RD Sharma XII Vol 2 2020 Solutions for Class 12 Humanities 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 Humanities 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 Humanities 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 12 Humanities 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)−1x (x − 1)

(d) 3x2ln x

Answer:

(c) (ln x)−1x (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 abx 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 xπ4π2+log 1+cos xπ4π2I=-log 1++log 2+1+log 1+0-log 1+12I=log 2+1-log 2+12I=log 22+12+1I=log2I=12log 2

Page No 19.16:

Question 12:

0π/4sec x dx

Answer:

Let I=0π4sec x dx. Then,I=0π4sec x sec x+tan xsec x+tan xdxI=0π4sec2 x+sec x tan xsec x+tan xdxPut u=sec x+tan xdu=sec2 x+sec x tan x dx0π4sec2 x+sec x tan xsec x+tan x dx=duuI=  log uI=log sec x+tan x0π4I=log secπ4+tanπ4-log sec 0+tan 0I=log (2+1)-log 1I=log (2+1)

Page No 19.16:

Question 13:

π/6π/4cosec x dx

Answer:

Let I=π6π4cosec x dx. Then,I=π6π4cosec x cosec x-cot xcosec x-cot x dxI=π6π4cosec2 x-cosec x cot xcosec x-cot x dxI=log cosec x-cot xπ6π4I=log 2-1-log2-3

Page No 19.16:

Question 14:

011-x1+x dx

Answer:

Let I=011-x1+x dx. Then,I=0111+x-1+x-11+x dxI=0111+x-1+x1+x dxI=log 1+x-x+log 1+x01I=log 2-1+log 2-log 1-0+log 1=2 log 2-1

Page No 19.16:

Question 15:

0π11+sin x dx

Answer:

Let I=0π11+sin x dx. Then, I=0π1-sin x1+sin x1-sin x dxI=0π1-sin x1-sin2 x dx            I=0π1-sin xcos2 x dx                             sin2 x+cos2 x=1I=0πsec2 x-sec x tan x dxI=tan x-sec x0πI=tan π-sec π-tan 0-sec 0I=0+1-0-1I=1+1I=2

Page No 19.16:

Question 16:

-π/4π/411+sin x dx

Answer:

Let I=-π4π411+sin x dx. Then,I=-π4π411+sin x×1-sin x1-sin x dxI=-π4π41-sin x1-sin2 x dxI=-π4π41-sin xcos2 x dxI=-π4π41cos2 x-sin xcos2 x dxI=-π4π4sec2 x-sec x tan x dxI=tan x-sec x-π4π4I=1-2--1-2I=2

Page No 19.16:

Question 17:

0π/2cos2 x dx

Answer:

Let I=0π2cos2 x dx. Then,I=0π2cos2 x dxI=120π21+cos 2x dx           cos 2x=2 cos2 x-1I=x2+sin 2x40π2I=π4+0-0I=π4

Page No 19.16:

Question 18:

0π/2cos3 x dx

Answer:

Let I=0π2cos3 x dx. Then,I=0π2cos2 x cos x dxI=0π21-sin2 x cos x dxLet u= sin x, du= cos x dxI=1-u2 duI=u-u33I=sin x-sin3 x30π2I=1-13-0I=23

Page No 19.16:

Question 19:

0π/6cos x cos 2x dx

Answer:

Let I=0π6cos x cos 2x dx. Then,I=0π6cos x cos2 x-sin2 x dxI=0π62 cos3 x-cos x dxI=0π62 cos x 1-sin2 x - cos x dxI=2sin x-sin3 x3-sin x0π6I=212-124-12-0I=512

Page No 19.16:

Question 20:

0π/2sin x sin 2x dx

Answer:

Let I=0π2sin x sin 2x dx. Then,I=0π22 sin2x cos x dxI=0π221-cos2 x cos x dxI=0π22 cos x -2 cos3 x dxI=2sinx-2sin x-sin3 x30π2I=2-21-13-0I=23

Page No 19.16:

Question 21:

π/3π/4tan x+cot x2 dx

Answer:

Let I=π3π4tan x+cot x2 dx. Then,I=π3π4tan2 x+cot2 x+2 tan x cot x dxI=π3π4tan2 x+cot2 x+2 dxI=π3π4sec2 x-1+cosec2 x-1+2 dxI=π3π4sec2 x+cosec2 x dxI=tan x-cot xπ3π4I=1-1-3-13I=-23

Page No 19.16:

Question 22:

0π/2cos4 x dx

Answer:

Let I=0π2cos4 x dx. Then,I=0π2cos2 x2 dxI=0π21+cos 2x24 dxI=140π21+cos2 2x+2 cos 2x dxI=140π21+2 cos 2x+1+cos 4x2 dxI=140π23+4 cos 2x+cos 4x2 dxI=143x2+2 sin 2x2+sin 4x80π2I=143π4+0I=3π16

Page No 19.16:

Question 23:

0π/2a2 cos2 x+b2 sin2 x dx

Answer:

Let I=0π2a2 cos2 x+b2 sin2 x dx. Then, I=0π2a2 cos2 x+b2 1-cos2 x dxI=0π2b2+a2-b2 cos2 x dxI=0π2b2+a2-b21+cos 2x2dxI=b2x+a2-b22x+sin 2x20π2I=b2π2+a2-b22π2+0I=π4a2+b2

Page No 19.16:

Question 24:

0π/21+sin x dx

Answer:

Let I=0π21+sin x dx. Then,I=0π21+sin x×1-sin x1-sin x dxI=0π21-sin2 x1-sin x dxI==0π2cos x1-sin x dxLet 1-sin x=u-cos x dx=duI=-duuI==-2uI==-21-sin x0π2I==0+2I==2

Page No 19.16:

Question 25:

0π/21+cos x dx

Answer:

Let I=0π21+cos x dx. Then,I=0π21+cos x×1-cos x1-cos x dxI=0π21-cos2 x1-cos x dxI=0π2sin x1-cos x dxLet 1-cos x=usin x dx=duI=duuI=2uI=21-cos x0π2I=2-0I=2

Page No 19.16:

Question 26:

Evaluate the following definite integrals:

0π2x2sinxdx                         [CBSE 2014]

Answer:


Let I=0π2x2sinx

Applying integration by parts, we have

I=x2-cosx|0π2-0π22x-cosxdxI=0-0+20π2xcosxdx                       cosπ2=0

Again applying integration by parts, we have

I=0+2xsinx|0π2-0π21×sinxdxI=2π2sinπ2-0-20π2sinxdxI=2π2-0-2-cosx|0π2I=π+2cosπ2-cos0I=π+20-1I=π-2



Page No 19.17:

Question 27:

0π/2x cos x dx

Answer:

Let I=0π2x cos x dx. Then,Integrating by partsI=x sin x0π2-0π21 sin x dxI=x sin x0π2+cos x0π2I=π2-1                   

Page No 19.17:

Question 28:

0π/2x2 cos x dx

Answer:

Let I=0π2x2 cos x dx. Then,Integrating by partsI=x2 sin x0π2-0π22x sin x dxI=x2 sin x0π2--2x cos x0π2+0π2-2 cos x dxI=x2 sin x0π2+2x cos x0π2-2 sin x0π2I=π24-2

Page No 19.17:

Question 29:

0π/4x2 sin x dx

Answer:

Let I=0π4x2 sin x dx. Then,Integrating by partsI=-x2 cos x0π4-0π4-2x cos x dxI=-x2 cos x0π4+2x sin x0π4-0π42 sin x dxI=-x2 cos x0π4+2x sin x0π4+2 cos x0π4I=-π2162+π22+22-2I=2+π22-π2162-2

Page No 19.17:

Question 30:

0π/2x2 cos 2x dx

Answer:

Let I=0π2x2 cos 2x dx. Then,Integrating by partsI=x2 sin 2x20π2-0π22x sin 2x2 dxI=x2 sin 2x20π2--x cos 2x20π2+0π2-1 cos 2x2 dxI=x2 sin 2x20π2+x cos 2x20π2-sin 2x40π2I=0-π4-0I=-π4

Page No 19.17:

Question 31:

0π/2x2 cos2 x dx

Answer:

Let I=0π2x2 cos2 x dx. Then,I=0π2x2 1+cos 2x2dxI=0π2x22+x2 cos 2x2 dxI=x360π2+x2 sin 2x40π2-0π2x2 sin 2x dxI=x360π2+x2 sin 2x40π2--x cos 2x40π2+0π2-1 cos2x2dxI=x360π2+x2 sin 2x40π2+x cos 2x40π2-sin 2x40π2I=π348-π8

Page No 19.17:

Question 32:

12log x dx

Answer:

Let I=12log x dx. Then,I=121 log x dxIntegrating by partsI=x log x12-121x x dxI=x log x12-12dxI=x log x12-x12I=2 log 2-2+1I=2 log 2-1

Page No 19.17:

Question 33:

13log xx+12 dx

Answer:

Let I=13log x1+x2 dx. Then,I=-11+x log x13-131x-1x+1 dxI=-11+x log x13+131xx+1 dxI=-11+x log x13+131x-1x+1 dxI=-11+x log x13+log x-log x+113I=-14 log 3+log 3-log 4+log 2I=34 log 3-log 2

Page No 19.17:

Question 34:

1eexx 1+x log x dx

Answer:

Let I=1eexx1+x log x dx. Then,I=1eexx+ex log x dxI=1eexx dx+1eex log x dxIntegrating first term by partsI=log x ex1e-1eex log x dx+1eex log x dxI=log e ee -0I=ee

Page No 19.17:

Question 35:

1elog xx dx

Answer:

Let I=1elog xx dxLet log x=u1x dx=duI=u duI=u22I=(log x)221eI=12-0I=12

Page No 19.17:

Question 36:

ee21log x-1log x2 dx

Answer:

Let I=ee21log x-1log x2 dx. Then,I=ee21 1log x dx-ee21log x2 dxIntegrating by partsI=xlog xee2-ee2-1xlog x2 x dx-ee21log x2 dxI=xlog xee2+ee21log x2 dx-ee21log x2 dxI=xlog xee2+0I=e2log e2-elog eI=e22 log e-elog eI=e22-e

Page No 19.17:

Question 37:

12x+3x x+2 dx

Answer:

Let I=12x+3xx+2 dx. Then,I=12xxx+2+3xx+2 dxI=12dxx+2+123xx+2 dxI=log x+212+32121x-1x+2 dxI=log x+212+32log x-log x+212I=log 4-log 3+32log 2-log 4-0+log 3I=log 4-log 3+32-log 2+log 3I=2 log 2-log 3+32 log 3-32 log 2I=12 log 2+12 log 3I=12log 2+ log 3I=12 log 6

