-an

Here, a=0, b=3, fx=x+4, h=3-0n=3nTherefore, I=03x+4dx    =limh0 hf0+f0+h+.......+f0+n-1h   =limh0 h0+4+h+4+.......+n-1h+4   =limh0 h4n+h1+2+.......+n-1  =limh0 h4n+hnn-12    =limn 3n4n+3n×nn-12   =limn12+921-1n  =12+92=332

Page No 19.110:

Question 2:

02x+3 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x+3, h=2-0n=2nTherefore,I=02x+3dx =limh0 hf0+f0+h+....................+f0+n-1h =limh0 h0+3+0+h+3+...............+0+n-1h+3 =limh0 h3n+h1+2+3.........+n-1 =limh0 h3n+hnn-12 =limn 2n3n+n-1 =limn24-1n =8

Page No 19.110:

Question 3:

133x-2 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an
Here a=1, b=3, fx=3x-2, h=3-1n=2nTherefore,I=133x-2dx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h3-2+3+3h-2+3+6h-2...............+3n-1h+3-2=limh0 hn+3h1+2+3.........+n-1=limh0 hn+3hnn-12=limn 2nn+3n-3=limn24-3n =8

Page No 19.110:

Question 4:

-11x+3 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=-1, b=1, fx=x+3, h=1+1n=2nTherefore,I=-11x+3dx=limh0 hf-1+f-1+h+....................+f-1+n-1h=limh0 h-1+3+-1+h+3+...............+-1+n-1h+3=limh0 h2n+h1+2+3.........+n-1=limh0 h2n+hnn-12=limn 2n2n+n-1=limn23-1n=6

Page No 19.110:

Question 5:

05x+1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here, a=0, b=5, fx=x+1, h=5-0n=5nTherefore,I=05x+1dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+1+h+1+...............+n-1h+1=limh0 hn+h1+2+3.........+n-1=limh0 hn+hnn-12=limn 5nn+5n-52=limn572-5n=352

Page No 19.110:

Question 6:

132x+3 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

 Here, a=1, b=3, fx=2x+3, h=3-1n=2nTherefore,I=132x+3dx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h2+3+2+2h+3+...............+2+2n-1h+3=limh0 h5n+2h1+2+3.........+n-1=limh0 h5n+2hnn-12=limn 2n5n+2n-2=limn27-2n=14

Page No 19.110:

Question 7:

352-x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h+...+fa+n-1hwhere h=b-an

Here a=3, b=5, fx=2-x, h=5-3n=2nTherefore,I=352-xdx  =limh0 hf2+f2+h+...+f2+n-1h  =limh0 h2-2+2-h-2+...+2-n-1h-2  =limh0 h-h1+2+3+...+n-1  =limh0 h-2hnn-12   =limn 2n-2n+2    =limn2-2+2n =-4

Page No 19.110:

Question 8:

02x2+1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+1, h=2-0n=2nTherefore,I=02x2+1dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+1+h2+1+...............+n-12h2+1=limh0 hn+h212+22+32.........+n-12=limh0 hn+h2nn-12n-16=limn 2nn+2n-12n-13n=limn21+231-1n2-1n=2+83=143

Page No 19.110:

Question 9:

12x2 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=1, b=2, fx=x2, h=2-1n=1nTherefore,I=12x2dx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h1+h+12+...............+n-1h+12=limh0 hn+h212+22+32.........+n-12+2h1+2+3+...........+n-1=limh0 hn+h2nn-12n-16+2hnn-12=limn 1nn+n-12n-16n+n-1=limn2+161-1n2-1n-1n=2+13=73

Page No 19.110:

Question 10:

232x2+1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=2, b=3, fx=2x2+1, h=3-2n=1nTherefore,I=232x2+1dx=limh0 hf2+f2+h+....................+f2+n-1h=limh0 h22.22+1+22+h2+1+...............+22+n-1h2+1=limh0 hn+222+2+h2+.............2+n-1h2=limh0 hn+8n+2h212+22+32.........+n-12+8h1+2+.......+n-1=limh0 h9n+h22nn-12n-16+8hnn-12=limn 1n9n+n-12n-13n+4n-4=limn13+131-1n2-1n-4n =13+23=413

Page No 19.110:

Question 11:

12x2-1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=1, b=2, fx=x2-1, h=2-1n=1nTherefore,I=12x2-1dx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h1-1+h2-1+...............+n-12h2-1=limh0 hn-1+h212+22+32.........+n-12=limh0 hn-1+h2nn-12n-16=limn 1nn-1+n-12n-16n=limn1-1n+161-1n2-1n=1+13=43

Page No 19.110:

Question 12:

02x2+4 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+4, h=2-0n=2nTherefore,I=02x2+4dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+4+h2+4+...............+n-12h2+4=limh0 h4n+h212+22+32.........+n-12=limh0 h4n+h2nn-12n-16=limn 2n4n+2n-12n-13n=limn24+231-1n2-1n=8+83=323



Page No 19.111:

Question 13:

14x2-x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=x2-x, h=4-1n=3nTherefore,I=14x2-xdx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h1-1+1+h2-1+h+...............+1+n-1h2-1+n-1h=limh0 hh212+22+32.........+n-12+1+2h1+2+......+n-1-n-h1+2+.....+n-1=limh0 hh2nn-12n-16+hn-12=limn 3n9n-12n-16n+3n-12=limn3321-1n2-1n+321-1n=9+92=272

Page No 19.111:

Question 14:

013x2+5x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=1, fx=3x2+5x, h=1-0n=1nTherefore,I=013x2+5xdx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+0+3h2+5h+...............+3n-12h2+5n-1h=limh0 h5h1+2+.........+n+3h212+22+32.........+n-12=limh0 h5hnn-12+h23nn-12n-16=limn 1n5n-12+n-12n-12n=limn521-1n+121-1n2-1n=52+1=72

Page No 19.111:

Question 15:

02ex dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=ex, h=2-0n=2nTherefore,I=02exdx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 he0+eh+e2h+.......+en-1h=limh0 hehn-1eh-1=limh0 e2-1eh-1h=e2-11=e2-1

Page No 19.111:

Question 16:

abex dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=a, b=b, fx=ex, h=b-anTherefore,I=abexdx=limh0 hfa+fa+h+....................+fa+n-1h=limh0 hea+ea+h+............+ea+n-1h=limh0 heaehn-1eh-1=limh0 heaeb-a-1eh-1=limh0eb-eaeh-1h=eb-ea1=eb-ea

Page No 19.111:

Question 17:

abcos x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h+...+fa+n-1hwhere h=b-an
Here a=a, b=b, fx=cos x, h=b-anTherefore,I=abcos x dx=limh0 hfa+fa+h+...+fa+n-1h=limh0 hcosa+cosa+h+...+cosa+n-1h=limh0 hcosa+n-1h2sinnh2sinh2=limh0h2sinh22cosa+b-a2-h2 sinb-a2               Using nh=b-a=limh0h2sinh2×limh02cosa+b2-h2sinb-a2=2cosa+b2sinb-a2=sin b-sin a                         Since, 2cosA sinB=sinA+B-sinA-B 

Page No 19.111:

Question 18:

0π/2sin x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h+...+fa+n-1hwhere h=b-an
Here a=0, b=π2, fx=sinx, h=π2-0n=π2nTherefore,I=0π2sinxdx  =limh0 hf0+f0+h+...+f0+n-1h  =limh0 hsin0+sinh+sin2h+...+sinn-1h  =limh0 hsinn-1h2sinnh2sinh2  =limh0 h2sinh2×2sinπ4-h2sinπ4            Using nh=π2   =limh0h2sinh2×limh02sinπ4-h2sinπ4 =2sinπ4sinπ4=2×12×12=1

Page No 19.111:

Question 19:

0π/2cos x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h+...+fa+n-1hwhere h=b-an
Here a=0, b=π2, fx=cosx, h=π2-0n=π2nTherefore,I=0π2cosx dx  =limh0 hf0+f0+h+...+f0+n-1h  =limh0 hcos0+cosh+cos2h+...+cosn-1h  =limh0 hcosn-1h2sinnh2sinh2  =limh0 hcosπ4-h2sinπ4sinh2                Using , nh=π2   =limh0h2sinh2×2cosπ4-h2sinπ4  =limh0h2sinh2×limh02cosπ4-h2sinπ4    =2cosπ4 sinπ4=2×12×12=1

Page No 19.111:

Question 20:

143x2+2x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=3x2+2x, h=4-1n=3nTherefore,I=143x2+2xdx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h3.12+2×1+31+h2+21+h+...............+31+n-1h2+21+n-1h=limh0 h312+1+h2+1+2h2+...........+1+n-1h2+21+1+h+..........+1+n-1h=limh0 h3n+3h212+22+32.........+n-12+6h1+2+.........n-1h+2n+2h1+2+..........+n-1h=limh0 h5n+3h2nn-12n-16+8hnn-12=limn 3n5n+9n-12n-12n+12n-12=limn317-12n+921-1n2-1n=51+27=78

Page No 19.111:

Question 21:

023x2-2 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=3x2-2, h=2-0n=2nTherefore,I=023x2-2dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0-2+3h2-2+...............+3n-12h2-2=limh0 h-2n+3h212+22+32.........+n-12=limh0 h-2n+3h2nn-12n-16=limn 2n-2n+2n-12n-1n=limn2-2+21-1n2-1n=-4+8=4

Page No 19.111:

Question 22:

02x2+2 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+2, h=2-0n=2nTherefore,I=02x2+2dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+2+h2+2+...............+n-12h2+2=limh0 h2n+h212+22+32.........+n-12=limh0 h2n+h2nn-12n-16=limn 2n2n+2n-12n-13n=limn22+231-1n2-1n=4+83=203

Page No 19.111:

Question 23:

04x+e2x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here, a=0, b=4, fx=x+e2x, h=4-0n=4nTherefore,I=04x+e2xdx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+e0+h+e2h+...............+n-1h+e2n-1h=limh0 hh1+2+.....+n-1h+e0+e2h+e4h+.........+e2n-1h=limh0 hhnn-12+e2hn-1e2h-1=limn 16n2×nn-12+limh0 e8-1e2h-1h=limn81-1n+limh0 e8-12(e2h-1)2h=8+e8-12=15+e82 

Page No 19.111:

Question 24:

02x2+x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+x, h=2-0n=2nTherefore,I=02x2+xdx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+0+h2+h+...............+n-12h2+h=limh0 hh212+22+32.........+n-12+h1+2+3........+n-1h=limh0 hh2nn-12n-16+hnn-12=limn 2n2n-12n-13n+n-1=limn2231-1n2-1n+1-1n=83+2=143

Page No 19.111:

Question 25:

02x2+2x+1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+2x+1, h=2-0n=2nTherefore,I=02x2+2x+1dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+0+1+h2+2h+1+...............+n-12h2+2n-1h+1=limh0 hn+h212+22+32.........+n-12+2h1+2+.........+n-1h=limh0 hn+h2nn-12n-16+2hnn-12=limn 2nn+2n-12n-13n+2n-2=limn23+231-1n2-1n-2n =6+83=263

Page No 19.111:

Question 26:

032x2+3x+5 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=3, fx=2x2+3x+5, h=3-0n=3nTherefore,I=032x2+3x+5dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+0+5+2h2+3h+5+...............+2n-12h2+3n-1h+5=limh0 h5n+2h212+22+32.........+n-12+3h1+2+.......+n-1h=limh0 h5n+2h2nn-12n-16+3hnn-12=limn 3n5n+3n-12n-1n+9n-12=limn35+31-1n2-1n+921-1n=15+18+272=933



Page No 19.112:

Question 27:

abx dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=a, b=b, fx=x, h=b-anTherefore,I=abx dx=limh0 hfa+fa+h+....................+fa+n-1h=limh0 ha+a+h+ a+2h+..........+a+n-1h=limh0 hna+h1+2+3+........+n-1=limh0 hna+hnn-12=limh0 b-anna+b-an-12=limh0b-aa+b-ab-a-h2=b-aa+b-a22=2ab-2a2+b2+a2-2ab2=b2-a22

Page No 19.112:

Question 28:

05x+1 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=5, fx=x+1, h=5-0n=5nTherefore,I=05x+1dx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0+1+h+1+...............+n-1h+1=limh0 hn+h1+2+3+.................+n-1h=limh0 hn+hnn-12=limn 5nn+5n-12=limn51+521-1n=5+252=352

Page No 19.112:

Question 29:

23x2 dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=2, b=3, fx=x2, h=3-2n=1nTherefore,I=23x2 dx=limh0 hf2+f2+h+....................+f2+n-1h=limh0 h22+2+h2+...........+2+n-1h2=limh0 h4n+h212+22+32.........+n-12+4h1+2+......+n-1h =limh0 h4n+h2nn-12n-16+4hnn-12=limn 1n4n+n-12n-16n+2n-2=limn6+161-1n2-1n-2n =6+13=193