Page No 19.17:

Question 38:

012x+35x2+1 dx

Answer:

Let I=012x+35x2+1 dx. Then,I=012x5x2+1 dx+0135x2+1 dxI=150110x5x2+1 dx+30115x2+12 dxI=15log 5x2+101+35tan-15x01I=15 log 6 +35tan-15

Page No 19.17:

Question 39:

0214+x-x2 dx

Answer:

Let I=0214+x-x2 dx.Then,I=-021x2-x-4 dxI=-021x2-x+14-14-4 dx=-021x-122-174 dx=-021x-122-1722 dx=021-2x-122+1722 dx=117log 17+2x-117-2x+102=117log 17+317-3-log 17-117+1=117log 26+6178-log 18-21716=117log 52+121718-217=117log 52+121718-217×18+21718+217I=117 log 1344+32017256I=117 log 21+5174

Page No 19.17:

Question 40:

0112x2+x+1 dx

Answer:

Let I=0112x2+x+1 dx. Then,I=12011x2+x2+12 dxI=12011x2+x2+116-116+12 dxI=12011x+142+716 dxI=12×47tan-1x+147401I=27tan-157-tan-117

Page No 19.17:

Question 41:

01x 1-x dx

Answer:

Let I=01x1-x dx. Then,I=0114-x-122  dxI=12011-x-122 14 dxI=12011-x-12 122 dxLet x-1212=sin u2 dx=cos u duI=14-π2π21-sin2 u cos u duI=14-π2π2cos2 u duI=14-π2π2cos 2u+12 duI=18sin 2u2+u-π2π2I=18π2+π2I=π8

Page No 19.17:

Question 42:

0213+2x-x2 dx

Answer:

Let I=0213+2x-x2 dx. Then,I=021-x2+2x-1+1+3 dxI=021-x-12+4 dxI=sin-1x-1202I=sin-112-sin-1-12I=2 sin-112I=2×π6=π3

Page No 19.17:

Question 43:

0414x-x2 dx

Answer:

Let I=0414x-x2 dx. Then,I=0414x-x2-4+4 dxI=041-x-22+4 dxI=sin-1x-2204I=sin-11-sin-1(-1)I=2 sin-11I=2 π2=π

Page No 19.17:

Question 44:

-111x2+2x+5 dx

Answer:

Let I=-111x2+2x+5 dx. Then, I=-111x2+2x+1+4 dxI=-111x+12+22 dxI=12tan-1x+12-11I=12tan-11-tan-10I=12π4I=π8

Page No 19.17:

Question 45:

14x2+x2x+1 dx

Answer:

Let I=14x2+x2x+1 dx. Let 2x+1=ux=u-12dx=du2I=u-122+u-12u du2I=18u2+1-2u+2u-2u du=18u2-1u du=18u32-u-12 du=182u525-2u121=1822x+1525-22x+112114=1825×243-6-25×93+23I=184565-835I=57-35

Page No 19.17:

Question 46:

01x 1-x5 dx

Answer:

Let I=01x1-x5 dx. Then,I=01x-1+11-x5 dxI=01-1-x6+1-x5 dxI=1-x7701-1-x6601I=-17+16I=142

Page No 19.17:

Question 47:

12x-1x2 ex dx

Answer:

Let I=12x-1x2ex dx. Then,I=12exx-exx2 dxI=12exx dx- 12exx2 dxIntegrating first term by partsI=exx12-12-1x2ex dx- 12exx2 dxI=exx12+12exx2 dx- 12exx2 dxI=exx12I=e22-e

Page No 19.17:

Question 48:

01xe2x+sinπx2 dx

Answer:

Let I=01x e2x+sin πx2 dx. Then,I=01x e2x dx+01sin πx2 dxIntegrating first term by partsI=x e2x201-011 e2x2 dx+-cos πx2π201I=x e2x201-e2x401-2πcos πx201I=e22-e24+14+2πI=e24+14+2π

Page No 19.17:

Question 49:

01xex+cosπx4 dx

Answer:

Let I=01x ex+cos πx4 dx. Then,I=01x ex dx+01cosπx4 dxIntegrating first term by partsI=x ex01-011 ex dx+sin πx4π401I=x ex01-ex01+sin πx4π401I=e-e+1+4π sin π4I=1+4π2I=1+22π

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

Page No 19.17:

Question 50:

π/2πex1-sin x1-cos x dx

Answer:

Let I=π2πex 1-sin x1-cos x dx. Then,I=π2πex1-2 sin x2 cos x22 sin2 x2 dx                                   As, sin A=2 sin A2 cos A2, cos A=1-2 sin2 A2I=π2πex 12 cosec2 x2-cot x2 dxI=π2π12 ex cosec2 x2 dx-π2πex cot x2 dxIntegrating second term by partsI=-ex cot x2π2π-π2π12 ex cosec2 x2 dx+π2π12 ex cosec2 x2 dxI=-0-eπ2I=eπ2

Page No 19.17:

Question 51:

02πex/2 sinx2+π4 dx

Answer:

Let I=02πex2sin x2+π4 dx. Then,Integrating by partsI=-2ex2 cos x2+π402π-02π-22 ex2 cos x2+π4 dxAgain, integrating second term by partsI=-2ex2 cos x2+π402π+2ex2sin x2+π402π-02π22 ex2sin x2+π4 dxI=-2ex2 cos x2+π402π+2ex2sin x2+π402π-I2I=22 eπ+22-22eπ-22=0I=0

Page No 19.17:

Question 52:

02πex cosπ4+x2 dx

Answer:

Let I=02πex cosπ4+x2 dx. Then,Integrating by partsI=2ex sin π4+x202π-02π2ex sin π4+x2 dxIntegrating second term by partsI=2ex sin π4+x202π+4ex cos π4+x202π+02π-4ex cos π4+x2 dxI=2ex sin π4+x202π+4ex cos π4+x202π-4I5I=-2e2π 12-2 12-4e2π 12-4 125I=-32 e2π-32I=-325e2π+1

Page No 19.17:

Question 53:

Evaluate 0πe2x· sinπ4+x dx

Answer:

Let I=0πe2xsin π4+x dx Integrating by parts, we getI=12e2x sin π4+x0π-120π e2x cos π4+x dxNow, integrating the second term by parts, we getI=12e2x sin π4+x0π-1212e2xcos π4+x0π+120π e2xsin  π4+x dxI=12e2x sin  π4+x0π-14e2xcos  π4+x0π-14I54I=12e2πsinπ+π4-sinπ4-14e2πcosπ+π4-cosπ454I=12-e2π×12-12-14-e2π×12-1254I=-122e2π-122+142e2π+142I=-152e2π+1

Page No 19.17:

Question 54:

0111+x-x dx

Answer:

Let I=0111+x-x dx. Then,I=0111+x-x×1+x+x1+x+x dxI=011+x+x1+x-x dxI=011+x+x dxI=231+x32+23x3201I=23×22+23-23I=423

Page No 19.17:

Question 55:

12xx+1 x+2 dx

Answer:

Let I=12xx+1x+2 dx. Then,I=12-1x+1+2x+2 dxI=-121x+1 dx+2121x+2 dxI=-log x+1+2 log x+212I=-log 3+2 log 4+log 2-2 log 3I=5 log 2-3 log 3I=log 25- log 33I=log 3227

Page No 19.17:

Question 56:

0π/2sin3 x dx

Answer:

Let I=0π2sin3 x dx. Then,I=0π2sin x sin2 x dxI=0π2sin x 1-cos2 x dxLet u =cos x, du= -sin x dxI=-1-u2 duI=u33-uI=cos3 x3-cos x0π2I=0-13+1I=23

Page No 19.17:

Question 57:

0πsin2x2-cos2x2 dx

Answer:

Let I=0πsin2 x2-cos2 x2 dx. Then,I=-0π cos x dx                      cos A=cos2 A2-sin2 A2I=-sin x0πI=0

Page No 19.17:

Question 58:

12e2x 1x-12x2 dx

Answer:

Let I12e2x1x-12x2 dx. Then,I=12e2x 1x-12e2x 12x2 dxIntegrating first term by partsI=e2x2x12-12-e2x 12x2-12e2x 12x2 dxI=e2x2x12I=e44-e22I=e4-2e24

Page No 19.17:

Question 59:

Evaluate the following definite integrals:

011x-12-xdx                                [NCERT EXEMPLAR]

Answer:


Let I = 121x-12-xdx

Put x=cos2θ+2sin2θ

dx=2cosθ-sinθdθ+4sinθcosθdθ=2sinθcosθdθ

Also,

x=cos2θ+2sin2θx=1+sin2θsinθ=x-1

When x1, sinθ0 or θ0

When x2, sinθ1 or θπ2

I = 121x-12-xdx
I=0π22sinθcosθdθcos2θ+2sin2θ-12-cos2θ-2sin2θI=0π22sinθcosθdθsin2θcos2θ                         sin2θ+cos2θ=1I=0π22sinθcosθdθsinθcosθI=20π2dθI=2θ|0π2
I=2π2-0=π



Page No 19.18:

Question 60:

If 0k12+8x2 dx=π16, find the value of k.

Answer:

We have,0k12+8x2 dx=π16180k114+x2 dx=π1614 tan-12x0k=π16tan-12k=π42k=tanπ42k=1k=12

Page No 19.18:

Question 61:

If 0a3x2 dx=8, find the value of a.