Page No 19.112:

Question 30:

14x2-x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=x2-x, h=4-1n=3nTherefore,I=14x2-xdx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h1-1+1+h2-1+h+...............+n-1h+12-n-1h+1=limh0 hh212+22+32.........+n-12-h1+2+.......+n-1=limh0 hh2nn-12n-16-hnn-12=limn 3n3n-12n-12n+3n-12=limn3321-1n2-1n+321-1n=9+93=383

Page No 19.112:

Question 31:

02x2-x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2-x, h=2-0n=2nTherefore,I=02x2-xdx=limh0 hf0+f0+h+....................+f0+n-1h=limh0 h0-0+h2-h+...............+n-12h2-n-1h=limh0 hh212+22+32.........+n-12-h1+2+.....+n-1h=limh0 hh2nn-12n-16-hnn-12=limn 2n2n-12n-13n-n+1=limn2231-1n2-1n-1+1n=83-2=23

Page No 19.112:

Question 32:

132x2+5x dx

Answer:

abfxdx=limh0 hfa+fa+h+fa+2h...............+fa+n-1hwhere h=b-an

Here, a=1, b=3, fx=2x2+5x, h=3-1n=2nTherefore,I=132x2+5xdx=limh0 hf1+f1+h+....................+f1+n-1h=limh0 h2+5+21+h2+51+h+...............+21+n-1h2+51+n-1h=limh0 h212+1+h2+............+1+n-1h2+51+1+h+1+2h+........+1+n-1h=limh0 h2n+2h212+22+32.........+n-12+4h1+2+......+n-1+5n+5h1+2+......+n-1=limh0 h7n+2h2nn-12n-16+9hnn-12=limn 2n7n+4n-12n-13n+9n-9=limn216+431-1n2-1n-9n=32+163=1123

Page No 19.112:

Question 33:

Evaluate the following integrals as limit of sums:

133x2+1dx                    [CBSE 2014]

Answer:


We have,

abfxdx=limh0fa+fa+h+fa+2h+ ...+ fa+n-1h

Here, a = 1, b = 3,  f(x) = 3x2 + 1 and h=3-1n=2nnh=2

133x2+1dx=limh0h f1+f1+h+f1+2h+ ...+ f1+n-1h=limh0h 3×12+1+3×1+h2+1+3×1+2h2+1+ ...+3×1+n-1h2+1=limh0h31+1+2h+h2+1+4h+22h2+ ... +1+2n-1h+n-12h2+n=limh0h3n+21+2+ ... +n-1h+12+22+ ... +n-12h2+n=limh0h4n+6×nn-12h+3×n-1n2n-16h2
=limh04nh+6×nhnh-h2+3×nh-hnh2nh-h6=limh04nh+3×nhnh-h+3×nh-hnh2nh-h6=limh04×2+3×2×2-h+3×2-h×2×2×2-h6=8+6×2-0+2-0×2×4-02=8+12+8=28

Page No 19.112:

Question 34:

13x2+3x+exdx

Answer:

13x2+3x+exdxa=1, b=3, fx=x2+3x+ex, h=2nI=13x2+3x+exdx=limh0hf1+f1+h+f1+2h+...+f1+n-1h=limh0h4+e+1+h2+31+h+e1+h+1+2h2+31+2h+e1+2h+...+1+n-1h2+3 1+n-1h+e1+n-1h=limh0h4+e+n-1·1+h21+22+32+...n-12+2h1+2+3+n-1+3n-1+3h1+2+3...n-1+e1+h+e1+2h+....e1+n-1h=limh0h4n+e+h2n-1n-22n-36+2hn-1 n2+3h n-1 n2+eeh·eh n-1-1eh-1=limn 4n·2n+e·2n+8n32n3...6+5.4n2 n2-n2+e2n+1e2nn-1-1=8+0+83+10+e e2-1=263+12+e3-e=623+e3-e



Page No 19.115:

Question 1:

04x4-x dx

Answer:

Let, I=04x4-x dx        =044-x4-4+x dx        =044-xx dx        =044x-x32 dx        =8x32304-2x52504        =643-645        =12815

Page No 19.115:

Question 2:

12x3x-2 dx

Answer:

12x3x-2dxLet, 3x-2=t, then 3dx=dtwhen, x=1 ; t=1 and x=2 ; t=4Therefore the integral becomes14t+23t dt3=1914t32+2t  dt=192t525+4t32314=19645+323-25-43=46135

Page No 19.115:

Question 3:

15x2x-1 dx

Answer:

Let  I=15x2x-1dxLet, 2x-1=t, then 2dx=dt,When, x1 ; t1 and x5 ; t9x=t+12I=1219t+1t×dt2=142t323+2t19=1418+6-23-2=163

Page No 19.115:

Question 4:

01cos-1 x dx

Answer:

01cos-1x dx=01cos-1x ×1 dx= cos-1x x01-01-x1-x2dx=x cos-1x01-221-x201=0+1=1

Page No 19.115:

Question 5:

01tan-1 x dx

Answer:

01tan-1x dx=01tan-1x×1 dx=tan-1x x01-01x1+x2dx=xtan-1x01-12log1+x201=π4-0-12log2+0=π4-12log2

Page No 19.115:

Question 6:

01cos-11-x21+x2 dx

Answer:

01cos-11-x21+x2dxLet, x= tanθ, dx= sec2θ dθWhen, x0 ; θ0and x1 ; θπ4Therefore, the integral becomes0π4cos-11-tan2θ1+tan2θ sec2θ dθ= 0π4cos-1cos2θ sec2θ dθ=20π4θ sec2θ dθ=2θ tanθ0π4-20π4tanθ dθ=2θ tanθ0π4+2logcosθ0π4=2π4-0+2log12-0=π2-log2

Page No 19.115:

Question 7:

01tan-12x1-x2 dx

Answer:

01tan-12x1-x2dxLet, x= tanθ, then dx= sec2θ dθWhen, x0 ; θ0And x1 ; θπ4Therefore the integral becomes0π4tan-12tanθ1-tan2θ  sec2θ dθ=0π4tan-1tan2θ  sec2θ dθ=20π4θ sec2θ dθ=2θ tanθ0π4-20π4tanθ dθ=2θ tanθ0π4-2-logcosθ0π4=2π4-0+2log12-0=π2-log2

Page No 19.115:

Question 8:

01/3tan-13x-x31-3x2 dx

Answer:

013tan-13x-x31-3x2dxLet x = tanθ, then dx= sec2θ  dθWhen, x0 ; θ0And x13 ; θπ6Therefore the integral becomes0π6tan-13tanθ-tan3θ1-3tan2θsec2θ dθ=0π6tan-1tan3θsec2θ  dθ=30π6θ sec2θ  dθ=3θ tanθ0π6-30π6tanθ dθ=3θ tanθ0π6-3-logcosθ0π6=3π6×13-0+3log32=π23+3log32=π23-32log43



Page No 19.116:

Question 9:

011-x1+x dx 

Answer:

011-x1+x dx=011-x-1+11+xdx=012-x+11+xdx=0121+x-011+x1+xdx=0121+x-01dx=2log1+x01-x01=2log2-1 

Page No 19.116:

Question 10:

0π/3cos x3+4 sin x dx

Answer:

0π3cosx3+4sinxdxLet, sin x = t  cosx dx = dtWhen, sinx 0 ; t0And sinx π3 ; t32=032dt3+4t=14log3+4t032=14loglog3+23-log3+0=14loglog23+3-log3=14log23+33

Page No 19.116:

Question 11:

0π/2sin2 x1+cos x2 dx

Answer:

0π2sin2x1+cosx2dx=0π21-cos2x1+cosx2dx=0π21+cosx1-cosx1+cosx2dx=0π21-cosx1+cosxdx=0π21-cosx-1+11+cosxdx=0π22-1+cosx1+cosxdx=0π221+cosxdx-0π2dx=0π221-cosx1+cosx1-cosxdx-0π2dx=20π21-cosxsin2xdx-x0π2=20π2cosec2x-cosecx cotx dx-x0π2=2-cotx+cosecx0π2-x0π2=2-π2

Page No 19.116:

Question 12:

0π/2sin x1+cos x dx

Answer:

0π2sinx1+cosxdxLet 1+cosx =t, then -sinx dx = dtWhen, x0, t2 and  xπ2, t1Therefore, the integral becomes21-1tdt=121tdt=2t12=22-1

Page No 19.116:

Question 13:

0π/2cos x1+sin2 x dx

Answer:

0π2cosx1+sin2xdxLet sinx =t, then cosx dx = dtWhen x0 ; t0And xπ2; t1Therefore the integral becomes01dt1+t2=tan-1x01=π4

Page No 19.116:

Question 14:

0πsin3 x1+2 cos x1+cos x2 dx

Answer:

We have,I=0πsin3x1+2cosx1+cosx2dx=0πsin2x1+2cosx1+cosx2sinx dx=0π1-cos2x1+2cosx1+cosx2sinx dxPutting cosx=t-sinx dx=dtWhen x0; t1and xπ; t-1I=-1-11-t21+2t1+t2dt=-111-t21+2t1+t2dt=-111+2t-t2-2t31+2t+t2dt=-111+2t+t2+2t+4t2+2t3-t2-2t3-t4-2t3-4t4-2t5dt=-111+4t+4t2-2t3-5t4-2t5dt=t+2t2+4t33-t42-t5-t63-11=1+2+43-12-1-13--1-2-12-4-133+-142+-15+-163=1+2+43-12-1-13+1-2+43+12-1+13=83

Page No 19.116:

Question 15:

0x1+x1+x2 dx

Answer:

I = 0x1+x1+x2 dx

using partial fraction,
 x(1+x)(1+x2)A1+x + Bx+C1+x2x=A(1+x2) + (Bx+C)(1+x)x=A+Ax2+Bx+Bx2+C+CxB+C=1A+C=0A+B=0so, A=-12, B=12, C=12

putting the values of A,B and C we get

-121+x+12x+121+x2=-1211+x + 12x+11+x2Therefore, I=0-1211+x + 12x+11+x2I=-12log1+x0 + 120x1+x2+11+x2I=-12log1+x0 + 12×202x1+x2 +12011+x2
I=-12log1+x0 + 14log1+x20 + 12tan-1x0I=-12log1+x0 + 12×12log1+x20 + 12tan-1x0I=12logx2+1x+10 + 12tan-1x0I=12log1+1x21+1x0 +  12tan-1x0I=120 + 12tan-1 - tan-10
I = π4
 

Page No 19.116:

Question 16:

0π/4sin 2x sin 3x dx

Answer:

Let, I=0π4sin2x sin3x dx      ...iI=-sin2xcos3x30π4+0π42cos2xcos3x3dxI=-sin2xcos3x30π4+23cos2xsin3x30π4+490π4sin2x sin3x dxI=-sin2xcos3x30π4+23cos2xsin3x30π4+49I    From i59I=-sin2xcos3x30π4+23cos2xsin3x30π459I=132+059I=132 I=352

Page No 19.116:

Question 17:

011-x1+x dx

Answer:

011-x1+xdx=011-x1+x×1-x1-xdx=011-x1-x2dx=0111-x2dx-01x1-x2dx=sin-1x01+1-x201=π2-1

Page No 19.116:

Question 18:

121x2 e-1/x dx

Answer:

121x2e-1xdxLet -1x=t, then 1x2 dx=dtWhen, x1 ; t-1And x2 ; t-12Therefore the integral becomes-1-12etdt=et-1-12=e-12-e-1=e-1e

Page No 19.116:

Question 19:

0π/4cos4 x sin3 x dx

Answer:

0π4cos4x sin3x dx=0π4cos4x sinx 1-cos2x dx=0π4cos4x sinx dx-0π4cos6x sinx dx=-cos5x50π4+cos7x70π4=-1202+15+1562-17=-240+235+2112=235-92560

Page No 19.116:

Question 20:

π/3π/21+cos x1-cos x5/2 dx

Answer:

π3π21+cosx1-cosx52dx=π3π21+cosx1-cosx52×1-cosx1-cosxdx=π3π2sinx1-cosx3dx=-121-cosx-2π3π2=-121-4=32

Page No 19.116:

Question 21:

0π/2x2 cos 2x dx

Answer:

0π2x2 cos2x dx=x2sin2x20π2-0π22x sin2x2dx=x2sin2x20π2-0π2x sin 2x dx=x2sin2x20π2--xcos2x20π2+-0π2cos2x2dx=x2sin2x20π2+xcos2x20π2-0π2cos2x2dx=x2sin2x20π2+xcos2x20π2-12sin2x20π2=0-π4-0=-π4

Page No 19.116:

Question 22:

01log1+x dx

Answer:

01log1+xdx=01log1+x×1 dx=log1+x x01-01x1+xdx=log1+x x01-011-11+xdx=xlog1+x01-x-log1+x01=log2-1+log2=2log2-1=log4-loge=log4e

Page No 19.116:

Question 23:

Evaluate the following integrals:

24x2+x2x+1dx

Answer:


Let I = 24x2+x2x+1dx

Put 2x + 1 = z2

2dx=2zdzdx=zdz


When x2, z5

When x4, z3

I=53z2-122+z2-12z×zdzI=53z4-2z2+1+2z2-24dzI=1453z4-1dzI=14×z55-z53
I=142435-3-2555-5I=14×2285-14×45I=575-5

Page No 19.116:

Question 24:

01xtan-1 x2 dx

Answer:

We have,I=01xtan-1x2dxPutting tan-1x=ux=tan udx=sec2u duWhen x0; u0and x1; uπ4I=0π4tan uu2sec2u du=0π4u2 tan u sec2u du=u2 tan2 u20π4-0π42u tan2 u2 du=u2 tan2 u20π4-0π4u sec2 u-1 du=u2 tan2 u20π4-0π4u sec2 u du+0π4u du=u2 tan2 u20π4-u tan u0π4+0π4 tan u du+u220π4=u2 tan2 u20π4-u tan u0π4+log sec u0π4+u220π4=π216 ×12-π4+log2+π232=π216-π4+log2=π216-π4+12log 2

Page No 19.116:

Question 25:

01cos-1 x2 dx

Answer:


I=01(cos-1x)2dxlet cos-1x = θx = cosθdx = -sinθ dθwhen x=0, θ=π2 and when x=1, θ=0therefore, I = π20θ2(-sinθ)dθ  I = -π20θ2(sinθ)dθI = 0π2θ2(sinθ)dθI = [θ2(-cosθ)]0π2 -0π22θ0π2sinθdθ]I = [θ2(-cosθ)]0π2 - 0π2[2θ(-cosθ)dθ]
=[-θ2cosθ]0π2 + 0π22θ(cosθ)dθ=[-θ2cosθ]0π2 + 2[θsinθ - 0π2sinθdθ]=[-θ2cosθ]0π2+  2[θsinθ + cosθ]0π2

 I = 2[(π2 + 0) - 1] I = π - 2

Page No 19.116:

Question 26:

12x+3xx+2 dx

Answer:

12x+3xx+2dx=12x+2+1xx+2dx=121xdx+121xx+2dx=121xdx+1212x+2-xxx+2dx=121xdx+12121xdx-12121x+2dx=32121xdx-12121x+2dx=32logx12-12logx+212=32log2-12log4+12log3=32log2- log2+12log3=12log2+12log3=12log6

Page No 19.116:

Question 27:

0π/4ex sin x dx

Answer:

Let,  I=0π4ex sinxdx     ...i          =-excosx0π4+0π4ex cosx dx           =-excosx0π4+exsinx0π4-0π4ex sinx dx  I  =-excosx0π4+exsinx0π4- I         Using i     2I=-excosx0π4+exsinx0π4              =-12eπ4+1+12eπ4-0         =1Hence I = 12

Page No 19.116:

Question 28:

0π/4tan4 x dx

Answer:

0π4tan4x dx=0π4tan2xsec2x-1 dx=0π4tan2x sec2x dx-0π4tan2x dx=tan3x30π4-tanx-x0π4=13-1+π4=π4-23

Page No 19.116:

Question 29:

012x-1 dx

Answer:

We have,2x-1=-2x-1,     0x12   2x-1,     12x1012x-1dx=012-2x-1 dx+1212x-1 dx=-x2+x012+x2-x121=-14+12+1-1-14+12=12

Page No 19.116:

Question 30:

13x2-2x dx

Answer:


We have,x2-2x=-x2-2x,     1x2   x2-2x,     2x313x2-2xdx=12-x2-2x dx+23x2-2x dx=-x33+x212+x33-x223=-83+4+13-1+9-9-83+4=2

Page No 19.116:

Question 31:

0π/2sin x-cos x dx

Answer:

0π2sinx-cosxdx=20π2sinx12-cosx12dx=20π2sinx cosπ4-cosx sinπ4dx=20π2sinx-π4dxWe have,sinx-π4=-sinx-π4,     0xπ4  sinx-π4,     π4xπ20π2sinx-cosxdx=20π4-sinx-π4dx+2π4π2sinx-π4dx                                        =2cosx-π40π4-2cosx-π4π4π2                                        =2cos 0-cos-π4-2cosπ4-cos 0                                        =21-12-12+1                                        =22-22                                        =  22-2                                        =  22-1

Page No 19.116:

Question 32:

01sin 2π x dx

Answer:


We have,sin2πx=sin2πx,     0x12-sin2πx,    12x101sin2πxdx =012sin2πx  dx+121-sin2πx  dx                               =-cos2πx2π012+cos2πx2π121                               =12π+12π+12π+12π                               =2π

Page No 19.116:

Question 33:

13x2-4 dx

Answer:

13x2-4dx=12-x2-4 dx+23x2-4  dx=-x33+4x12+x33-4x23=-83+8+13-4+9-12-83+8=4

Page No 19.116:

Question 34:

-π/2π/2sin9 x dx

Answer:

-π2π2sin9xdxLet fx=sin9xConsider, f-x=sin9-x=-sin9x=-fxThus fx is an odd functionTherefore,-π2π2sin9xdx=0

Page No 19.116:

Question 35:

-1/21/2cos x log1+x1-x dx

Answer:

-1212cosx log1+x1-xdxLet fx=cosx log1+x1-xConsider f-x=cos-x log1-x1+x                        =cosx-log1+x1-x=-cosx log1+x1-x=-fxThus fx is an odd functionTherefore,-1212cosx log1+x1-xdx=0

Page No 19.116:

Question 36:

-aax ex21+x2 dx

Answer:

-aax ex21+x2dxLet fx=x ex21+x2Consider f-x=-x ex21+x2=-fxThus fx is an odd functionTherefore,-aax ex21+x2dx=0ode is 4430.

Page No 19.116:

Question 37:

0π/211+cot7 x dx

Answer:

Let, I=0π211+cot7xdx        ... (i)       =0π211+cot7π2-xdx     =0π211+tan7xdx             ...(ii)Adding (i) and (ii)2I=0π211+cot7x+11+tan7xdx      =0π22+cot7x+tan7x1+cot7x1+tan7xdx   =0π22+cot7x+tan7x2+cot7x+tan7xdx   =0π2dx   =x0π2   =π2Hence, I=π4

Page No 19.116:

Question 38:

02πcos7 x dx

Answer:

Let, I=02πcos7xdx    ...i        =02πcos72π-xdx        =02π-cos7xdx I=-02πcos7xdx    ...iiAdding i and ii we get,      2I=02πcos7xdx-02πcos7xdx2I=0I = 0

Page No 19.116:

Question 39:

0axx+a-x dx

Answer:

Let, I=0axx+a-xdx                 ...(i)         =0aa-xa-x+a-a+xdx         =0aa-xa-x+xdxI  =0aa-xx+a-xdx                 ...(ii)Adding (i) and (ii)2I=0axx+a-x+a-xx+a-xdx   =0adx   =x0a   =aHence, I=a2

Page No 19.116:

Question 40:

0π/211+tan3 x dx

Answer:

Let, I=0π211+tan3xdx              ... (i)       =0π211+tan3π2-xdx       =0π211+cot3xdx              ....(ii)Adding (i) and (ii)2I=0π211+tan3x+11+cot3xdx   =0π22+tan3x+cot3x1+tan3x1+cot3xdx   =0π22+tan3x+cot3x2+tan3x+cot3xdx    =0π2dx     =x0π2    =π2Hence, I=π4

Page No 19.116:

Question 41:

0πx sin x1+cos2 x dx

Answer:

Let, I=0πx sinx1+cos2xdx            ...(i)       =0ππ-x sinπ-x1+cos2π-xdx       =0ππ-x sinx1+cos2xdx        ...(ii)Adding (i) and (ii)2I=0πx sinx1+cos2x+π-x sinx1+cos2x dx    =  0ππ sinx1+cos2xdx       =π-tan-1cosx0π    =-πtan-1-1-tan-11    =-π-π4-π4    =π22Hence, I=π24



Page No 19.117:

Question 42:

0πx sin x cos4 x dx

Answer:

Let, I=0πx sinx cos4x dx              ...(i)       =0ππ-x sinπ-x cos4π-x dx       =0ππ-x sinx cos4x dx       ...(ii) Adding (i) and (ii)2I=0πx sinx cos4x +π-x sinx cos4x dx    =0πx+π-x  sinx cos4x      dx    = π0πsinx cos4x      dx    =π-cos5x50π  =π15+15  =2π5Hence, I=π5

Page No 19.117:

Question 43:

0πxa2 cos2 x+b2 sin2 x dx

Answer:

We have,I =0πxa2cos2x+b2sin2xdx        .....1=0ππ-xa2cos2π-x+b2sin2π-xdx=0ππ-xa2cos2x+b2sin2xdx        .....2Adding 1 and 22I=0πx+π-xa2cos2x+b2sin2xdx=π0π1a2cos2x+b2sin2xdx=π0πsec2xa2+b2tan2xdx         Dividing numerator and denominator by cos2x=2π0π2sec2xa2+b2tan2xdx         Using 02afxdx=0afxdx+0af2a-xdxPutting tan x=tsec2x dx=dtWhen x0; t0and xπ2; t2I=2π0π2dta2+b2t2I=πb20π2dta2b2+t2=πb2×batan-1bta0=πabπ2-0=πab×π2=π22ab    Hence I=π22ab         

Page No 19.117:

Question 44:

-π/4π/4tan x dx

Answer:


-π4π4tanxdx=-π40-tanx  dx+0π4 tanx dx=log cosx-π40+-log cosx0π4=-log12-log12=2log2=log2

Page No 19.117:

Question 45:

015x2 dx

Answer:

We have,I=01.5x2 dx=01x2 dx+12x2 dx+21.5x2 dx=010 dx+121 dx+21.52 dx        x2=0      where, 0<x<11      where, 1<x<22        where, 2<x<1.5=0+x12+2x21.5=x12+2x21.5=2-1+21.5-2=2-1+3-22=2-2

Page No 19.117:

Question 46:

0πx1+cos α sin x dx

Answer:

We have,I=0πx1+cos α sin x dx     .....1I=0ππ-x1+cos α sin π-x dx           0afxdx=0afa-xdxI=0ππ-x1+cos α sin x dx     .....2Adding 1 and 2, we get

2I=0ππ1+cos α sin x dx I=π20π11+cos α sin x dx =π20π11+cos α 2tan x21+tan2x2 dx =π20π1+tan2x21+tan2x2+cos α 2tan x2 dxPutting tanx2=t12sec2x2dx=dtWhen x0 ; t0and xπ ; tNow, integral becomes

I=π0dt1+t2+2t cos α =π0dtt+cos α2+1-cos2α=π0dtt+cos α2+sin2α=π1sin αtan-1t+cos αsin α0=πsin αtan-1t+cos αsin α0=πsin απ2-tan-1cot α=πsin απ2-tan-1tanπ2-α=πsin απ2-π2-α=παsin α

Page No 19.117:

Question 47:

0π/2x sin x cos xsin4 x+cos4 x dx

Answer:

Let, I=0π2xsinx cosxsin4x+cos4xdx       ...(i)=0π2π2-xsinπ2-x cosπ2-xsin4π2-x+cos4π2-xdx=0π2π2-xcosx sinxcos4x+sin4xdx         ...(ii)Adding (i) and (ii)2I=0π2x+π2-xsinx cosxsin4x+cos4xdx         =π20π2sinx cosxsin2x+cos2x2-2sin2x cos2xdx      = π20π2sinx cosx1-2sin2x cos2x dx      = π20π2sinx cosx1-2sin2x 1-sin2xdx     =π20π2sinx cosx1-2sin2x+2sin4xdxLet, sin2x =t, then 2sinxcosx dx = dt When, x0 ; t0 and xπ2 ; t1   2I=π40111-2t+2t2dt       =π8011t-122+14     =π82 tan-12t-101     =π4tan-11-tan-1-1     =π4π4+π4     =π28Hence, I=π216

Page No 19.117:

Question 48:

0π/2cos2 xsin x+cos x dx

Answer:

We have,I=0π2cos2xsinx+cosxdx          .....1=0π2cos2π2-xsinπ2-x+cosπ2-xdx=0π2sin2xcosx+sinx dx           .....2

Adding 1 and 22I=0π2cos2xsinx+cosx+sin2xcosx+sinxdx=  0π21sinx+cosxdx=0π212tanx21+tan2x2+1-tan2x21+tan2x2dx