Answer:

We have,0a3x2 dx=83 x330a=8a3=8a=2

Page No 19.18:

Question 62:

π3π21-cos2xdx

Answer:


 π3π21-cos2xdx
=π3π22sin2xdx=2π3π2sinxdx=-2π3π2sinxdx                         sinx<0 for πx2π
=-2-cosx|π3π2=2cos3π2-cosπ=2 0--1=2×1=2

Page No 19.18:

Question 63:

02π1+sinx2dx

Answer:


I=02π1+sinx2dx=02πcos2x4+sin2x4+2sinx4cosx4dx=02πcosx4+sinx42dx=02πcosx4+sinx4dx

When 0x2π, 0x4π2
sinx40, cosx40cosx4+sinx40cosx4+sinx4=cosx4+sinx4

I=02πcosx4+sinx4dx=sinx41402π+-cosx41402π=4sinπ2-sin0-4cosπ2-cos0=41-0-40-1=4+4=8

Page No 19.18:

Question 64:

0π4tanx+cotx-2dx

Answer:


0π4tanx+cotx-2dx=0π41tanx+cotx2dx=0π41sinxcosx+cosxsinx2dx=0π41sin2x+cos2xsinxcosx2dx=0π4sin2xcos2xdx
=140π42sinxcosx2dx=140π4sin22xdx=140π41-cos4x2dx=180π4dx-180π4cos4xdx=18x0π4-18sin4x40π4
=18π4-0-132sinπ-sin0=π32-132×0-0=π32

Page No 19.18:

Question 65:

01xlog1+2xdx

Answer:


Let I = 01xlog1+2xdx

Applying integration by parts, we have

I=log1+2xx2201-0121+2x×x22dx=12log3-0-01x21+2xdx=12log3-14014x2-1+11+2xdx=12log3-14012x+12x-11+2xdx-140111+2xdx=12log3-14012x-1dx-140111+2xdx
=12log3-14×2x-122×201-14×log1+2x201=12log3-1161-1-18log3-log1=12log3-0-18log3                              log1=0=38log3

Page No 19.18:

Question 66:

π6π3tanx+cotx2dx

Answer:


π6π3tanx+cotx2dx=π6π3tan2x+cot2x+2tanxcotxdx=π6π3sec2x-1+cosec2x-1+2dx=π6π3sec2xdx+π6π3cosec2xdx
=tanxπ6π3+-cotxπ6π3=tanπ3-tanπ6-cotπ3-cotπ6=3-13-13-3=23-23=43

Page No 19.18:

Question 67:

0π4a2cos2x+b2sin2xdx

Answer:


0π4a2cos2x+b2sin2xdx=0π4a21+cos2x2+b21-cos2x2dx=0π4a2+b22+a2-b22cos2xdx=a2+b220π4dx+a2-b220π4cos2xdx
=a2+b22×x0π4+a2-b22×sin2x20π4=a2+b22π4-0+a2-b24sinπ2-sin0=a2+b22π4+a2-b241-0=a2+b2π8+14a2-b2

Page No 19.18:

Question 68:

0111+2x+2x2+2x3+x4dx

Answer:


0111+2x+2x2+2x3+x4dx=011x2+12+2xx2+1dx=011x2+1x2+1+2xdx=011x2+1x+12dx

Let
1x+12x2+1=Ax+1+Bx+12+Cx+Dx2+11=Ax+1x2+1+Bx2+1+Cx+Dx+12

Putting x = −1, we have

1 = 2B B=12           .....(1)

Putting x = 0, we have

A + B + D = 1              .....(2)

Equating coefficient of x3 on both sides, we have

A + C = 0                    .....(3)

Equating coefficient of x2 on both sides, we have

A + B + 2C + D = 0               .....(4)

2C = −1               [Using (1)]

C=-12

A=12                  [Using (3)]

Putting A=12,B=12 and C=-12 in (4), we have

D = 0

011x+12x2+1dx=0112x+1dx+0112x+12dx+01-12xx2+1=12logx+101+12×-1x+101-14012xx2+1dx=12log2-log1-1212-1-14logx2+101=12log2+14-14log2-log1                            log1=0        
=12log2+14loge-14log2=14log2+14loge=14log2+loge=14log2e



Page No 19.38:

Question 1:

24xx2+1 dx

Answer:

Let x2=t.  Then, 2x dx=dtWhen x=2, t=4 and x=4, t=16. I=24xx2+1 dxI=41612dtt+1I=12 log t+1416I=12 log 17-12 log 5I=12 log 175

Page No 19.38:

Question 2:

121x 1+log x2 dx

Answer:

Let 1+log x=t. Then, 1x dx=dtWhen x=1, t=1 and x=2, t=1+log 2I=121x1+log x2 dxI=11+log 21t2 dtI=-1t11+log 2I=-11+log 2+1I=log 2log 2+log eI=log 2log 2e

Page No 19.38:

Question 3:

123x9x2-1 dx

Answer:

Let x2=t. Then, 2x dx=dtWhen x=1, t=1 and x=2, t=4I=123x9x2-1 dxI=3214dt9t-1I=318log 9t-114I=318log 35-log 8I=log 35-log 86

Page No 19.38:

Question 4:

0π/215 cos x+3 sin x dx

Answer:

Let I=0π215 cos x+3 sin x dx. Then,I=0π2151-tan2x21+tan2x2+32 tan x21+tan2 x2 dx                                  sin A=2 tan A21+tan2 A2, cos A=1-tan2 A21+tan2 A2I=0π21+tan2 x25-5 tan2 x2+6 tan x2 dxI=0π2sec2 x25-5 tan2 x2+6 tan x2 dxLet tan x2 =t. Then, 12 sec2 x2 dx=dtAlso, x=0, t=0 and x=π2, t=1I=012dt5-5t2+6tI=15012dt1-t2+65t+36100-36100=2501dt-t-6102+136100=25×10136-log t-610-13610t-610+1361001=134-log 4-2344+234+log -6-234-6+234=134 log 6+2346-234×4+2344-234=134 log 160+2034160-2034=134 log 8+348-34

Page No 19.38:

Question 5:

0axa2+x2 dx 

Answer:

Let x= a tan t. Then, dx=a sec2 t dtWhen x=0, t=0 and x=a, t=π4I=0axa2+x2 dxI=0π4a tan ta2+a2 tan2 ta sec2 t dt=0π4a tan t a sec2 ta sec t dt=0π4a tan t sec t dt=asec t0π4=a2-1 

Page No 19.38:

Question 6:

01ex1+e2x dx

Answer:

Let ex=t. Then, ex dx=dtWhen x=0, t=1 and x=1, t=eI=01ex1+e2x dxI=1edt1+t2I=tan-1x1eI=tan-1e-tan-11I=tan-1e-π4

Page No 19.38:

Question 7:

01xex2 dx 

Answer:

Let I=01x ex2dx.Let x2=t. Then, 2x dx=dtWhen x=0, t=0 and  x=1, t=1 I=1201et dtI=12et01I=12e-1 

Page No 19.38:

Question 8:

13cos log xx dx

Answer:

Let I=13cos log xx dx.Let log x=t. Then, 1x dx=dtWhen x=1, t=0 and x=3, t =log 3I=0log 3cos t dt=sin t0log 3=sin log 3

Page No 19.38:

Question 9:

012x1+x4 dx

Answer:

Let I=012x1+x4 dx.Let x2=t. Then, 2x dx=dtWhen x=0, t=0 and x=1, t=1 I=012x1+x4 dxI=0111+t2 dtI=tan-1t01I=tan-11- tan-10I=π4

Page No 19.38:

Question 10:

0aa2-x2 dx

Answer:

Let I=0aa2-x2 dx.Let x= a sin t. Then, dx= a cos t dtWhen x=0, t=0 and x=a, t=π2 I=0aa2-x2 dxI=0π2a2-a2 sin2 t a cos t dtI=0π2a2 cos2 t dtI=a20π21+cos 2t2 dtI=a22t+sin 2t20π2I=a22π2-0I=πa24



Page No 19.39:

Question 11:

0π/2sin ϕcos5 ϕ dϕ

Answer:

0π2sin ϕ cos5 ϕ dϕLet sin ϕ=t. Then, cos ϕ dϕ=dtWhen ϕ=0, t=0 and ϕ=π2, t=1Also, cos5 ϕ=cos4 ϕ cos ϕ=1-sin2 ϕ2 cos ϕ I=0π2sin ϕ cos5 ϕ dϕI=01t 1-t22 dtI=01t1+t4-2t2 dtI=01t+t92-2t52  dtI=2t323+2t11211-4t72701I=23+211-47I=64231

Page No 19.39:

Question 12:

0π/2cos x1+sin2 x dx 

Answer:

Let I=0π2cos x1+sin2 x dx.Let sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1I=0π2cos x1+sin2 x dxI=0111+t2 dtI=tan-1 t01I=π4

Page No 19.39:

Question 13:

0π/2sin θ1+cos θ dθ

Answer:

Let I=0π2sin θ1+cos θ dθ.Let cos θ=t. Then, -sin θ dθ=dtWhen θ=0, t=1 and θ=π2, t=0 I=0π2sin θ1+cos θ dθ=10-dt1+t=01dt1+t=21+t01=22-1

Page No 19.39:

Question 14:

0π/3cos x3+4 sin x dx

Answer:

Let I=0π3cos x3+4 sin x dx.Let sin x =t. Then, cos x dx=dtWhen x=0, t=0 and x=π3, t=32 I=0π3cos x3+4sin xdx=03213+4tdt=14log 3+4t032=14log 3+23-log 3=14 log 3+233

Page No 19.39:

Question 15:

01tan-1 x1+x2 dx

Answer:

Let I=01tan-1x1+x2 dx.Let tan-1x =t. Then, 11+x2 dx=dtWhen x=0, t=0 and x=1, t=π4 I=01tan-1x1+x2 dxI=0π4t dtI=2t3230π4I=23π432I=112π32

Page No 19.39:

Question 16:

02xx+2 dx

Answer:

Let I=02xx+2 dx.Let x+2=t2. Then, dx=2t dtWhen x=0, t=2 and x=2, t=2 I=22t2-2 t 2t dtI=222t4-2t2 dtI=2t55-23t322I=2323-163-425-423I=21615+8215I=16152+2

Page No 19.39:

Question 17:

01tan-12x1-x2 dx

Answer:

01tan-12x1-x2dx=012tan-1x=2x tan-1x01-201x1+x2dx=2x tan-1x01-log1+x201=2π4-0-log2+0=π2-log2

Page No 19.39:

Question 18:

0π/2sin x cos x1+sin4 x dx

Answer:

Let I=0π2sin x cos x1+sin4 x dx.Let sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1 I=01t1+t4 dtLet  t2=u. Then, 2t dt=duSo, I=01t1+t4dt I=120111+u2duI=12tan-1u01I=π8

Page No 19.39:

Question 19:

0π/2dxa cos x+b sin xa, b>0

Answer:

0π21acosx+b sinxdx=0π21a1-tan2x21+tan2x2+b2tanx21+tan2x2dx=0π21+tan2x2a-atan2x2+2b tanx2dx=0π2sec2x2a-atan2x2+2b tanx2dxLet tanx2=t, Then, 12sec2x2dx=dtWhen x=0, t=0, x=π2, t=1Therefore the integral becomesI= 012dta-at2+2bt=012dt-at2-2bta-1=2a01dt-t-ba2-1-b2a2=2a01dtb2a2+1-t-ba2=2a12a2+b2a2loga2+b2a2+t-baa2+b2a2-t-ba01
1a2+b2loga+b+a2+b2a+b-a2+b2

Page No 19.39:

Question 20:

0π/215+4 sin x dx

Answer:

Let I=0π215+4 sin x dx. Then,I=0π215+42 tan x21+tan2 x2 dxI=0π21+tan2x251+tan2x2+8 tan x2 dxI=0π2sec2 x25 tan2 x2+8 tan x2+5 dxLet tan x2=t. Then, 12sec2 x2 dx=dtWhen x=0, t=0 and x=π2, t=1 I=20115t2+8t+5 dtI=20115t2+8t+5+452-452 dtI=20115t+452+95 dtI=23tan-15t+453501I=23tan-13-tan-143I=23tan-13-431+3×43I=23 tan-113

Page No 19.39:

Question 21:

0πsin xsin x+cos x dx

Answer:

0πsinxsinx+cosxdx=120π2sinxsinx+cosxdx=120πsinx+cosx-cosx-sinxsinx+cosxdx=120πdx-120πcosx-sinxsinx+cosxdx=12x0π-12logsinx+cosx0π=12π-0-12log1-log1=π2

Page No 19.39:

Question 22:

0π13+2 sin x+cos x dx

Answer:

Let I=0π13+2 sin x+cos x dx. Then,I=0π13+22 tanx21+tan2x2+1-tan2x21+tan2x2 dxI=0π1+tan2 x22 tan2 x2+4 tan x2+4 dxLet tan x2=t. Then, 12 sec2 x2 dx=dtWhen x=0, t=0 and x=π, t= I=02 dt2t2+4t+4I=0dtt+12+1I=tan-1t+10I=π2-π4I=π4

Page No 19.39:

Question 23:

01tan-1 x dx

Answer:

Let I=01tan-1x dx. Then,I=011 tan-1x dxIntegrating by partsI= x tan-1x01-01x1+x2 dxI=x tan-1x01-12log x2+101I=π4-0-12 log 2+0I=π4-12 log 2

Page No 19.39:

Question 24:

012xsin-1x1-x2dx

Answer:


Let I = 012xsin-1x1-x2dx

Put x=sinθ

dx=cosθdθ

When x0, θ0

When x12,θπ6

I=0π6sinθsin-1sinθcosθcosθdθ     =0π6θsinθdθ

Applying integration by parts, we have

I=θ-cosθ0π6-0π61×-cosθdθ=-π6cosπ6-0+0π6cosθdθ=-π6×32+sinθ0π6=-π43+sinπ6-sin0=-π43+12-0=12-π43

Page No 19.39:

Question 25:

0π/4tanx+cotx dx

Answer:

Let I=0π4tan x+cot x dx. Then,I=0π4sin xcos x+cos xsin x dx I=0π4sin x+cos xsin x cos x dxI=20π4sin x+cos x2 sin x cos x dxI=20π4sin x +cos x1-sin x-cos x2 dxLet sin x-cos x =t. Then, cos x+ sin x dx= dtWhen x=0, t=1 and x=π4, t=0 I=2-10dt1-t2I=2 sin-1 t-10I=π2I=

Page No 19.39:

Question 26:

0π/4tan3 x1+cos 2x dx

Answer:

Let I=0π4tan3 x1+cos 2x dx. Then,I=0π4tan3 x2 cos2 x dxI=120π4tan3 x sec2 x dxLet tan x=t. Then, sec2 x dx=dtWhen x=0, t=0 and x=π4, t=1 I=1201t3 dtI=12t4401I=1214-0I=18

Page No 19.39:

Question 27:

0π15+3 cos x dx

Answer:

Let I=0π15+3 cos x dx. Then,I=0π15+31-tan2 x21+tan2 x2 dxI=0π1+tan2 x25+5 tan2 x2-3 tan2 x2 dxI=0πsec2 x25+2 tan2 x2 dxLet tan x2=t. Then, 12 sec2 x2 dx=dtWhen x=0, t=0 and x=π, t= I=0dt5+2t2I=120dt52+t2I=12tan-15t20I=12π2-0I=π4

Page No 19.39:

Question 28:

0π/21a2 sin2 x+b2 cos2 x dx

Answer:

Let I=0π21a2 sin2 x+b2 cos2 x dx. Then,Dividing the numerator and denominator by cos2x, we getI=0π2sec2xa2 tan2x+b2 dxLet tan x =t. Then, sec2 x dx= dtWhen x=0, t=0 and x=π2 , t= I=01a2 t2+b2  dtI=1a201t2+b2a2 dtI=1a2×abtan-1atb0I=1abπ2I=π2ab

Page No 19.39:

Question 29:

0π/2x+sin x1+cos x dx

Answer:

Let, I=0π/2x+sin x1+cos x dx       =0π/2x+sin x2 cos2 x2 dx       =0π/2 x2 cos2 x2 +sin x2 cos2 x2dx       =120π/2x sec2 x2dx+0π/22sin x2cosx22 cos2 x2dx       =12x tanx2120π/2-120π/2 tanx212dx+0π/2tanx2dx       =x tanx20π/2-0π/2 tanx2 dx+0π/2tanx2dx       =π2 tanπ4       =π2 ×1       =π2 

Page No 19.39:

Question 30:

01tan-1 x1+x2 dx 

Answer:

Let I=01tan-1x1+x2 dx. Then,Let tan-1x=t. Then, 11+x2 dx=dtWhen x=0, t=0 and x=1, t=π4 I=0π4 t dtI=t220π4I=π232 

Page No 19.39:

Question 31:

0π4sinx+cosx3+sin2xdx

Answer:


Let I=0π4sinx+cosx3+sin2xdx
=0π4sinx+cosx4-1-sin2xdx=0π4sinx+cosx4-sin2x+cos2x-2sinxcosxdx=0π4sinx+cosx4-sinx-cosx2dx

Put sinx-cosx=z

cosx+sinxdx=dz

When x0, z-1                                           z=sin0-cos0=0-1=-1

When xπ4, z0                                  z=sinπ4-cosπ4=12-12=0

I=-10dz22-z2=12×2log2+z2-z-10=14log1-log13=140-log1-log3=-140-log3=14log3

Page No 19.39:

Question 32:

01x tan-1 x dx

Answer:

Let I=01x tan-1x dx. Then,Integrating by partsI=x2 tan-1x201-1201x21+x2 dxI=x2 tan-1x201-12011+x21+x2-11+x2 dxI=x2 tan-1x201-12x-tan-1x01I=π8-0-121-π4-0I=π4-12 

Page No 19.39:

Question 33:

011-x2x4+x2+1 dx

Answer:

Let, I=1-x2x4+x2+1 dx=-x2-1x4+x2+1 dx=-1-1x2x2+1+1x2 dx=-1-1x2x2+2+1x2-1 dx=-1-1x2x+1x2-1 dxLet, x+1x=t1-1x2dx=dtThen integral becomes,I=-1t2-1 dt=-12logt-1t+1=12logt+1t-1=12logx+1x+1x+1x-1=12logx2+x+1x2-x+1i.e., 1-x2x4+x2+1 dx=12logx2+x+1x2-x+1011-x2x4+x2+1 dx=12logx2+x+1x2-x+1  01                                          =12log 3

Page No 19.39:

Question 34:

0124 x31+x24 dx

Answer:

Let I=0124x31+x24 dx. Then,Let x2 =t. Then, 2x dx=dtWhen x=, t=0 and x=1, t=1 I=0112t1+t4 dtIntegrating by partsI=12t-31+t301+1201131+t3dtI=12t-31+t301-161+t201I=12-124-0-124+16I=12×112I=1

Page No 19.39:

Question 35:

412x x-41/3 dx

Answer:

Let I=412xx-413 dx.Let x-4=t. Then, dx=dtWhen x=4, t=0 and x=12, t=8 I=08t+4t13 dtI=08 t43+4t13  dtI=37t73+31t4308I=3847+48I=7207

Page No 19.39:

Question 36:

0π/2x2 sin x dx

Answer:

Let I=0π2x2 sin x dx. Then,Integrating by partsI=-x2 cos x0π2-0π2-2x cos x dxAgain, integratting by partsI= -x2 cosx0π2+2x sin x0π2-0π21  sin x dxI= -x2 cos x0π2+2x sin x0π2--cos x0π2I=π24 0-0+2π2-0+0-2I=π-2

Page No 19.39:

Question 37:

011-x1+x dx

Answer:

Let I=011-x1+x dx. Then,I=011-x1+x×1-x1-x dxI=011-x1-x2 dxI=0111-x2 dx-01x1-x2 dxI=sin-1x01+1201-2x1-x2 dxI=sin-1x01+1221-x201I=π2-0+0-1I=π2-1

Page No 19.39:

Question 38:

011-x21+x22 dx 

Answer:

Let I=011-x21+x22 dx. Then,I=011x2-1x+1x2 dxLet x+1x=t. Then, 1-1x2 dx=dtWhen x=0, t= and x=1, t=2I=2-dtt2I=1t2I=12-0I=12

Page No 19.39:

Question 39:

-115x4 x5+1 dx

Answer:

Let I=-115x4x5+1 dx. Then,Let x5+1=t. Then, 5x4 dx=dtWhen x=-1, t=0 and x=1, t=2 I=02t dtI=23t3202I=238I=423

Page No 19.39:

Question 40:

0π2cos2x1+3sin2xdx                              [CBSE 2015]

Answer:


Let I=0π2cos2x1+3sin2xdx=0π2cos2x1+31-cos2xdx=0π2cos2x4-3cos2xdx=-130π24-3cos2x-44-3cos2xdx
=-130π2dx+430π214-3cos2xdx=-13x0π2+430π2sec2x4sec2x-3dx                   Dividing numerator and denominator by cos2x=-13π2-0+430π2sec2x41+tan2x-3dx=-π6+430π2sec2x4tan2x+1dx

Put tanx = z

sec2xdx=dz

When x0, z0

When xπ2, z

I=-π6+430dz4z2+1=-π6+430dz2z2+1=-π6+43×tan-12z20=-π6+23tan-1-tan-10=-π6+23π2-0=-π6+π3=π6

Page No 19.39:

Question 41:

0π/4sin3 2t cos 2t dt

Answer:

Let I=0π4sin3 2t cos 2t dt. Then,Let sin 2t =u. Then, 2 cos 2t dt=duWhen t=0, u=0 and t=π4, u=1 I=1201u3 duI=12u4401I=1214-0I=18

Page No 19.39:

Question 42:

0π5 5-4 cos θ1/4 sin θ d θ 

Answer:

Let I=0π55-4cos  θ14 sin θ dθ. Let 5-4 cos θ=t . Then, 4 sin θ dθ=dtWhen θ=0, t=1 and θ=π, t=9I=5419t14  dtI=544t54519I=93-1 

Page No 19.39:

Question 43:

0π/6cos-3 2 θ sin 2 θ d θ

Answer:

Let I=0π6cos-3 2θ sin 2θ dθ. Then,I=0π6sin 2θcos3 2θ dθLet cos 2θ=t. Then, -2 sin 2θ dθ=dtWhen θ=0, t=1 and θ=π6, t=12 I=-12112dtt3I=1212t2112I=122-12I=34

Page No 19.39:

Question 44:

0π2/3xcos2 x3/2 dx 

Answer:

Let I=0π23x cos2 x32 dx. Then,Let x32=t. Then, 32x dx= dtWhen x=0, t=0 and x=π23, t=π I=230πcos2 t dtI=230π1+cos 2x2 dxI=13x+sin 2x20πI=13π+0I=π3 



Page No 19.40:

Question 45:

121x 1+log x2 dx

Answer:

Let I=121x1+log x2 dx. Then, Let 1+log x=t. Then, 1x dx=dtWhen x=1, t=1 and x=2, t=1+log 2 I=11+log 21t2 dtI=-1t11+log 2I=-11+log 2+1I=log 21+log 2

Page No 19.40:

Question 46:

0π/2cos5 x dx

Answer:

Let I=0π2cos5 x dx. Then,I=0π2cos4 x cos x dxI=0π21-sin2 x2 cos x dxI=0π21-2 sin2 x+sin4 x cos x dx Let sin x =t. Then, cos dx= duWhen x=0, t=0 and x=π2, t=1 I=011-2t2+t4 dtI=t-2t33+t5501I=1-23+15I=815

Page No 19.40:

Question 47:

49x30-x3/22 dx

Answer:

Let I=49x30-x322 dx. Then,Let 30-x32=t. Then, -32x dx= dtWhen, x=4, t=22 and x=9, t=3 I=223-231t2 dtI=231t223I=2313-122I=1999

Page No 19.40:

Question 48:

0πsin3 x1+2 cos x1+cos x2 dx

Answer:

Let I=0πsin3 x 1+2 cos x1+cos x2 dx. Then,I=0πsin x sin2 x 1+2 cos x1+cos x2 dxI=0πsin x 1-cos2 x1+2 cos x1+cos x2 dxI=0πsin x 1-cos x1+2 cos x1+cos x3 dxLet  cos x =t. Then, - sin x dx = dtWhen x=0, t=1 and x=π, t=-1 I=-1-11-t1+2t1+t3 dtI=-111+t-2t21+t3+3t+3t2 dtI=-111+t3+3t+3t2+t+t4+3t2+3t3-2t2-2t5-6t3-6t4 dtI=-111+4t+4t2-2t3-5t4-2t5 dtI=t+2t2+4t33-t42-t5-t63-11I=1+2+43-12-1-13+1-2+43+12-1+13I=83

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Question 49:

0π/22 sin x cos x tan-1 sin x dx

Answer:

Let I=0π22 sin x cos x tan-1sin x dx. Then,Let sinx =t. Then, cos x dx = dtWhen x=0, t=0 and x=π2, t=1 I=01 2t tan-1 t dtI=2t2 tan-1 t201-201t21+t2 dtI=2t2 tan-1 t201-2011+t21+t2-11+t2 dtI=2t2 tan-1 t201-t-tan-1 t+01I=1 tan-1 1 -0-1+tan-11+0I=π4-1+π4I=π2-1

Page No 19.40:

Question 50:

0π/2sin 2x tan-1 sin x dx

Answer:

Let I=0π2sin 2x tan-1sin x dx. Then,I=0π22 sin x cos x tan-1sin x dxLet sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1 I=201 t tan-1t  dtI=2t22 tan-1t01-201t1+t2 dtI=2t22 tan-1t01-log 1+t201     I=2π4-1I=π2-1

Page No 19.40:

Question 51:

01cos-1 x2 dx

Answer:

Let I=01cos-1x2 dx. Then,I=011cos-1x2dxIntegrating by partsI=xcos-1x201-012x cos-1x -11-x2 dxAgain, integrating second term by partsI=xcos-1x201+21-x2  cos-1x01-20111-x21-x2 dxI=xcos-1x201+21-x2  cos-1x01-2x01I=0+2π2-2I=π-2

Page No 19.40:

Question 52:

0asin-1xa+x dx

Answer:

Let, I=0asin-1xa+x dxLet, x=a tan2θ θ=tan-1xaWhen, xx ; θ0 and xa ; θπ4and  dx=2a tanθ sec2θ dθThen,I=0π4sin-1a tan2θa+a  tan2θ 2a tanθ sec2θ dθI= 2a 0π4sin-1sinθ tanθ sec2θ dθI= 2a 0π4θ tanθ sec2θ dθLet, tan θ=tθ=tan-1tsec2θ dθ =dtwhen, θ0 ; t0 and θπ4 ; t1Then, I=2a01tan-1t  t dt             =2a01tan-1t  t dt             =2atan-1t  t2201-2a201t21+t2   dt             =2aπ4×12-0-a011-11+t2   dt             =2aπ8-at-tan-1t01             =πa4-a1-π4             =πa4-a+πa4             =πa2-a            =aπ2-1

Page No 19.40:

Question 53:

π/3π/21+cos x1-cos x3/2 dx

Answer:

π3π21+cosx1-cosx32dx=π3π21+cosx1-cosx32×1-cosx1-cosxdx=π3π21-cos2x1-cosx2dx=π3π2sinx1-cosx2dxLet 1-cosx=t, Then sinx dx=dtWhen x=π3, t=12 and x=π2, t=1Therefore the integral becomes=121dtt2=-1t121=-1+2=1

Page No 19.40:

Question 54:

0ax a2-x2a2+x2 dx

Answer:

Let, I=0ax a2-x2a2+x2 dxConsider, x2=a2cos2θ   2x dx=-2a2 sin2θ dθ   x dx=-a2 sin2θ dθWhen, x0 ; θπ4 and xa ;θ0Now, integral becomes,I=π40-a2 sin2θ  a2-a2cos2θa2+a2cos2θ dθ =π40-a2 sin2θ tanθ dθ =a2 0π42 sinθ cosθ sinθcosθ dθ =a2 0π42sin2θ dθ =a2 0π41-cos 2θ dθ =a2  θ -sin2θ20π4 =a2  π4 -12

Page No 19.40:

Question 55:

-aaa-xa+x dx

Answer:

Let, I= -aaa-xa+x dxConsider, x=a cos 2y Then y=12cos-1xa  dx=-2a sin 2y dyWhen, x-a ;yπ2 and xa ;y0Now, integral becomes, I= π20 -2a sin 2ya-a cos 2y a+a cos 2y  dy   =0π2 2a sin 2y tan y dy   =2a0π2 2sin y cos y sin ycos y dy   =2a0π2 2sin2 y dy   =2a0π2 1-cos 2y dy   =2a y-sin 2y20π2   =2a π2-sin 2y20π2   =πa

Page No 19.40:

Question 56:

0π/2sin x cos xcos2 x+3 cos x+2 dx

Answer:

Let I=0π2sin x cos xcos2 x+3 cos x+2 dx. Then,Let cos x =t. Then, - sin x dx= dtWhen x=0, t=1 and x =π2, t=0 I=-10t dtt2+3t+2I=10-t dtt+2t+1I=101t+1-2t+2 dtI=log t+1-2 log t+210I=log t+1t+2201I=log 14-log 2901I=log 98

Page No 19.40:

Question 57:

0π2tanx1+m2tan2xdx

Answer:


Let I = 0π2tanx1+m2tan2xdx
        =0π2sinxcosx1+m2sin2xcos2xdx=0π2sinxcosxcos2x+m2sin2xdx

Put cos2x+m2sin2x=z

2cosx-sinxdx+m2×2sinxcosxdx=dz2m2-1sinxcosxdx=dzsinxcosxdx=dz2m2-1

When x0, z1                                       z=cos2x+m2sin2x=1+m2×0=1

When xπ2, zm2                                 z=cos2x+m2sin2x=0+m2×1=m2

I=12m2-11m2dzz=12m2-1logz1m2=12m2-1logm2-log1=12m2-12logm-0=logmm2-1

Page No 19.40:

Question 58:

01211+x21-x2dx

Answer:


Let I = 01211+x21-x2dx

Put x=sinθ

dx=cosθdθ

When x0, θ0

When x12, θπ6

I=0π611+sin2θcosθ×cosθdθ=0π611+sin2θdθ

Dividing numerator and denominator by cos2θ, we have

I=0π6sec2θsec2θ+tan2θdθ=0π6sec2θ1+2tan2θdθ

Now, put tanθ=u

sec2θdθ=du

When θ0, u0

When θπ6, u13
I=013du1+2u2=013du1+2u2=tan-12u2013=12tan-123-0=12tan-123

Page No 19.40:

Question 59:

131x-x313x4dx

Answer:


Let I = 131x-x313x4dx
=131x3xx3-113x4dx=131x1x2-113x4dx=1311x2-113x3dx

Put 1x2-1=z

-2x3dx=dzdxx3=-dz2

When x13, z8

When x1, z0

I=-1280z13dz=-12×z434380=-380-843=-38×-16=6

Page No 19.40:

Question 60:

0π4sin2xcos2xsin3x+cos3x2dx

Answer:


Let I=0π4sin2xcos2xsin3x+cos3x2dx
=0π4sin2xcos2xcos6xtan3x+12dx=0π4tan2xsec2xtan3x+12dx

Put tan3x+1=z

3tan2xsec2xdx=dztan2xsec2xdx=dz3

When x0, z1

When xπ4, z2

I=1312dzz2=13×-1z12=-1312-1=-13×-12=16

Page No 19.40:

Question 61:

0π2cosx-cos3xsec2x-1cos2xdx

Answer:


Let I = 0π2cosx-cos3xsec2x-1cos2xdx
=0π2cosx1-cos2x-tan2xcos2xdx=-0π2cosxsin2xsin2xdx=-0π2cosxsinxsin2xdx=-0π2cosx1-cos2xsinxdx                sinx=sinx for 0xπ2

Put cosx = z2

-sinxdx=2zdz

When x0, z1

When xπ2, z0

I=-10z1-z42zdz=-210z2dz+210z6dz=-2×z3310+2×z7710=-230-1+270-1=23-27=821

Page No 19.40:

Question 62:

0π2cosxcosx2+sinx2ndx

Answer:


Let I = 0π2cosxcosx2+sinx2ndx
=0π2cos2x2-sin2x2cosx2+sinx2ndx=0π2cosx2+sinx2cosx2-sinx2cosx2+sinx2ndx=0π2cosx2-sinx2cosx2+sinx2n-1dx

Put cosx2+sinx2=z

-sinx2×12+cosx2×12dx=dzcosx2-sinx2dx=2dz

When x0, z1

When xπ2, z2                     z=cosπ4+sinπ4=12+12=22=2

I=212dzzn-1=2×z2-n2-n12=22-n22-n-1=22-n22-n2-1=22-n21-n2-1



Page No 19.55:

Question 1:

(i) 14fx dx, where fx=4x+3, if 1x23x+5, if 2x4

(ii) 09fx dx, where fx sin x,0xπ/21,π/2x3ex-3,3x9

(iii) 14fx dx, where fx=7x+3,if 1x38x,if 3x4

(iv) -12xxdx

Answer:

(iv) -12xxdx


Let I=-12xxdxWe know,x=x          , x0-x       , x<0Thus,I=-10xxdx+02xxdx  =-10-xxdx+02xxdx  =-10-1dx+021dx  =-x-10+x02  =-0-1+2-0  =-1+2  =1


Hence, -12xxdx = 1.
 