=-0π21+tan2x2tan2x2-2tanx2-1  dx=-0π2sec2x2tan2x2-2tanx2-1  dxPutting tanx2=t12sec2x2dx=dtsec2x2dx=2dtWhen x0; t0and xπ2; t1

2I=-201dtt2-2t-1I=-01dtt-12-22=-122logt-1-2t-1+2 01=-122log-1-log-1-2-1+2 =-122log 1-log2+12-1 

=-122-log2+12-1 =122log2+12+12-12+1=122log2+122-1=122log2+12=122×2 log2+1=12log2+1

Page No 19.117:

Question 49:

0πcos 2x log sin x dx

Answer:

0πcos2x logsinx dx=logsinx sin2x20π-0πcosxsinxsin2x2 dx=logsinx sin2x20π-0πcos2x dx=logsinx sin2x20π-0π1+cos2x2dx=logsinx sin2x20π-12x+sin2x20π=0-12π+0=-π2

Page No 19.117:

Question 50:

0πxa2-cos2 x dx, a>1

Answer:

Let I=0πxa2-cos2xdx      ... (i)       =0ππ-xa2-cos2π-xdx        =0ππ-xa2-cos2xdx        ...(ii)Adding (i) and (ii)2I=0ππa2-cos2xdx    =π2a0π1a-cosx+1a+cosx dx  =π2a0πsec2x2a-1+a+1tan2x2+sec2x2a+1+a-1tan2x2dxLet, tanx2=t, then  12sec2x2 dx=dt2I=πa01a-1+a+1t2+1a+1+a-1t2 dt   =πaa2-1tan-1a+1a-1t+tan-1a-1a+1t0  =πaa2-1π2+π2=π2aa2-1I=π22aa2-1

Page No 19.117:

Question 51:

0πx tan xsec x+tan x dx

Answer:

Let I=0πx tanxsecx +tanxdx         ...(i)       =0ππ-x tanπ-xsecπ-x +tanπ-xdx      =0ππ-x tanxsecx +tanxdx            ...(ii)Adding (i) and (ii) we get2I=0ππ tanxsecx +tanxdx    =π0πsinx1+sinxdx    =π0π1+sinx-11+sinxdx    =π0π1-11+sinxdx   =πx0π-π0π11+2tanx21+tan2x2dx   =π2-π0πsec2x21+tan2x2+2tanx2dx =π2-π0πsec2x21+tanx22dx  =π2+π21+tanx20π  =π2+π0-2  =π2-2π  =ππ-2Hence I=π2π-2

Page No 19.117:

Question 52:

23x5-x+x dx

Answer:

Let, I=23x5-x+xdx      ...(i)       =235-x5-5+x+5-xdx        =235-xx+5-xdx          ...(ii)Adding (i) and (ii)  2I=23x5-x+x+5-xx+5-xdx    =235-x+x5-x+x dx    =23dx    =x23    =3-1=1Hence, I=12

Page No 19.117:

Question 53:

0π/2sin2 xsin x+cos x dx

Answer:

We have,I=0π2sin2xsinx+cosxdx     .....1=0π2sin2π2-xsinπ2-x+cosπ2-xdx=0π2cos2xcosx+sinx dx     .....2Adding 1 and 22I=0π2sin2xsinx+cosx+cos2xcosx+sinx  dx=0π21sinx+cosx dx=0π21+tan2x22tanx2+1-tan2x2 dx=0π2sec2x22tanx2+1-tan2x2 dxPutting  tanx2=t 12sec2x2dx=dt sec2x2dx=2 dtWhen x0; t0and xπ2; t12I=012dt2t+1-t2 dx=201dt22-t-12=222log2+t-12-t+1 01=12log22 - log2-12+1  =120-log2-12+1 =-12log2-12+1=12log2+12-1=12log2+12+12-12+12I=12log2+122-12I=22log2+1Hence I=12log2+1

Page No 19.117:

Question 54:

0π/2xsin2 x+cos2 x dx

Answer:

Let, I=0π2xsin2x+cos2xdx    =0π2x1dx    =0π2x dx    =x220π2  I=π28

Page No 19.117:

Question 55:

-ππx10 sin7 x dx

Answer:

-ππx10 sin7x dxLet fx=x10 sin7xConsider f-x=-x10 sin7-x=-x10 sin7x=-fxHence fx is an odd functionTherefore -ππx10 sin7x dx=0

Page No 19.117:

Question 56:

01cot-11-x+x2 dx

Answer:

01cot-11-x+x2dx=01cot-1xx-1+1dx=01cot-1xx-1+1x-x-1dx=01cot-1x-cot-1x-1  dx=xcot-1x01+01x1+x2dx-x-1cot-1x-101-01x-11+x-12dx=xcot-1x01+12log1+x201-x-1cot-1x-101-12log1+1-x201=π4-12log2+π4-12log2=π2-log2

Page No 19.117:

Question 57:

0πdx6-cos x

Answer:

0π16-cosxdx=0π1+tan2x26+6tan2x2-1+tan2x2dx=0πsec2x25+7tan2x2dxLet, tanx2=t, then 12sec2x2 dx=dtTherefore the integralbecomes02dt5+7t2 =270dt57+t2 =235tan-17t50=π35

Page No 19.117:

Question 58:

0π/212 cos x+4 sin x dx

Answer:

We have,I=0π212cosx+4sinxdx=0π21+tan2x22-2tan2x2+8tanx2dxPutting tanx2=t 12sec2x2dx=dtWhen x0; t0and xπ2; t1I=201dt2-2t2+8t=-2201dt t2-4 t-1=-01dtt-22-5=01dt52-t-22=125log5+t-25-t+2 01= 125log5-15 +1 -log5 -25 +2 = 125log5-15 +1 ×5 +25 -2=125log5+25-5-25-25+5-2 =125log5+3-5+3 
I = 125log 3 + 53 - 5×3 + 53 + 5 I = 125log 3 + 522I =225log 3 + 52 I = 15log 3 + 52

Page No 19.117:

Question 59:

π/6π/2cosec x cot x1+cosec2 x dx

Answer:

π6π2cosecx cotx1+cosec2xdx=π6π2cosx1+sin2xdx=tan-1sinxπ6π2=tan-11-tan-112=tan-11-121+1×12=tan-113 

Page No 19.117:

Question 60:

0π/2dx4 cos x+2 sin x

Answer:

0π214cosx+2sinxdx=0π21+tan2x24-4tan2x2+4tanx2dxLet tanx2=t, then 12sec2x2 dx=dtWhen x=0, t=0, x=π2, t=1=-1401dtt-122-54=-14×-45log2t-1-52t-1+501=15log5+15-1

Page No 19.117:

Question 61:

04x dx

Answer:

04xdx=x2204=8-0=8

Page No 19.117:

Question 62:

022x2+3 dx

Answer:

022x2+3dx=2x33+3x02=163+6=343

Page No 19.117:

Question 63:

14x2+x dx

Answer:

Here a =1,b=4, fx=x2+x, h=4-1n=3nTherefore,14x2+xdx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h                          =limh0 hf1+f1+h+..........+f1+n-1h                         =limh0 h1+1+1+h2+1+h+1+2h2+1+2h+.........+1+n-1h2+1+n-1h                         =limh0 h2n+h212+22+..............n-12+2h1+2+......+n-1+h1+2+......+n-1                         =limh0 h2n+h2nn-12n-16+3hnn-12                        =limn06+921-1n2-1n+921-1n                        =6+9+92=272

Page No 19.117:

Question 64:

-11e2x dx

Answer:

Here a =-1,b=1, fx=e2x, h=1+1n=2nTherefore,-11e2xdx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf-1+f-1+h+..........+f-1+n-1h           =limh0 he-2+e2-1+h+e2-1+2h+.......+e2-1+n-1h           =limh0 he-2e2hn-1e2h-1            =limh0 e-2 e4-1e2h-12h×12             Since, nh=2             =12e2-e-2

Page No 19.117:

Question 65:

23e-x dx

Answer:

Here a =2,b=3, fx=e-x, h=3-2n=1nTherefore,23e-xdx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf2+f2+h+..........+f2+n-1h           =limh0 he-2+e-2+h+e-2+2h+.......+e-2+n-1h           =limh0 he-2e-hn-1e-h-1           =limh0 e-2 e-1-1e-h-1-h×-1             Since nh=1           =e-2-e-3

Page No 19.117:

Question 66:

132x2+5x dx

Answer:

Here, a =1,b=3, fx=2x2+5x, h=3-1n=2nTherefore,132x2+5xdx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf1+f1+h+..........+f1+n-1h           =limh0 h2+5+21+h2+51+h+21+2h2+51+2h+.........+2n-1h2+5n-1h           =limh0 h2n+2h212+22+..............n-12+4h1+2+............n-1+5n+5h1+2+............n-1           =limh0 h7n+2h2nn-12n-16+9hnn-12           =limn014+831-1n2-1n+181-1n           =14+163+18           =1123

Page No 19.117:

Question 67:

13x2+3x dx

Answer:

Here a =1,b=3, fx=x2+3x, h=3-1n=2nTherefore,13x2+3xdx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf1+f1+h+..........+f1+n-1h           =limh0 h1+3+1+h2+31+h+1+2h2+31+2h+.........+n-1h2+3n-1h           =limh0 hn+h212+22+..............n-12+2h1+2+............n-1+3n+3h1+2+............n-1           =limh0 h4n+h2nn-12n-16+5hnn-12           =limn08+431-1n2-1n+101-1n           =8+83+10           =623

Page No 19.117:

Question 68:

02x2+2 dx

Answer:

Here a =0,b=2, fx=x2+2, h=2-0n=2nTherefore,02x2+2dx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf0+f0+h+..........+f0+n-1h           =limh0 h0+2+0+h2+2+0+2h2+2+.........+n-1h2+2           =limh0 h2n+h212+22+..............n-12           =limh0 h2n+h2nn-12n-16           =limn04+431-1n2-1n           =4+83          =203

Page No 19.117:

Question 69:

03x2+1 dx

Answer:

Here, a =0, b=3, fx=x2+1, h=3-0n=3nTherefore,03x2+1dx =limh0 hfa+fa+h+fa+2h+............+fa+n-1h           =limh0 hf0+f0+h+..........+f0+n-1h           =limh0 h0+1+h2+1+2h2+1+.........+n-1h2+1           =limh0 hn+h212+22+..............n-12           =limh0 hn+h2nn-12n-16           =limh03+921-1n2-1n           =3+9=12



Page No 19.118:

Question 1:

01x 1-x dx equals

(a) π/2
(b) π/4
(c) π/6
(d) π/8

Answer:

(d) π/8

Let, I=01x1-xdx  =01x-x2dx  =0114-x2-x+14dx  =01122-x-122 dx  =x-122x-x2+12×14sin-12x-101 =18 sin-11-sin-1-101=18π2+π2  =π8

Page No 19.118:

Question 2:

0π11+sin x dx equals

(a) 0
(b) 1/2
(c) 2
(d) 3/2

Answer:

(c) 2

0π11+sinxdx=0π11+sinx×1-sinx1-sinxdx=0π1-sinx1-sin2xdx=0π1-sinxcos2xdx =0πsec2x-secx tanx  dx=tanx-secx0π=0+1-0+1=2

Page No 19.118:

Question 3:

The value of 0πx tan xsec x+cos x dx is
(a) π24

(b) π22

(c) 3π22

(d) π23

Answer:

a π24π24

Let I=0πxtanxsecx+cosxdx             (i)=0ππ-xtanπ-xsecπ-x+cosπ-xdx=0ππ-xtanxsecx+cosx dx                  (ii)Adding (i) and (ii)2I=0πxtanxsecx+cosx+π-xtanxsecx+cosxdx   =0ππ tanxsecx+cosxdx  =π0πsinx1+cos2x dx    =-πtan-1cosx0π    =-π-π4-π4=π22Hence  I=π24We have, I=0πx tanxsecx+cosxdx             .....1=0ππ-xtanπ-xsecπ-x+cosπ-xdx=0ππ-xtanxsecx+cosx dx               .....2Adding 1 and 2, we get2I=0πxtanxsecx+cosx+π-xtanxsecx+cosxdxI=120ππ tanxsecx+cosxdx  =π20πsinx1+cos2x dxPutting cos x=t-sinx dx=dtsinx dx=-dtWhen x0; t1and xπ; t-1I=π21-1-dt1+t2=π2-11dt1+t2=π2tan-1t-11=π2tan-11-tan-1-1=π2π4--π4=π2×π2=π24Hence  I=π24



Page No 19.119:

Question 4:

The value of 02π1+sinx2dx is
(a) 0
(b) 2
(c) 8
(d) 4

Answer:

(c) 8

02π1+sinx2dx=02πsin2x4+cos2x4+2sinx4cosx4  dx=02πsinx4+cosx4dx=-cosx414+sinx41402π=4sinx4-cosx402π=4sin2π4-cos2π4-sin 0+cos 0=4sinπ2-cosπ2-0+1=41-0-0+1=4×2=8