Page No 19.56:

Question 2:

Evaluate the following integrals:

 -44x+2 dx

Answer:

-44x+2 dxWe know that, x+2=-x+2 , -4x-2x+2,   -2<x4I=-44x+2 dxI=-4-2-x+2 dx+-24x+2 dxI=-x22-2x-4-2+x22+2x-24I=-2+4-8-8+8+8-2+4I=20

Page No 19.56:

Question 3:

Evaluate the following integrals:
-33x+1 dx

Answer:

I=-33x+1 dxWe know that, x+1=-x+1 , -3x-1x+1,   -1<x3 I=-3-1-x+1 dx+-13x+1 dxI=-x+122-3-1+x+122-13I=0+2+8-0I=10

Page No 19.56:

Question 4:

Evaluate the following integrals:
-112x+1 dx

Answer:

-112x+1 dxWe know that, 2x+1=-2x+1, -1x-122x+1,   -12<x1I=-1-12-2x+1 dx+-1212x+1 dxI=-x2+x-1-12+x2+x-121I=-14+12+1-1+1+1-14+12I=52

Page No 19.56:

Question 5:

Evaluate the following integrals:
-222x+3 dx

Answer:

-222x+3 dxWe know that, 2x+3=-2x+3, -2x-322x+3,   -32<x2I=-2-32-2x+3 dx+-3222x+3 dxI=-x2+3x-2-32+x2+3x-322I=-94+92+4-6+4+6-94+92I=252

Page No 19.56:

Question 6:

Evaluate the following integrals:
02x2-3x+2 dx

Answer:

02x2-3x+2 dxWe know that, x2-3x+2=-x2-3x+2, x-1x-20 or, 1x2x2-3x+2,   x2-3x+20 or, x-, 12, I=02x2-3x+2 dxI=01x2-3x+2 dx-12x2-3x+2 dxI=x33-3x22+2x01-x33-3x22+2x12I=13-32+2-83-6+4-13+32-2I=13-32+2-83+6-2+13-32I=1

Page No 19.56:

Question 7:

Evaluate the following integrals:
033x-1 dx

Answer:

033x-1 dxWe know that, 3x-1=-3x-1, 0x133x-1, 13<x3I==013-3x+1 dx+1303x+1 dxI=-3x22-x013+3x22+x133I=-16+13-0+272+3-16-13I=656

Page No 19.56:

Question 8:

Evaluate the following integrals:
-66x+2 dx

Answer:

-66x+2dxWe know that, x+2=-x+2 , -6x-2x+2,   -2<x6I=-66x+2 dxI=-6-2-x+2 dx+-26x+2 dxI=-x22-2x-6-2+x22+2x-26I=-2+4+18-12+18+12-2+4I=40

Page No 19.56:

Question 9:

Evaluate the following integrals:
-22x+1 dx

Answer:

-22x+1 dxWe know that, x+1=-x+1 , -2x-1x+1,   -1<x2I=-22x+1 dxI=-2-1-x+1 dx  + -12x+1 dxI=-x22-x-2-1+x22+x-12I=-12+1+2-2+2+2-12+1I=5

Page No 19.56:

Question 10:

Evaluate the following integrals:
12x-3 dx

Answer:

12x-3 dxWe know that, x+1=-x+1 , 1x3x+1,   x>3I=12x-3 dxI=12-x-3 dxI=-x22-3x12I=-2-6+12+3I=32

Page No 19.56:

Question 11:

Evaluate the following integrals:
0π/2cos 2x dx

Answer:

0π2cos 2x dxWe know that, cos 2x=-cos 2x ,π4 xπ2cos 2x,   0<xπ4I=-22cos 2x dxI=0π4cos 2x dx- π4π2 cos 2x dxI=sin 2x20π4-sin 2x2π4π2I=12-0-0+12I=1

Page No 19.56:

Question 12:

Evaluate the following integrals:
02πsin x dx

Answer:

02πsin x dxWe know that, sin x=- sin x ,π x2πsin x,   0<xπI=02πsin x dxI=0π sin x dx+π2π- sin x dxI=-cos x0π+cos xπ2πI=1+1+1--1I=4

Page No 19.56:

Question 13:

Evaluate the following integrals:
-π/4π/4sin x dx

Answer:

-π4π4sin x dxWe know that, sin x=- sin x ,-π4 x0sin x,   0<xπ4I=-π4π4sin x dxI=-π40-sin x dx +0π4 sin x dxI=cos x-π40-cos x0-π4I=1-12-12+1I=2-22I=2-2

Page No 19.56:

Question 14:

Evaluate the following integrals:
28x-5 dx

Answer:

28x-5 dxWe know that, x-5=-x-5 , 2x5x-5,   5<x8I=28x-5 dxI=25-x-5 dx+58 x-5 dxI=-x22-5x25+x22-5x58I=-252+25+2-10+32-40-252+25I=9

Page No 19.56:

Question 15:

Evaluate the following integrals:
-π/2π/2sin x+cos x dx

Answer:

-π2π2sin x+ cos x dxSince,  f-x=sin -x + cos -x=sin x + cos x=fxSo, fx is an even function.I=20π2sin x+cos x dxI=2-cos x+sin x0π2I=20+1+1-0I=4

Page No 19.56:

Question 16:

Evaluate the following integrals:
04x-1 dx

Answer:

04x-1 dxWe know that, x-1=-x-1 , 0x1x-1,   1<x4I=04x-1 dxI=01-x-1 dx+14x-1 dxI=-x22+x01+x22-x14I=-12+1-0+8-4-12+1I=5

Page No 19.56:

Question 17:

Evaluate the following integrals:
14x-1+x-2+x-4 dx

Answer:

I=14x-1+x-2+x-4 dxI=14x-1 dx+14x-2 dx+14x-4 dxWe know that, x-1=-x-1 , x1x-1,   1<x4x-2=-x-2 , 1x2x-2,  2<x4x-4=-x-4 , 1x4x-4,  x>4I=14x-1 dx-12x-2 dx+24x-2 dx-14x-4 dxI=x22-x14-x22-2x12+x22-2x24-x22-4x14I=8-4-12+1-2-4-12+2+8-8-2+4-8-16-12+4I=232

Page No 19.56:

Question 18:

Evaluate the following integrals:
-50fx dx, where fx=x+x+2+x+5

Answer:

I=-50x+x+2+x+5 dxI=-50x dx+-50x+2 dx+-50x+5 dxWe know that, x=-x , -5x0x,   x>0x+2=-x+2 , -5x-2x+2,  -2<x0x+5=-x+5 , -5x0x+5,  x>-5I=--50x dx--5-2x+2 dx+-20x+2 dx+-50x+5 dxI=-x22-50-x22+2x-5-2+x22+2x-20+x22+5x-50I=252-2-4-252+10-2+4+-252+25I=632

Page No 19.56:

Question 19:

Evaluate the following integrals:
04x+x-2+x-4 dx

Answer:

I=04x+x-2+x-4 dxI=04x dx+04x-2 dx+04x-4 dxWe know that, x=-x , -5x0x,   x>0x-2=-x-2 , 0x2x-2,  2<x4x-4=-x-4 , 0x4x-4,  x>4I=04x dx-02x-2 dx+24x-2 dx-04x-4 dxI=x2204-x22-2x02+x22-2x24-x22-4x04I=8-2-4+8-8-2+4-8-16I=20

Page No 19.56:

Question 20:

-12x+1+x+x-1dx

Answer:

We know that

x+1=x+1,if x+10-x+1,if x+1<0=x+1,if x-1-x+1,if x<-1

x=x,if x0-x,if x<0

x-1=x-1,if x-10-x-1,if x-1<0=x-1,if x1-x-1,if x<1

When -1x0,

x+1+x+x-1=x+1+-x+-x-1=2-x

When 0x1,

x+1+x+x-1=x+1+x+-x-1=x+2

When 1x2,

x+1+x+x-1=x+1+x+x-1=3x

-12x+1+x+x-1dx=-102-xdx+01x+2dx+123xdx=2-x22×-1-10+x+22201+3×x2212=-124-9+129-4+324-1=52+52+92=192

Page No 19.56:

Question 21:

-22xexdx

Answer:


Consider fx=xex.

Now,

f-x=-xe-x=-xex=-fx

f(x) is an odd function.