Page No 19.119:

Question 5:

The value of the integral 0π/2cos xcos x+sin x dx is
(a) 0
(b) π/2
(c) π/4
(d) none of these

Answer:

(c) π/4

Let I=0π2cosxcosx+sinxdx        ... (i)      =0π2cosπ2-xcosπ2-x+sinπ2-xdx        =   0π2sinxsinx+cosxdx      =   0π2sinxcosx+sinxdx          ... (ii)Adding (i) and (ii)2I=0π2cosxcosx+sinx+sinxcosx+sinxdx   =0π2dx   =x0π2=π2Hence I=π4

Page No 19.119:

Question 6:

011+ex dx equals

(a) log 2 − 1
(b) log 2
(c) log 4 − 1
(d) − log 2

Answer:

(b) log 2

We have,I=011+exdxPutting ex=t exdx= dtdx= dttWhen x0; t1and x; tI=11t1+tdt=11t+t2dt=11t+122-122dt

=12×12logt+12-12t+12+121=logtt+11=logtttt+1t1=log11+1t1=log11+0-log11+1=log1-log12=0--log2=log2

Page No 19.119:

Question 7:

0π2/4sinxx dx equals
(a) 2
(b) 1
(c) π/4
(d) π2/8

Answer:

(a) 2

0π24sinxxdxLet x=t, then12xdx=dtWhen x=0, t=0, x=π24, t=π2Therefore the integral becomes0π22 sint  dt=-2cost0π2=2

Page No 19.119:

Question 8:

0π/2cos x2+sin x1+sin x dx equals

(a) log23

(b) log32

(c) log34

(d) log43

Answer:

(d) log43

Let, I= 0π2cosx2+sinx1+sinxdxLet sinx , then cosx dx =dtWhen x=0, t=0, x=π2, t=1Therefore the integral becomesI=01dt2+t1+t=01-12+t+11+t dt=-log2+t+log1+t01=log1+t-log2+t01=log2-log3-log1+log2=log43

Page No 19.119:

Question 9:

0π/212+cos x dx equals

(a) 13tan-113

(b) 23tan-113

(c) 3 tan-13

(d) 23 tan-13

Answer:

b 23tan-113

We have,I=0π212+cosxdx=0π212+1-tan2x21+tan2x2dx=0π21+tan2x22+2 tan2x2+1-tan2x2dx=0π2sec2x23+tan2x2dx

Putting tan x2=t12sec2x2dx=dtsec2x2dx=2dtWhen, x0; t0and xπ2; t1I=0123+t2dt=201132+t2dt=23tan-1t301=23tan-113-tan-103=23tan-113
3tan1(13)

Page No 19.119:

Question 10:

0π1-x1+xdx=

(a) π2

(b) π2-1

(c) π2+1

(d) π + 1

Answer:

Disclaimer: None of the given option is correct.

We have,I=0π1-x1+xdx=0π1-x1+x×1-x1-xdx=0π1-x1-x2dx=0π11-x2dx-0πx1-x2dxPutting 1-x2=t-2x dx=dtx dx=-dt2When x0; t1and xπ; t1-π2I=0π11-x2dx-11-π2  -dt2t=sin-1x0π+22 t11-π2=0-0+ 1-π2-1=1-π2-1

Page No 19.119:

Question 11:

0π1a+b cos x dx=

(a) πa2-b2

(b) πab

(c) πa2+b2

(d) (a + b) π

Answer:

a πa2-b2We have,I=0π1a+bcosxdx=0π1a+b1-tan2x21+tan2x2dx

=0π1+tan2x2a1+tan2x2+b1-tan2x2dx=0π1+tan2x2a+b+a-btan2x2dx=0πsec2x2a+b+a-btan2x2dx

Putting tanx2=t12sec2x2dx=dtsec2x2dx=2 dtWhen x0; t0and xπ; tI=02dta+b+a-bt2=2a-b01a+ba-b+t2dt

=2a-b01a+ba-b2+t2dt=2a-b×a-ba+btan-1ta+ba-b0=2a2-b2π2-0=2a2-b2π2=πa2-b2

Page No 19.119:

Question 12:

π/6π/311+cotx dx is

(a) π/3
(b) π/6
(c) π/12
(d) π/2

Answer:

(c) π12Let, I=π6π311+cotxdx     ... (i)=π6π311+cotπ3+π6-xdx             Using abfxdx=abfa+b-xdx=π6π311+tanxdx    ... (ii)Adding (i) and (ii) we get 2I=  π6π3   11+cotx+11+tanxdx   =π6π32+cotx+tanx1+cotx1+tanxdx  =π6π32+cotx+tanx2+cotx+tanxdx   =π6π3dx   =xπ6π3  =π3-π6  =π6Hence, I=π12

Page No 19.119:

Question 13:

Given that 0x2x2+a2x2+b2x2+c2 dx=π2a+bb+cc+a, the value of 0dxx2+4x2+9, is

(a) π60

(b) π20

(c) π40

(d) π80

Answer:


(b) π6001x2+4x2+9dx=1501x2+4-1x2+9dx=1512tan-1x2-13tan-1x30=1512×π2-13×π2=15×π12=π60



Page No 19.120:

Question 14:

1elog x dx=

(a) 1
(b) e − 1
(c) e + 1
(d) 0

Answer:

(a) 1

1elogx dx=1elogx x0 dx=x logx1e-1e1xx dx=x logx1e-x1e=e-0-e-1=e-e+1=1

Page No 19.120:

Question 15:

1311+x2 dx is equal to

(a) π12

(b) π6

(c) π4

(d) π3

Answer:

(a) π12

1311+x2dx=tan-1x13=π3-π4=π12
30.

Page No 19.120:

Question 16:

033x+1x2+9 dx=

(a) π12+log22

(b) π2+log22

(c) π6+log22

(d) π3+log22

Answer:

a π12+log22

We have,I=033x+1x2+9dxI=033xx2+9dx+031x2+9dxI1=033xx2+9dx and I2=031x2+9dxPutting x2+9=t in I12x dx=dtx dx=dt2When x0; t9and x3; t18I=9183 dt2 t+031x2+9dx=32918dtt+031x2+32dx=32logt918+13tan-1x303=32log18-log9+13π4-0=32log189+π12=32log 2+π12=log8+π12=log22+π12=π12+log22

Page No 19.120:

Question 17:

The value of the integral 0x1+x1+x2 dx is

(a) π2

(b) π4

(c) π6

(d) π3

Answer:

b π4

We have,I=0x1+x1+x2 dxPutting x=tan θdx=sec2θ dθWhen x0 ; θ0and x ; θπ2Now, integral becomes

I=0π2tan θ1+tan θ sec2θ sec2θ dθ=0π2tan θ1+tan θ dθ=0π2sin θcos θ1+sin θcos θdθI=0π2sin θsin θ+cos θdθ     .....1I=0π2sinπ2-θsinπ2-θ+cosπ2-θdθ        0afxdx=0afa-xdxI=0π2cos θcos θ+sin θdθI=0π2cosθsinθ+cosθdθ    .....2

Adding 1 and 2, we get2I=0π2sinθ+cosθsinθ+cosθ dθ2I=0π2dθ2I=π2I=π40x1+x1+x2 dx=π4




 

Page No 19.120:

Question 18:

-π/2π/2sinx dx  is equal to

(a) 1
(b) 2
(c) − 1
(d) − 2

Answer:

(b) 2

-π2π2sinxdx=--π20sinx dx+0π2sinx  dx=--cosx-π20+-cosx0π2=1-0-0+1=2 

Page No 19.120:

Question 19:

0π/211+tan x dx  is equal to

(a) π4 

(b) π3 

(c) π2 

(d) π

Answer:

(a) π4 


Let, I=0π211+tanxdx     ... (i)       =0π211+tanπ2-xdx         = 0π211+cot xdx         ... (ii)Adding (i) and (ii) we get2I=0π211+tanx+11+cotx dx   =0π21+cotx+1+tan x1+tan x1+cotx dx   =0π22+ tan x+cot x1+tan x+cotx +tan x cot x dx   =0π22+ tan x+cot x2+ tan x+cot x dx   =0π2dx   =x0π2=π2Hence, I=π4

Page No 19.120:

Question 20:

The value of 0π/2cos x esin x dx is
(a) 1
(b) e − 1
(c) 0
(d) − 1  

Answer:

(b) e − 1


Let, I=0π2cosx esinx dxLet sinx =t, then cosx dx =dtWhen x=0, t=0 and x=π2, t=1Therefore the integral becomesI=01 et dt=et01=e-1

Page No 19.120:

Question 21:

If 0a11+4x2 dx=π8, then a equals

(a) π2

(b) 12

(c) π4

(d) 1

Answer:

(b) 12


0α11+4x2dx=π80α11+2x2dx=π812 tan-12x0α=π812tan-12α=π82α=tanπ42α=1 α=124430.

Page No 19.120:

Question 22:

If 01fx dx=1,01xfx dx=a,01x2fx dx=a2, then01a-x2 fx dx equals

(a) 4a2
(b) 0
(c) 2a2
(d) none of these

Answer:

(b) 0

01a-x2 fxdx=a201fxdx+01x2 fx dx-2a01x fxdx=a2×1+a2-2aa              As per given values=2a2-2a2=0

Page No 19.120:

Question 23:

The value of -ππsin3 x cos2 x dx is

(a) π42

(b) π44

(c) 0

(d) none of these

Answer:

(c) 0

-ππsin3x cos2 xdx=-ππsinx1-cos2x cos2x  dxLet cos x =t, then -sin x dx =dt,When, x=-π, t=-1, x=π,t=-1Therefore the integral becomes-1-1-1-t2t2  dt=0



Page No 19.121:

Question 24:

π/6π/31sin 2x dx is equal to

(a) loge 3

(b) loge3

(c) 12log-1

(d) log (−1)

Answer:

(b) loge3

π6π31sin2xdx=π6π3cosec2x  dx=12π6π32cosec2x  dx=-12logcosec2x+cot2xπ6π3=-12-2log3=log3

Page No 19.121:

Question 25:

-111-x dx is equal to

(a) −2
(b) 2
(c) 0
(d) 4

Answer:

(b) 2

-111-xdx=-101-x  dx+011-x  dx=x-x22-10+x-x2201=0+1+12+1-12-0=2

Page No 19.121:

Question 26:

The derivative of fx=x2x31loge t dt, x>0, is

(a) 13 ln x

(b) 13 ln x-12 ln x

(c) (ln x)−1 x (x − 1)

(d) 3x2ln x

Answer:

(c) (ln x)−1 x (x − 1)

Using Newton Leibnitz formula

f'(x)=1logex3(3x2)1logex2(2x)=3x23lnx2x2lnx=x2lnxxlnx=1lnxx(x1)=(lnx)1x(x1)

Page No 19.121:

Question 27:

If I10=0π/2x10 sin x dx, then the value of I10 + 90I8 is

(a) 9π29

(b) 10π29

(c) π29

(d) 9π28

Answer:

(b) 10π29We have,I10=0π/2x10 sin x dx=x10 -cos x0π2-0π/210 x9 sin x dxdx=-x10cos x0π2-100π/2 x9 -cos x dx=-x10 cos x0π2+100π/2 x9 cos x dx=-x10 cos x0π2+10x9 sin x0π2-100π/2 9x8 sin x dx=-π210 ×0-010 cos 0+10π29 ×1-09 ×0-900π/2 x8 sin x dx=10π29 ×1-90 I8=10π29-90 I8I10+90 I8=10π29

Page No 19.121:

Question 28:

01x1-x54 dx=

(a) 1516

(b) 316

(c) -316

(d) -163

Answer:

Disclaimer: The question given is not correct because the function provided does not converge in the given domain.

Page No 19.121:

Question 29:

0π21-sin2x dx is equal to

(a) 22

(b) 22+1

(c) 2

(b) 22-1

Answer:

Let I=0π21-sin2x dx      =0π2cos2x+sin2x-2sinxcosx dx        using the identity: cos2x+sin2x=1 and 2sinxcosx=sin2x      =0π2cosx-sinx2 dx      =0π2cosx-sinx dx      =0π4cosx-sinx dx+π4π2cosx-sinx dx      =0π4cosx-sinx dx+π4π2-cosx-sinx dx      =0π4cosx-sinx dx+π4π2sinx-cosx dx      =sinx+cosx0π4+-cosx-sinxπ4π2      =sinπ4+cosπ4-sin0-cos0+-cosπ2-sinπ2--cosπ4-sinπ4      =12+12-0-1+-0-1--12-12      =12+12-1+-1+12+12      =22-1+-1+22      =2-1+-1+2      =22-2      =22-1


Hence, the correct option is (d).

 

Page No 19.121:

Question 30:

The value of the integral -221-x2 dx is
(a) 4
(b) 2
(c) −2
(d) 0

Answer:

(a) 4

We have,I=-221-x2dx1-x2=-1-x2,      -2<x<-11-x2,      -1<x<1-1-x2,        1<x<2I=-2-11-x2dx+-111-x2dx+121-x2dx=-2-1-1-x2dx+-111-x2dx+12-1-x2dx=--2-11-x2dx+-111-x2dx-121-x2dx=-x-x33-2-1+x-x33-11-x-x3312=--1+13+2-83+1-13+1-13-2-83-1+13=-1-73+2-23-1-73=-1+73+2-23-1+73=4

Page No 19.121:

Question 31:

0π/211+cot3 x dx is equal to

(a) 0
(b) 1
(c) π/2
(d) π/4

Answer:

(d) π/4

We have, I=0π211+cot3xdx        .....1=0π211+cot3π2-xdx  I=0π211+tan3xdx        .....2Adding 1 and 2 we get2I=0π211+cot3x+11+tan3xdx

=0π21+tan3x+1+cot3x1+cot3x1+tan3x  dx=0π22+tan3x+cot3x1+tan3x+cot3x+cot3x tan3xdx=0π22+tan3x+cot3x1+tan3x+cot3x+1dx=0π22+tan3x+cot3x2+tan3x+cot3x dx=0π2[1]dx=x0π2=π2Hence I=π4

Page No 19.121:

Question 32:

0π/2sin xsin x+cos x dx equals to

(a) π
(b) π/2
(c) π/3
(d) π/4

Answer:

(d) π/4

We have, I=0π2sinxsinx+cosxdx            .....1I=0π2sinπ2-xsinπ2-x+cosπ2-xdxI=0π2cosxcosx+sinx dx I=0π2cosxsinx+cosx dx            .....2Adding 1 and 2, we get2I=0π2sinxsinx+cosx+cosxcosx+sinxdx=0π2sinx+cosxsinx+cosxdx  =0π2dx=x0π2=π2Hence I=π4

Page No 19.121:

Question 33:

01ddxsin-12x1+x2 dx is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

Answer:

(c) π/2

We have,I=01ddxsin-12x1+x2dxWe know since f'(x) = f(x)f(x) =sin-12x1+x2 and f'(x)=ddxsin-12x1+x2  Therefore, I=sin-12x1+x201=sin-11-sin-10=π2



Page No 19.122:

Question 34:

0π/2x sin x dx is equal to

(a) π/4
(b) π/2
(c) π
(d) 1

Answer:

(d) 1

We have, I=0π2x sinx dx =-x cosx0π2-0π21-cosx dx=-x cosx0π2+0π2cosx dx=-x cosx0π2+sinx0π2=-0-0+ 1-0=1

Page No 19.122:

Question 35:

0π/2sin 2x log tan x dx is equal to

(a) π
(b) π/2
(c) 0
(d) 2π

Answer:

(c) 0

I=0π2sin2x log tanx dx     .....1I=0π2sinπ-2x log tanπ2-x dxI=0π2sin2x log cotx dx         .....2Adding 1 and 2, we get,2I=0π2sin2x log tanx +log cotx dx2I=0π2sin2x log tanx cotx dx2I=0π2sin2x log1 dxI=0

Page No 19.122:

Question 36:

The value of 0π15+3 cos x dx  is

(a) π/4
(b) π/8
(c) π/2
(d) 0

Answer:

(a) π/4


0π15+3 cosxdx=0π15+3 1-tan2x21+tan2x2dx=0π1+tan2x25+5tan2x2+3-3tan2x2dx=0πsec2x28+2tan2x2dxLet tanx2=t, then sec2x2 dx=2dtWhen x=0, t=0, x=π, t=Therefore the integral becomes120dt4+t2=12tan-1t20=12π2-0=π4 

Page No 19.122:

Question 37:

0logx+1x 11+x2 dx=

(a) π ln 2
(b) −π ln 2
(c) 0
(d) -π2ln 2

Answer:

(a) π ln 2

0log x+1x 11+x2dx
Substitute x = tan θ
dx = sec2 θ dθ.
when,
x = 0  ⇒ θ = 0
x=θ=π20π2 tan θ+1tan θ11+tan2θ×sec2θ dθ0π2log tan2θ+1tanθ 11+tan2θ×sec2θdθ0π2log sec2θtan θ1sec2θ×sec2θdθ             1+tan2θ=sec2θ0π2log sec2θtan θdθ0π2log 1sin θ.cos θdθ-0π2log sin θ.cos θdθ-0π2 log sin θ+log cos θdθ-0π2log sin θdθ-0π2log cos θ dθ
Let us consider,
0π2log sin θdθ=I         .....(i)I=0π2log sin π2-θdθ=0π2log cos θdθ          .....iiAdding i and ii2I=0π2log sin θdθ+0π2log cos θdθ    =0π2log sin θ.cos θdθ    =0π2log sin 2θdθ-0π2log 2dθLet us consider 2θ=t2dθ=dt2I=120πlog sin tdt-π2log 22I=220π2log sin tdt-π2log 2             sin θ is positive in both 1st and 2nd quadrants2I=I-π2log 22I-I=-π2log 2I=-π2log 2, where I=0π2log sin θdθNow,-0π2logsin θdθ-0π2log cos θdθ-20π2log sin θdθ=-2×I=-2×-π2log 2                  where I=-π2log2=π log 2

Page No 19.122:

Question 38:

02afx dx is equal to

(a) 20afx dx

(b) 0

(c) 0afx dx+0af2a-x dx

(d) 0afx dx+02af2a-x dx

Answer:

c 0afx dx+0af2a-x dx

According to the additivity property of integrals,abf(x)dx=acf(x)dx+cbf(x)dx, where a<c<busing this property, 02af(x)dx=0af(x)dx +02af(x)dx   ......(1)Now, consider the integral, 02af(x)dxLet x=2a-t. Then dx=d(2a-t)dx=-dtAlso, x=at=a and x=2at=0Therefore, a2af(x)dx=-a0f(2a-t)dta2af(x)dx=0af(2a-t)dta2af(x)dx=0af(2a-x)dxSubstituting this in equation (1) we get,02af(x)dx=0af(x)dx +0af(2a-x)dx


Page No 19.122:

Question 39:

If f (a + bx) = f (x), then ab x f (x) dx is equal to

(a) a+b2 ab fb-x dx

(b) a+b2 ab fb+x dx

(c) b-a2 ab fx dx

(d) a+b2 ab fx dx

Answer:

(d) a+b2 ab fx dx


Let, I=abx fxdx           ...(i)       =aba+b-x fa+b-xdx      =aba+b-x fx  dx          ...(ii)   Adding (i) and (ii)2I=abx+a+b-x fxdx   =a+bab fxdx Hence I=a+b2ab fxdx 

Page No 19.122:

Question 40:

The value of 01tan-12x-11+x-x2 dx, is

(a) 1
(b) 0
(c) −1
(d) π/4

Answer:

(b) 0

Let, I=01tan-12x-11+x-x2dx         ... (i)=01tan-121-x-11+1-x-1-x2dx=01tan-11-2x2-x-1-x2+2x dx=01tan-11-2x1+x-x2 dx=-01tan-12x-11+x-x2 dx            ... (ii)Adding (i) and (ii)2I=01tan-12x-11+x-x2 dx-01tan-12x-11+x-x2 dx   =0Hence, I=0

Page No 19.122:

Question 41:

The value of 0π/2log4+3 sin x4+3 cos x dx is

(a) 2

(b) 34

(c) 0

(d) −2

Answer:

(c) 0

Let I=0π2log4+3sinx4+3cosxdx         ...(i)      =0π2log4+3sinπ2-x4+3cosπ2-x  dx      =0π2log4+3 cosx4+3sinxdx        ...(ii)Adding (i) and (ii)2I=0π2log4+3sinx4+3cosx+log4+3 cosx4+3sinx  dx    =   0π2log4+3sinx4+3cosx×4+3 cosx4+3sinx  dx    =   0π2log1 dx=0Hence I=0 

Page No 19.122:

Question 42:

The value of -π/2π/2x3+x cos x+tan5 x+1 dx, is

(a) 0
(b) 2
(c) π
(d) 1

Answer:

(c) π

-π2π2x3+xcosx+tan5x+1dx=x44-π2π2+x sinx-π2π2--π2π2sinx dx+-π2π2tan3x sec2x-1dx+x -π2π2=π464-π464+π2-π2--cosx-π2π2+-π2π2tan3x sec2x dx--π2π2tan3x dx+π2+π2=π+0+tan4x4-π2π2--π2π2tanx sec2x dx--π2π2tanx dx=π-tan2x2-π2π2--logcosx-π2π2=π



Page No 19.123:

Question 43:

-π4π4 11+cos 2xdx is equal to

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

Let I=-π4π4 11+cos2xdxLet fx=11+cos2xf-x=11+cos-2x        =11+cos2x        =fxWe know,-aafxdx=0                          ,if f-x=-fx20afxdx              ,if f-x=fxThus,I=20π4 11+cos2xdx  =20π4 12cos2xdx           using the identity: cos2x=2cos2x-1 =0π4 1cos2xdx =0π4 sec2x dx =tanx0π4 =tanπ4-tan0 =1-0 =1


Hence, the correct option is (a).
 

Page No 19.123:

Question 44:

a+cb+cfx dx is equal to

(a) abfx-c dx

(b) abfx+c dx

(c) abfx dx

(d) a-cb-cfx dx

Answer:

Let I=a+cb+cfx dxLet x=t+cdx=dtAlso, if x=a+c, t=aif x=b+c, t=bThus, I=abft+c dt  =abfx+c dx       abfx dx=abft dt


Hence, the correct option is (b).

Page No 19.123:

Question 45:

If f and g are continuous functions in [0, 1] satisfying f(x) = f(ax) and g(x) = g(ax) = a, then 0afx gx dx is equal to

(a) a2

(b) a20afx dx

(c) 0afx dx

(d) a0bfx dx

Answer:

Given: f(x) = f(a – x) and g(x) + g(a – x) = a

Let I=0afxgx dx       ...1I=0afa-xga-x dx          0afx dx=0afa-x dxI=0afxa-gx dx              fx=fa-x and ga-x=a-gxI=a0afx dx-0afxgx dxI=a0afx dx-I                     From 12I=a0afx dxI=a20afx dx


Hence, the correct option is (b).

Page No 19.123:

Question 1:

If 0a11+4x2dx=π8, then a = _______________.

Answer:

Given: 0a11+4x2dx=π8          ...1Let I=0a11+4x2dxI=140a114+x2dxI=140a1122+x2dxI=14112tan-1x120a              1a2+x2dx=1atan-1xa+cI=142tan-12x0aI=142tan-12a-2tan-100aI=12tan-12aSince, it is given that 0a11+4x2dx=π8Therefore, 12tan-12a=π8tan-12a=π42a=tanπ42a=1a=12


Hence, a = 12.

Page No 19.123:

Question 2:

The value of -ππsin3x cos2 x dx is _______________.

Answer:

Let I=-ππsin3x cos2x dxLet,fx=sin3x cos2xf-x=sin3-x cos2-x        =-sin3x cos2x        =-fxWe know,-aafxdx=0                          ,if f-x=-fx20afxdx              ,if f-x=fxThus,-ππsin3x cos2x dx=0


Hence, the value of -ππsin3x cos2x dx is 0.

Page No 19.123:

Question 3:

The value of 0π2esin x cos x dx is _______________.

Answer:

Let I=0π2esinxcosx dxLet,sinx=tdsinx=dtcosx dx=dtAlso, if x=0, t=0if x=π2, t=1Thus,I=01et dt =et01 =e1-e0 =e-1


Hence, the value of 0π2esinx cosx dx is e-1.

Page No 19.123:

Question 4:

0axa2+x2dx= _______________.

Answer:

Let I=0axa2+x2dxLet,a2+x2=tda2+x2=dt2x dx=dtAlso, if x=0, t=a2if x=a, t=2a2Thus,I=12a22a21t dt =12t-12+1-12+1a22a2 =12t1212a22a2 =122ta22a2 =1222a2-2a2 =122a2-2a =a2-a =a2-1


Hence, 0axa2+x2dx=a2-1.

Page No 19.123:

Question 5:

The value of the integral 1π2πsin 1πx2dx is _______________.

Answer:

Let I=1π2πsin1xx2dxLet,1x=td1x=dt-1x2 dx=dtAlso, if x=1π, t=πif x=2π, t=π2Thus,I=ππ2-sint dt =costππ2 =cosπ2-cosπ =0--1 =1


Hence, the value of the integral 1π2πsin1xx2dx is 1.

Page No 19.123:

Question 6:

The value of the integral 01tan-1 x1+x2dx is _______________.

Answer:

Let I=01tan-1x1+x2dxLet,tan-1x=tdtan-1x=dt11+x2 dx=dtAlso, if x=0, t=0if x=1, t=π4Thus,I=0π4t dt =t220π4 =12π42-02 =12π216 =π232


Hence, the value of the integral 01tan-1x1+x2dx is π232.

Page No 19.123:

Question 7:

The value of the integral 12ex1x-1x2dx is _______________.

Answer:

Let I=12ex1x-1x2dxWe know,exfx+f'xdx=exfx+cHere,fx=1xf'x=-1x2Thus,I=ex1x12 =e212-e111 =e22-e =e2-2e2 =ee-22


Hence, the value of the integral 12ex1x-1x2dx is ee-22.



Page No 19.124:

Question 8:

1eexx1+log x dx=________________.

Answer:

Let I=1eexx1+xlogx dx      =1eex1x+logx dx      =1eexlogx+1x dxWe know,exfx+f'xdx=exfx+cHere,fx=logxf'x=1xThus,I=exlogx1e =eeloge-e1log1 =ee1-e0 =ee


Hence, 1eexx1+xlogx dx=ee.

Page No 19.124:

Question 9:

The value of the integral -22ax5+bx3+cx+d dx, where a, b, c, d are constants, depends only on ________________.

Answer:

Let I=-22ax5+bx3+cx+d dx       =-22ax5+bx3+cx dx+-22d dxLet,fx=ax5+bx3+cxf-x=a-x5+b-x3+c-x        =-ax5-bx3-cx        =-ax5+bx3+cx        =-fxWe know,-aafxdx=0                          ,if f-x=-fx20afxdx              ,if f-x=fxThus,I=-22ax5+bx3+cx dx+-22d dx =0+-22d dx =dx-22 =2d--2d =2d+2d =4d

​Hence, the value of the integral -22ax5+bx3+cx+d dx, where abcd are constants, depends only on d.

Page No 19.124:

Question 10:

0π11+sin x dx=________________.

Answer:

Let I=0π11+sinx dx       =0π11+sinx×1-sinx1-sinxdx       =0π1-sinx1-sin2xdx       =0π1-sinxcos2xdx       =0π1cos2x-sinxcos2xdx       =0πsec2x-secxtanxdx       =tanx-secx0π       =tanπ-secπ-tan0-sec0       =0--1-0-1       =1--1       =1+1       =2

​Hence, 0π11+sinx dx=2.

Page No 19.124:

Question 11:

0π2sin x cos x1+sin4xdx=________________.

Answer:

Let I=0π2sinx cosx1+sin4xdxLet sin2x=tdsin2x=dt2sinxcosx dx=dtAlso, if x=0, t=0if x=π2, t=1Thus,I=0111+t2dt2 =120111+t2dt =12tan-1t01 =12tan-11-tan-10 =12π4-0 =π8

​Hence, 0π2sinx cosx1+sin4xdx=π8.

Page No 19.124:

Question 12:

0π4tan6x sec2x dx=________________.

Answer:

Let I=0π4tan6x sec2x dxLet tanx=tdtanx=dtsec2x dx=dtAlso, if x=0, t=0if x=π4, t=1Thus,I=01t6 dt =t7701 =177-07 =17

​Hence, 0π4tan6x sec2x dx=17.

Page No 19.124:

Question 13:

The value of 0π41+tan x1-tan xdx is ________________.

Answer:

Let I=0π41+tanx1-tanxdx       =0π4tanπ4+tanx1-tanπ4 tanxdx       =0π4tanπ4+x dx       =-log cosπ4+x0π4       =-log cosπ4+π4+log cosπ4+0       =-log cos2π4+log cosπ4       =-log cosπ2+log 12       =-log 0+log 2-12       =-12log 2

​Hence, the value of 0π41+tanx1-tanxdx is -12log 2.

Page No 19.124:

Question 14:

The value of 023xxdx is ________________.

Answer:

Let I=023xxdxLet x=tdx=dt12x dx=dtAlso, if x=0, t=0if x=2, t=2Thus,I=023t 2dt =2023t dt =23tlog302 =232log3-30log3 =232-1log3

​Hence, the value of 023xxdx is 232-1log3.

Page No 19.124:

Question 15:

The value of 0π2sin x1+cos2xdx is ________________.

Answer:

Let I=0π2sinx1+cos2xdxLet cosx=tdcosx=dt-sinx dx=dtAlso, if x=0, t=1if x=π2, t=0Thus,I=10-11+t2dt =-tan-1t10 =-tan-10-tan-11 =-0-π4 =π4

​Hence, the value of 0π2sinx1+cos2xdx is π4.

Page No 19.124:

Question 16:

If f(ax) = x and 0afxdx=k0a2fxdx, then k = _____________.

Answer:

0afxdx=k0a2fxdx

Given:
f(x) = f(a – x)                ...(1)
0afxdx=k0a2fxdx        ...(2)

Let I=0afxdx        ...3We know,02afx dx=20afx dx        ,if f2a-x=fx0                     ,if f2a-x=-fxHere, fa-x=fx     From 1Thus,I=20a2fxdx         ...4From 2 and 4,k0a2fxdx=20a2fxdxk=2


​Hence, k = 2.

Page No 19.124:

Question 17:

The value of the integral 02πcos7x sin4x dx is ________________.

Answer:

Let I=02πcos7x sin4x dxLet fx=cos7x sin4xf2π-x=cos2π-x7sin2π-x4             =cosx7-sinx4             =cos7x sin4x             =fxWe know,02afx dx=20afx dx            ,if f2a-x=fx0                         ,if f2a-x=-fxThus, I=20πcos7x sin4x dxAgain,Let fx=cos7x sin4xfπ-x=cosπ-x7sinπ-x4             =-cosx7sinx4             =-cos7x sin4x             =-fxThus,I=0

​Hence, the value of the integral 02πcos7x sin4x dx is 0.

Page No 19.124:

Question 18:

The value of the integral -11xx dx is ________________.

Answer:

Let I=-11xx dxLet fx=xxf-x=-x-x        =-xx        =-fxWe know,-aafx dx=20afx dx            ,if f-x=fx0                         ,if f-x=-fxThus, I=0

​Hence, the value of the integral -11xx dx is 0.

Page No 19.124:

Question 19:

The value of the integral -11log 2-x2+x dx is ________________.

Answer:

Let I=-11log 2-x2+x dxLet fx=log2-x2+xf-x=log2--x2+-x        =log2+x2-x        =log2-x2+x-1        =-log2-x2+x        =-fxWe know,-aafx dx=20afx dx            ,if f-x=fx0                         ,if f-x=-fxThus, I=0

​Hence, the value of the integral -11log 2-x2+x dx is 0.

Page No 19.124:

Question 20:

The value of the integral -111-x dx is ________________.

Answer:

Let I=-111-x dxWe know,1-x=1-x              ,if 1-x0-1-x           ,if 1-x<0         =1-x              ,if x1-1-x           ,if x>1Thus, I=-111-x dx  =x-x22-11  =1-122--1--122  =1-12--1-12  =2-12--2-12  =12--32  =12+32  =42  =2

​Hence, the value of the integral -111-x dx is 2.

Page No 19.124:

Question 21:

The value of the integral 0π211+tan3x dx is ________________.

Answer:

Let I=0π211+tan3x dx      =0π211+sin3xcos3x dx      =0π21cos3x+sin3xcos3x dx      =0π2cos3xcos3x+sin3x dx                     ...1We know,0afx dx=0afa-x dxThus, I=0π2cos3π2-xcos3π2-x+sin3π2-x dx  =0π2sin3xsin3x+cos3x dx                  ...2Adding 1 and 2, we get2I=0π2cos3xcos3x+sin3x dx+0π2sin3xsin3x+cos3x dx  =0π2cos3x+sin3xsin3x+cos3x dx   =0π21 dx   =x0π2  =π2I=π4

​Hence, the value of the integral 0π211+tan3x dx is π4.

Page No 19.124:

Question 22:

If -aaa-xa+x dx=kπ, then k = ________________.

Answer:

Given: -aaa-xa+x dx=kπ                  ...1Let I=-aaa-xa+x dxWe know,-aafx dx=0afx+f-x dxThus,I=0aa-xa+x+a--xa+-x dx =0aa-xa+x+a+xa-x dx =0aa-xa+x+a+xa-x dx   =0aa-x2+a+x2a+xa-x dx  =0aa-x+a+xa2-x2 dx  =0a2aa2-x2 dx =2a0a1a2-x2 dx =2asin-1xa0a =2asin-1aa-sin-10 =2asin-11-0 =2aπ2 =aπ           ...2From 1 and 2,aπ=kπk=a

​Hence, k = a.

Page No 19.124:

Question 23:

If f(x) = f(ax) and 0ax fx dx=k 0afx dx, then k = ________________.

Answer:

Given:
f(x) = f(a – x)                ...(1)
0axfx dx=k0afx dx        ...(2)

Let I=0axfx dx        ...3We know,0afx dx=0afa-x dxThus,I=0aa-xfa-x dxI=0aa-xfx dx             From 1I=0aafxdx-0axfx dx I =0aafxdx-I                From 32I =0aafxdx I =120aafxdxI =a20afxdx         ...4From 2 and 4,k0afxdx=a20afxdxk=a2

​Hence, k = a2.



Page No 19.125:

Question 24:

The value of the integral 010x1010-x10+x10 dx is ________________.

Answer:

Let I=010x1010-x10+x10 dx                     ...1We know,0afx dx=0afa-x dxThus, I=01010-x1010-10-x10+10-x10 dx   =01010-x1010-10+x10+10-x10 dx  =01010-x10x10+10-x10 dx                  ...2Adding 1 and 2, we get2I=010x1010-x10+x10 dx+01010-x10x10+10-x10 dx  =010x10+10-x10x10+10-x10 dx  =0101 dx   =x010  =10I=5

​Hence, the value of the integral 010x1010-x10+x10 dx is 5.

Page No 19.125:

Question 25:

 -11ex dx= ________________.

Answer:

Let I=-11ex dxLet fx=exf-x=e-x        =ex        =fxWe know,-aafx dx=20afx dx            ,if f-x=fx0                         ,if f-x=-fxThus, I=201ex dx  =201ex dx  =2ex01  =2e1-e0  =2e-1


​Hence,  -11ex dx=2e-1.

Page No 19.125:

Question 1:

0π/2sin2 x dx.

Answer:

0π2sin2x dx=0π21-cos2x2 dx=120π21-cos2x dx=12x-sin2x20π2=12π2-0=π4

Page No 19.125:

Question 2:

0π/2cos2 x dx.

Answer:

0π2cos2x dx=0π21+cos2x2 dx=120π21+cos2x dx=12x+sin2x20π2=12π2+0=π4

Page No 19.125:

Question 3:

-π/2π/2sin2 x dx.

Answer:

-π2π2sin2x dx=-π2π21-cos2x2 dx=12-π2π21-cos2xdx=12x-sin2x2-π2π2=12π2-0+π2-0=π2

Page No 19.125:

Question 4:

-π/2π/2cos2 x dx.

Answer:

-π2π2cos2 xdx=-π2π21+cos2x2 dx=12-π2π21+cos2x  dx=12x+sin2x2-π2π2=12π2+0+π2-0=π2

Page No 19.125:

Question 5:

-π/2π/2sin3 x dx.

Answer:

Let I=-π2π2sin3xdx=-π2π2sinx sin2x  dx=-π2π2sinx1-cos2x dxLet cosx =t, then -sinx dx =dt,When, x-π2 ; t0 and xπ2 ; t0I=00-1+t2  dt=0

Page No 19.125:

Question 6:

-π/2π/2x cos2 x dx.

Answer:

We have,I=-π2π2x cos2x dxLet fx=x cos2x f-x=-x cos2-x=-x cos2xf-x=-fxi.e., fx is odd functionWe know that -aafx dx=0 , if fx is odd function.I=-π2π2x cos2x dx=0

Page No 19.125:

Question 7:

0π/4tan2 x dx.

Answer:


0π4tan2x dx=0π4sec2x-1 dx=tanx-x0π4=1-π4-0=1-π44430.

Page No 19.125:

Question 8:

011x2+1 dx.

Answer:

0111+x2dx=tan-1x01=π4-0=π4ode is 4430.

Page No 19.125:

Question 9:

-21xx dx.

Answer:

Let, I=-21xxdxWe have,x=x                0x1-x          -2x<0xx=1                0x1-1          -2x<0Therefore,I=-20-1dx+01  1  dx =-x-20+x01 =0-2+1-0 =-1

Page No 19.125:

Question 10:

0e-x dx.

Answer:

0e-xdx=-e-x0=-0-1=0+1=1 4430.

Page No 19.125:

Question 11:

04116-x2 dx.

Answer:

04116-x2dx=04142-x2dx=sin-1x404=π2-0=π2s 4430.

Page No 19.125:

Question 12:

031x2+9 dx.

Answer:

031x2+9dx=031x2+32dx =13 tan-1x303=13tan-11- tan-10=13π4- 0=π12

Page No 19.125:

Question 13:

0π/21-cos 2x dx.

Answer:

0π21-cos2xdx=0π22sin2x  dx =0π22 sinx  dx=-2 cosx0π2=-0-2=2 

Page No 19.125:

Question 14:

0π/2log tan x dx.

Answer:

Let, I=0π2log tanxdx         ... (i)=0π2log tanπ2-x dx              Using, 0a fx dx=0a fa-x dx=0π2log cotx dx                ... (ii)Adding (i) and (ii) we get2I=0π2log tanx dx+0π2log cotx dx   =0π2logtanx×cotxdx   =0π2log1 dx=0Hence, I=0

Page No 19.125:

Question 15:

0π/2log 3+5 cos x3+5 sin x dx.

Answer:

Let, I=0π2log3+5cosx3+5sinxdx                   ... (i)=0π2log3+5cosπ2-x3+5sinπ2-x  dx=0π2log3+5sinx3+5cosx  dx                      ... (ii)Adding (i) and (ii)2I=0π2log3+5cosx3+5sinx+log3+5sinx3+5cosx  dx    =0π2log3+5cosx3+5sinx×3+5sinx3+5cosx  dx     =0π2log1  dx=0Hence I=0

Page No 19.125:

Question 16:

0π/2sinn xsinn x+cosn x dx, nN.

Answer:

Let I=0π2sinnxsinnx+cosnxdx                ...(i)=0π2sinnπ2-xsinnπ2-x+cosnπ2-xdx=0π2cosnxcosnx+sinnxdx=0π2cosnxsinnx+cosnxdx                    ...(ii)Adding (i) and (ii)2I=0π2sinnxsinnx+cosnx+cosnxsinnx+cosnxdx  = 0π2sinnx+cosnxsinnx+cosnx  dx=0π2dx=x0π2=π2Hence I=π4



Page No 19.126:

Question 17:

0πcos5 x dx.

Answer:

Let I=0πcos5x dx       =0πcosxcos2x2 dx       =0πcosx1-sin2x2 dx  Let sinx =t, then cosx dx = dtWhen, x0 ; t0 and xπ ; t0 Therefore,I=001-t22 dt       =0

Page No 19.126:

Question 18:

-π/2π/2loga-sin θa+sin θ dθ

Answer:

Let, I=-π2π2loga-sinθa+sinθdθHere, fθ=loga-sinθa+sinθConsider, f-θ=loga-sin-θa+sin-θ=-loga-sinθa+sinθ=-fθi.e., fθ is odd function.Therefore, I=0

Page No 19.126:

Question 19:

-11xx dx.

Answer:


x = -x , -1<x<0  x , 0<x<1xx = -x2 , -1<x<0  x2 , 0<x<1Now, -11xxdx=-10- x2 dx+01 x2 dx=--10 x2 dx+01 x2 dx=-x33-10+x3301=-0+13+13-0=0-13+13-0=0

Page No 19.126:

Question 20:

abfxfx+fa+b-x dx.

Answer:

Let I=abfxfx+fa+b-xdx           ... (i)      =abfa+b-xfa+b-x+fa+b-a-b+xdx      =abfa+b-xfa+b-x+fxdx I=abfa+b-xfx+fa+b-xdx            ...(ii)Adding (i) and (ii) we get2I=abfxfx+fa+b-x+fa+b-xfx+fa+b-xdx   =abfx+fa+b-xfx+fa+b-x dx   =xab   =b-aHence, I=b-a2

Page No 19.126:

Question 21:

0111+x2 dx

Answer:

0111+x2dx=tan-1x01=tan-11-tan-10=π4-0=π4

Page No 19.126:

Question 22:

Evaluate each of the following integrals:

0π4tanxdx

Answer:


0π4tanxdx=logsecx0π4=logsecπ4-logsec0=log2-log1=log212-0=12log2

Page No 19.126:

Question 23:

231xdx

Answer:

231xdx=logex23=loge3-loge2=loge32

Page No 19.126:

Question 24:

024-x2 dx

Answer:

024-x2dx=0222-x2dx=x24-x2+12×22sin-1x202=x24-x202+2sin-1x202=0+2π2-0=π

Page No 19.126:

Question 25:

012x1+x2 dx

Answer:

We have,I=012x1+x2dxPutting 1+x2=t2x dx=dtWhen x0; t1And x1; t2I=12dtt=loge t12=loge2-loge1=loge2-0=loge2 4430.

Page No 19.126:

Question 26:

Evaluate each of the following  integrals:

01xex2dx                 [CBSE 2014]

Answer:


I=01xex2dx=1201ex22xdx

Put x2=z

2xdx=dz

When x0, z0

When x1, z1

I=1201ezdz=12×ez01=12e-e0=12e-1

Page No 19.126:

Question 27:

Evaluate each of the following integrals:

0π4sin2xdx              [CBSE 2014]

Answer:


0π4sin2xdx=-cos2x20π4=-12cosπ2-cos0=-12×0-1=12

Page No 19.126:

Question 28:

Evaluate each of the following integrals:

ee21xlogxdx               [CBSE 2014]

Answer:


ee21xlogxdx=ee21xlogxdx=loglogxee2                                           f'xfxdx=logfx+C=logloge2-logloge=log2loge-logloge  =log2-log1                                            loge=1=log2-0=log2                

Page No 19.126:

Question 29:

Evaluate each of the following integrals:

0π2exsinx-cosxdx                 [CBSE 2014]

Answer:


Disclaimer: The solution has been provided by taking the lower limit of integral as 0.

0π2exsinx-cosxdx=-0π2excosx+-sinxdx=-excosx0π2                                   exfx+f'xdx=exfx+C=-eπ2cosπ2-e0cos0=-eπ2×0-1×1=-0-1=1

Page No 19.126:

Question 30:

Solve each of the following integrals:

24xx2+1dx                 [CBSE 2014]

Answer:


24xx2+1dx=12242xx2+1dx=12×logx2+124                     f'xfxdx=logfx+C=12log17-log5=12log175                             loga-logb=logab

Page No 19.126:

Question 31:

If 013x2+2x+k dx=0, find the value of k.

Answer:

We have,013x2+2x+kdx=0x3+x2+kx01=01+1+k-0=0k=-2.

Page No 19.126:

Question 32:

If 0a3x2 dx=8, write the value of a.

Answer:

We have,0a3x2dx=83x330a=8x30a=8a3-0=8a=83      =20.

Page No 19.126:

Question 33:

If fx=0xtsintdt, the write the value of f'x.                       [CBSE 2014]

Answer:


fx=0xtsintdtfx=t-cost0x-0xddtt×-costdtfx=-xcosx-0+0xcostdtfx=-xcosx+sint0x
fx=-xcosx+sinx-0fx=-xcosx+sinx

Differentiating both sides with respect to x, we get

f'x=-x×-sinx+cosx×1+cosxf'x=--xsinx-cosx+cosxf'x=xsinx

Thus, the value of f'x is x sinx.

Page No 19.126:

Question 34:

If 0a14+x2dx=π8, find the value of a.                                   [CBSE 2014]

Answer:


0a14+x2dx=π812tan-1x20a=π8                           1a2+x2dx=1atan-1xa+C12tan-1a2-tan-10=π8tan-1a2-0=π4
tan-1a2=π4a2=tanπ4=1a=2

Thus, the value of a is 2.

Page No 19.126:

Question 35:

Write the coefficient a, b, c of which the value of the integral -33ax2+bx+c dx is independent.

Answer:

-33ax2+bx+cdx=ax33+bx22+cx-33=9a+92b+3c+9a-92b+3c=18a+6c 

Hence, the given integral is independent of b

Page No 19.126:

Question 36:

Evaluate : 233x dx.

Answer:

I =233x =3xlog323 +C                 (Use:ax =axloga+C)=33log3-32log3+C

=1log3(33-32)+C=1log3(27-9)+C=1log3(18)+C

Page No 19.126:

Question 37:

02x dx. 

Answer:

We have,I=02xdxWe know that,x=0,        0<x<11,        1<x<2I=02xdx=01xdx+12xdx=010dx+121dx=0+x12=2-1=1

Page No 19.126:

Question 38:

015x dx.

Answer:

We have,I=01.5x dx=01x dx+11.5x dx=010 dx+11.51dx        x=0        0x<11       1x<1.5=0+x11.5=1.5-1=0.5=12

Page No 19.126:

Question 39:

01x dx, where {x} denotes the fractional part of x.  

Answer:

We have,I=01x dxWe know x=x,     0<x<1I=01x dx=x2201=12-02=12
s 4430.

Page No 19.126:

Question 40:

01ex dx.

Answer:

We have,I=01exdxWe know that,x=x,      when 0<x<1I=01exdx=ex01=e1-e0=e-1



Page No 19.127:

Question 41:

02xx dx.

Answer:

We have,I=02xx dxWe know that,xx=x×0,        0<x<1x×1,        1<x<2i.e.,xx=0,        0<x<1x,        1<x<2I=02xx dx=01xx dx+12xx dx=010 dx+12x dx=0+x2212=222-122=42-12=32

Page No 19.127:

Question 42:

012x-x dx

Answer:

We have,I=012x-x dx=012x-0 dx          x=0   where, 0<x<1=012x dx=2xloge201=21loge2-20loge2=2loge2-1loge2=1loge2

Page No 19.127:

Question 43:

12loge x dx.

Answer:

We have,I=12loge x dxWe know that,x=1,    when 1<x<2I=12loge 1 dxI=120 dx=0

Page No 19.127:

Question 44:

02x2 dx.

Answer:

We have,I=02x2 dx=01x2 dx+12x2 dx=010dx+121dx         x2=0        0< x<11         1<x<2=0+x12=2-1

Page No 19.127:

Question 45:

If · and · denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
0π/4sin x dx

Answer:

We have,I=0π/4sin x dxWe know that,x=x, when 0<x<π4       As π=3.14  π4=0.785<1I=0π/4sin x dx=-cos x0π4=-cos π4-cos 0=cos 0-cos π4=1-12=2-12



Page No 19.16:

Question 1:

491x dx

Answer:

Let I=491x dx. Then,I=24912x dxI=2x49I=23-2I=2

Page No 19.16:

Question 2:

-231x+7 dx

Answer:

Let I=-231x+7 dx. Then,I=log x+7-23I=log 10-log 5I=log 105                         log a- log b=logabI=log 2

Page No 19.16:

Question 3:

01/211-x2 dx

Answer:

Let I=01211-x2 dx. Then,I=sin-1x012I=sin-112-sin-10I=π6-0I=π6

Page No 19.16:

Question 4:

0111+x2 dx

Answer:

Let I=0111+x2dx. Then,I=tan-1x01I=tan-11-tan-10I=π4-0I=π4

Page No 19.16:

Question 5:

23xx2+1 dx

Answer:

Let I=23xx2+1dx. Then,I=12232xx2+1I=12log x2+123I=12log 10-log 5I=12log 105                  log a-log b=log abI=12log 2

Page No 19.16:

Question 6:

01a2+b2 x2 dx

Answer:

Let I=01a2+b2x2 dx. Then,I=1a2011+b2x2a2 dxI=1a2011+bxa2 dxI=aba2tan-1bxa0I=1abtan-1-tan-10I=π2ab

Page No 19.16:

Question 7:

-1111+x2 dx

Answer:

Let I=-1111+x2dx. Then,I=tan-1x-11I=tan-11-tan-1-1I=π4--π4I=π2

Page No 19.16:

Question 8:

0e-x dx

Answer:

Let I=0e-x dx. Then,I=-e-x0I=-e-+e0I=0+1I=1

Page No 19.16:

Question 9:

01xx+1 dx

Answer:

Let I=01xx+1 dx. Then,I=011-1x+1 dxI=x-log x+101I=1-log 2-(0-log 1)I=log e-log 2I=log e2

Page No 19.16:

Question 10:

0π/2sin x+cos x dx

Answer:

Let I=0π2sin x+cos x dx. Then,I=-cos x+sin x0π2I=0+1--1+0I=2

Page No 19.16:

Question 11:

π/4π/2cot x dx

Answer:

Let I=π4π2cot x dx. Then,I=-π4π2cot x-(cosec x+cot x)cosec x+cot x dxI=-π4π2-cosec x cot x-cot2 xcosec x+cot x dxI=-π4π2-cosec x cot x-cosec2x+1 cosec x+cot x dx          cosec2x=1+cot2xI=-π4π2-cosec x cot x-cosec2x cosec x+cot x dx-π4π21cosec x+cot xdxI=-π4π2-cosec x cot x-cosec2x cosec x+cot x dx-π4π2sin x1+cos xdxI=-log cosec x+cot x</