-22xexdx=0                                       -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx

Page No 19.56:

Question 22:

-π4π2sinxsinxdx

Answer:


-π4π2sinxsinxdx=-π40sinxsinxdx+0π2sinxsinxdx=-π40sinx-sinxdx+0π2sinxsinxdx                 sinx=sinx,0xπ2-sinx,-π4x0=--π40sin2xdx+0π2sin2xdx
=--π401-cos2x2dx+0π21-cos2x2dx=-12-π40dx+12-π40cos2xdx+120π2dx-120π2cos2xdx=-12×x-π40+12×sin2x2-π40+12×x0π2-12×sin2x20π2=-120+π4+140+sinπ2+12×π2-0-14sinπ-0=-π8+140+1+π4-140-0=π8+14

Page No 19.56:

Question 23:

0πcosxcosxdx

Answer:


Consider fx=cosxcosx

Now,

fπ-x=cosπ-xcosπ-x=-cosx-cosx=-cosxcosx=-fx

0πcosxcosxdx=0                         02afxdx=20afxdx,if f2a-x=fx0,if f2a-x=-fx

Page No 19.56:

Question 24:

-π2π22sinx+cosxdx

Answer:


Consider fx=2sinx+cosx

Now,

f-x=2sin-x+cos-x=2sinx+cosx=fx

-π2π22sinx+cosxdx=20π22sinx+cosxdx                            -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx=20π22sinx+cosxdx                                  x=x,if x0-x,if x<0=40π2sinxdx+20π2cosxdx
=4×-cosx0π2+2×sinx0π2=-4cosπ2-cos0+2sinπ2-sin0=-40-1+21-0=4+2=6

Page No 19.56:

Question 25:

-π2πsin-1sinxdx

Answer:


-π2πsin-1sinxdx=-π2π2sin-1sinxdx+π2πsin-1sinxdx=-π2π2xdx+π2ππ-xdx                    π2xπ-π-x-π20π-xπ2=x22-π2π2+π-x22×-1π2π=12π24-π24-120-π24
=0+π28=π28

Page No 19.56:

Question 26:

-π2π2-π2cosxsin2xdx

Answer:


Let I=-π2π2-π2cosxsin2xdx=-π2-π2π21cosxsin2xdx=-π2-π2π21cosxsinxdx=-π2×20π21cosxsinxdx                     f-x=cos-xsin-x=cosx-sinx=cosxsinx=fx
=-π0π21cosxsinxdx                            sinx=sinx, 0xπ2=-π0π2sinxcosx1-cos2xdx

Put cosx = z2

-sinxdx=2zdz

When x0, z1

When xπ2, z0

I=2π10zdzz1-z4=2π10dz1-z4=2π10dz1-z1+z1+z2
Now,

11-z1+z1+z2=A1-z+B1+z+Cz+D1+z21=A1+z1+z2+B1-z1+z2+Cz+D1-z1+z

Putting z = 1, we get

A=14

Putting z = −1, we get

B=14

Putting z = 0, we get

1=A+B+DD=1-14-14=12

Equating coefficient of z3 on both sides, we get

A-B+C=014-14+C=0C=0

I=2π10dz1-z1+z1+z2=2π10141-zdz+2π10141+zdz+2π10121+z2dz=2π4×log1-z-110+2π4×log1+z10+2π2×tan-1z10=-π2log1-log0+π2log1-log2+πtan-10-tan-11=-π20--+π20-log2+π0-π4=--π2log2-π24=-


Disclaimer: The answer does not matches with the answer provided for the question.

Page No 19.56:

Question 27:

022xxdx

Answer:


022xxdx=012xxdx+122xxdx=012x×0dx+122x×1dx             x=0,0x<11,1x<2=0+212xdx=2×x2212=4-1=3

Page No 19.56:

Question 28:

02πcos-1cosxdx

Answer:


02πcos-1cosxdx=0πcos-1cosxdx+π2πcos-1cosxdx=0πxdx+π2π2π-xdx                               πx2π-2π-x-π02π-xπ
=x220π+2π-x22×-1π2π=12π2-0-120-π2=π22+π22=π2



Page No 19.61:

Question 1:

Evaluate each of the following integrals:

02πesinxesinx+e-sinxdx

Answer:


Let I = 02πesinxesinx+e-sinxdx                     .....(1)

Then,

I=02πesin2π-xesin2π-x+e-sin2π-xdx                       0afxdx=0afa-xdx=02πe-sinxe-sinx+esinxdx                           .....2

Adding (1) and (2), we get

2I=02πesinx+e-sinxesinx+e-sinxdx2I=02πdx2I=x02π2I=2π-0I=π

Page No 19.61:

Question 2:

Evaluate each of the following integrals:

02πlogsecx+tanxdx

Answer:


Let I = 02πlogsecx+tanxdx                         .....(1)

Then,

I=02πlogsec2π-x+tan2π-xdx                           0afxdx=0afa-xdx=02πlogsecx-tanxdx                                 .....2

Adding (1) and (2), we get

2I=02πlogsecx+tanx+logsecx-tanxdx2I=02πlogsec2x-tan2xdx2I=02πlog1dx                               1+tan2x=sec2x2I=0                                                 log1=0I=0

Page No 19.61:

Question 3:

Evaluate each of the following integrals:

π6π3tanxtanx+cotxdx

Answer:


Let I = π6π3tanxtanx+cotxdx                             .....(1)

Then,

I=π6π3tanπ3+π6-xtanπ3+π6-x+cotπ3+π6-xdx                                  abfxdx=abfa+b-xdx=π6π3tanπ2-xtanπ2-x+cotπ2-xdx=π6π3cotxcotx+tanxdx                                        .....2

Adding (1) and (2), we get

2I=π6π3tanx+cotxtanx+cotxdx2I=π6π3dx2I=xπ6π32I=π3-π6=π6I=π12

Page No 19.61:

Question 4:

Evaluate each of the following integrals:

π6π3sinxsinx+cosxdx

Answer:


Let I = π6π3sinxsinx+cosxdx                             .....(1)
Then,
I=π6π3sinπ3+π6-xsinπ3+π6-x+cosπ3+π6-xdx                                  abfxdx=abfa+b-xdx=π6π3sinπ2-xsinπ2-x+cosπ2-xdx=π6π3cosxcosx+sinxdx                                        .....2

Adding (1) and (2), we get

2I=π6π3sinx+cosxsinx+cosxdx2I=π6π3dx2I=xπ6π32I=π3-π6=π6I=π12

Page No 19.61:

Question 5:

Evaluate each of the following integrals:

-π4π4tan2x1+exdx

Answer:


Let I = -π4π4tan2x1+exdx                               .....(1)
Then,
I=-π4π4tan2π4+-π4-x1+eπ4+-π4-xdx                        abfxdx=abfa+b-xdx=-π4π4tan2-x1+e-xdx=-π4π4tan2x1+1exdx=-π4π4extan2xex+1dx                                   .....2

Adding (1) and (2), we get

2I=-π4π4tan2x1+ex+extan2x1+exdx2I=-π4π41+extan2x1+exdx2I=-π4π4tan2xdx2I=-π4π4sec2x-1dx
2I=-π4π4sec2xdx--π4π4dx2I=tanx-π4π4-x-π4π42I=tanπ4-tan-π4-π4--π42I=1+1-2π42I=2-π2I=1-π4


Disclaimer: This answer does not matches with the given answer in the book.

Page No 19.61:

Question 6:

Evaluate each of the following integrals:

-aa11+axdx , a > 0  

Answer:


Let I = -aa11+axdx                 .....(1)
Then,
I=-aa11+aa+-a-xdx=-aa11+a-xdx=-aaaxax+1dx                              .....2

Adding (1) and (2), we get

2I=-aa1+ax1+axdx2I=-aadx2I=x-aa2I=a--a=2aI=a

Page No 19.61:

Question 7:

Evaluate each of the following integrals:

-π3π311+etanxdx

Answer:


Let I = -π3π311+etanxdx                      .....(1)
Then,
I=-π3π311+etanπ3+-π3-xdx                            0afxdx=0afa-xdx=-π3π311+etan-xdx=-π3π311+e-tanxdx=-π3π3etanxetanx+1dx                                      .....2

Adding (1) and (2), we get

2I=-π3π31+etanx1+etanxdx2I=-π3π3dx2I=x-π3π32I=π3--π3=2π3I=π3

Page No 19.61:

Question 8:

Evaluate each of the following integrals:

-π2π2cos2x1+exdx

Answer:


Let I = -π2π2cos2x1+exdx                        .....(1)
Then,
I=-π2π2cos2π2+-π2-x1+eπ2+-π2-xdx                        abfxdx=abfa+b-xdx=-π2π2cos2-x1+e-xdx=-π2π2excos2xex+1dx                           .....2

Adding (1) and (2), we get

2I=-π2π2cos2x1+ex+excos2x1+exdx2I=-π2π2cos2x1+ex1+exdx2I=-π2π2cos2xdx2I=-π2π21+cos2x2dx
2I=12-π2π2dx+12-π2π2cos2xdx2I=12×x-π2π2+12×sin2x2-π2π22I=12π2--π2+14sinπ-sin-π2I=12×π+140+0                                               sin-π=-sinπ=02I=π2I=π4

Page No 19.61:

Question 9:

Evaluate each of the following integrals:

-π4π4x11-3x9+5x7-x5+1cos2xdx

Answer:


Let I=-π4π4x11-3x9+5x7-x5+1cos2xdx=-π4π4x11-3x9+5x7-x5cos2xdx+-π4π41cos2xdx=-π4π4x11-3x9+5x7-x5cos2xdx+-π4π4sec2xdx=I1+I2

Now,

Consider fx=x11-3x9+5x7-x5cos2x.

f-x=-x11-3-x9+5-x7--x5cos2-x=-x11+3x9-5x7+x5cos2x=-x11-3x9+5x7-x5cos2x=-fx

I1=-π4π4x11-3x9+5x7-x5cos2xdx=0                          -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx

Let gx=sec2x.

g-x=sec2-x=sec2x=gx

I2=-π4π4sec2xdx=20π4sec2xdx                         -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx=2×tanx0π4=2tanπ4-tan0=2×1-0=2

I=I1+I2=0+2=2

Page No 19.61:

Question 10:

Evaluate each of the following integrals:

abx1nx1n+a+b-x1ndx,  nN, n2

Answer:


Let I = abx1nx1n+a+b-x1ndx                            .....1
Then,
I=aba+b-x1na+b-x1n+a+b-a+b-x1ndx                       abfxdx=abfa+b-xdx=aba+b-x1na+b-x1n+x1ndx                                           .....2

Adding (1) and (2), we get

2I=abx1n+a+b-x1nx1n+a+b-x1ndx2I=abdx2I=xab=b-aI=b-a2

Page No 19.61:

Question 11:

0π/22 log cos x-log sin 2x dx

Answer:

Let I =0π22 log cosx-logsin2xdx    =0π22 log cosx-log2sinx cosxdx    =0π22logcosx-log2-logsinx-logcosxdx    =0π2logcosx-log2-logsinxdx    =0π2logcosx dx-0π2log2  dx -0π2logsinx  dx    =0π2logcosx dx-0π2log2  dx -0π2logsinπ2-x  dx                  Using0afx dx=0afa-x dx      =0π2logcosx dx-0π2log2  dx-0π2logcosx dx    =-log2 x0π2   =-π2 log2

Page No 19.61:

Question 12:

0axx+a-x dx

Answer:

Let I=0axx+a-xdx         ... (i)          I=0aa-xa-x+xdx                                    Using, 0afx dx=0afa-x dx    I=0aa-xx+a-xdx     ... (ii)Adding (i) and (ii)2I=0ax+a-xx+a-x dx     =0a dx=x0a=aHence I=a2

Page No 19.61:

Question 13:

05x+44x+44+9-x4 dx

Answer:

Let I=05x+44x+44-9-x4dx       ... (i)I=059-x49-x4-x+44dx                        Using 0afxdx=0afa-xdxI=-059-x4x+44-9-x4dx             ... (ii)Adding (i) and (ii)2I=05x+44x+44-9-x4-9-x4x+44-9-x4dx    =05x+44-9-x4x+44-9-x4dx   =05dx   =x05   =5Hence I= 52

Page No 19.61:

Question 14:

07x3x3+73-x dx 

Answer:

Let  I=07x3x3+7-x3dx     ...  (i)        =077-x37-x3+x3 dx                                    Using  0afx dx=0afa-x dx        =077-x3x3+7-x3 dx       ...    (ii)          Adding (i) and (ii) we get2I=07x3+7-x3x3+7-x3dx      =     07dx   =x07=7Hence I =72

Page No 19.61:

Question 15:

π/6π/311+tan x dx 

Answer:

Let  I =π6π311+tanxdx            ... (i)         =π6π311+tanπ3+π6-xdx         =π6π311+cotxdx          ... (ii)Adding (i) and (ii)2I=π6π311+tanx+11+cotx  dx           =π6π31+cotx+1+tanx1+cotx+tanx+tanx cotx dx      =π6π32+cotx+tanx2+cotx+tanx dx    =π6π3 dx  =xπ6π3     =π3-π62I=π6Hence I=π12

Page No 19.61:

Question 16:

If fa+b-x=fx, then prove that abxfxdx=a+b2abfxdx.

Answer:


abxfxdx=aba+b-xfa+b-xdx                      abfxdx=abfa+b-xdx=aba+b-xfxdx                                   fa+b-x=fxabxfxdx=aba+bfxdx -abxfxdx 
abxfxdx+abxfxdx=a+babfxdx2abxfxdx=a+babfxdxabxfxdx=a+b2abfxdx



Page No 19.95:

Question 1:

0π/211+tan x

Answer:

Let I=0π211+tanxdx              ...  (i)=0π211+tanπ2-xdx       Using 0afxdx= 0afa-xdx=0π211+cotxdx          ...   (ii)Adding (i) and (ii)  2I =0π211+tanx+11+cotxdx     =0π21+cotx+1+tanx1+tanx1+cotx dx     =0π22+tanx+cotx1+tanx+cotx+tanxcotxdx     =0π22+tanx+cotx2+tanx+cotxdx     =0π2 dx     =x0π2  2I =π2 I =π4   

Page No 19.95:

Question 2:

0π/211+cot x dx

Answer:

Let I=0π211+cotxdx              ... (i)= 0π211+cotπ2-xdx            Using 0afxdx=0afa-xdx=0π211+tanxdx                 ... (ii)   Adding (i)  and (ii)2I=0π211+cotx+11+tanx    dx       =0π21+tanx+ 1+cotx1+cotx1+tanx dx    =0π22+tanx+ cotx1+tanx +cotx + tanx cotxdx     =0π22+tanx+ cotx2+tanx+ cotx dx    =0π2dx     = x0π2=π2Hence , I=π4

Page No 19.95:

Question 3:

0π/2cot xcot x+tan x dx

Answer:

Let   I=0π2cotxcotx+tanxdx          ...(i) =0 π2cotπ2-xcotπ2-x+tanπ2-xdx              Using 0afx dx=0afa-x dx=  0π2tanxtanx+cotx dx                ...(ii) Adding (i) and (ii)2I=0π2cotxcotx+tanx+tanxtanx+cotx  dx        =0π2dx         =x0π2      =π2Hence, I =π4​

Page No 19.95:

Question 4:

0π/2sin3/2 xsin3/2 x+cos3/2 x dx

Answer:

Let   I=0π2sinnxsinnx+ cosnxdx      ...  (i)= 0π2sinnπ2-xsinnπ2-x+ cosnπ2-xdx        Using 0afx dx=0afa-x dx= 0π2cosnxcosnx+ sinnx dx = 0π2cosnxsinnx+ cosnx dx              ... (ii)Adding (i) and (ii) we get2I  =0π2sinnxsinnx+ cosnx+cosnxsinnx+ cosnx dx    =0π2sinnx+ cosnxsinnx+ cosnx dx= 0π2 dx =x0π2=π2Hence I=π4i.e.,0π2sinnxsinnx+ cosnxdx=π40π2sin3/2xsin3/2x+ cos3/2xdx=π4

Page No 19.95:

Question 5:

0π/2sinn xsinn x+cosn x dx

Answer:

Let  I =0π2sinnxsinnx+ cosnxdx            ... (i)  =0π2sinnπ2-xsinnπ2-x +cosnπ2-x dx=0π2 cosnx cosnx+sinnx dx =0π2 cosnx sinnx+ cosnx dx          ... (ii)Adding (i) and (ii) we get2I   =0π2sinnxsinnx+ cosnx+ cosnx sinnx+ cosnxdx    =0π2sinnx + cosnx sinnx+ cosnx dx   =0π21 dx  =0π2 dx=x0π2=π2Hence I =π4

Page No 19.95:

Question 6:

0π/211+tan x dx

Answer:

Let I = 0π211+tanxdx         ...(i)= 0π211+tanπ2-xdx                   Using   0afx dx=0afa-x dx=0π211+cotxdx         ...(ii)    Adding (i) and (ii) we get2I =  0π211+tanx+11+cotxdx      =0π21+cotx+1+tanx1+tanx 1+cotx dx      =0π21+cotx+1+tanx1+cotx+tanx+tanx cotx dx    =0π22+cotx+tanx 2+cotx+tanx dx=  0π2 dx =x0π2=π2Hence   I = π4

Page No 19.95:

Question 7:

0a1x+a2-x2 dx

Answer:

We have, I=0a1x+a2-x2dxPutting x=a sin θdx=a cos θ dθWhen x0; θ0 And xa; θπ2I=0π2a cos θ a sin θ+a2-a sin θ2dθ=0π2a cos θ a sin θ+a cos θdθI=0π2cos θ sin θ+cos θdθ     .....1I=0π2cos π2-θ sin π2-θ +cos  π2-θ dθ=0π2sin θcos θ+sin  θdθI=0π2sin θ sin θ+cos θdθ     .....2By adding 1 and 2, we get2I=0π2cos θ +sin θsin θ+cos θdθ 2I=0π2dθ 2I=θ0π22I=π2I=π4

Page No 19.95:

Question 8:

0log x1+x2 dx

Answer:

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

I=0π2log tan θ1+tan2 θ sec2θ dθI=0π2log tan θ dθ     .....1I=0π2log tan π2-θ dθ        0afxdx=0afa-xdxI=0π2log cot θ dθ     .....2Adding 1 and 2, we get

2I=0π2log tan θ dθ+0π2log cot θ dθ=0π2log tan θ+log cot θ dθ=0π2log tan θ×cot θ  dθ=0π2log 1  dθ=0π20  dθ2I=0I=00log x1+x2 dx=0

Page No 19.95:

Question 9:

01log1+x1+x2 dx

Answer:

We have,I=01log 1+x1+x2 dxPutting x=tan θdx=sec2 θ dθWhen x0 ; θ0and x1 ; θπ4Now, integral becomes

I=0π4log 1+tan θsec2 θ sec2 θ dθI=0π4log 1+tan θ dθ     .....1I=0π4log1+tan π4-θ dθ        0afxdx=0afa-xdx=0π4log1+tanπ4-tan θ1+tanπ4 tan θ dθ=0π4log1+1-tan θ1+tan θ dθ=0π4log21+tan θ dθI=0π4log 2-log 1+tan θ dθ     .....2

Adding 1 and 2, we get2I=0π4log 2 dθ2I=log 2θ0π42I=π4log 2I=π8log 201log1+x1+x2dx=π8log 2

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Question 10:

0x1+x1+x2 dx

Answer:

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

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Question 11:

0πx tan xsec x cosec x dx

Answer:

Let  I =0πx tanxsecx cosecxdx                     ...(i)=0ππ-x tanπ-xsecπ-x cosecπ-xdx           Using 0afxdx=0afa-xdx=0π-π-xtanx-secx cosecxdx=0ππ-xtanxsecx cosecxdx        ...(ii)Adding (i) and (ii)2I=0πx tanxsecx cosecx+π-xtanxsecx cosecxdx      =0πx+π-x tanxsecx cosecxdx         =0π πtanxsecx cosecxdx         =0ππsin2x dx           =π0π1- cos2x  dx          =π x0π-π20π1+cos2x   dx                        =π2x0π-π2sin2x20π                   =π22Hence,   I =  π24        

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Question 12:

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) we get2I= 0πx+π-x sinx cos4x dx    = π0π sinx cos4x dx        Let cos x= t, Then -sinx dx = dt,       When x=0, t=1, x=π, t=-1Therefore, 2I=-π1-1t4 dt                    =π-11t4 dt                    =πt55-11                    =π5+π5                    =2π5Hence  I =π5

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Question 13:

0πx sin3 x dx

Answer:

Let  I=0πx sin3x dx          ...(i)        =0ππ-x sin3π-x dx    =0ππ-x sin3x  dx      ...(ii)Adding (i) and (ii)  we get2I=0πx+π-xsin3x dx    =0ππ sin3x dx    =0ππ 3 sin x -sin 3x4 dx    =π40π 3 sin x -sin 3x dx    =π4-3 cos x+cos 3x30π    =π4-3 cos π+3cos 0+cos 3π3-cos 03    =π43+3+-13-13    =π23-13    =π2×83    =4π3I = 2π3

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Question 14:

0πx log sin x dx

Answer:

0πx log sinxdxLet I =0πx logsinxdx               .....(i)     I=0ππ-x log sinπ-x dx    I=0ππ-x logsin x  dx          .....(ii)Adding (i) and (ii)2I=π0π log sinx dx   =2π0π2 log sinx dx I=π0π2log sinx dx           .....(iii)Let0π2log sinx dx= I2I2=0π2log sinπ2-x dx   =0π2log cosx dx2I2=0π2log sinx+log cosx dx=0π2logsinx cosx dx=0π2logsin2x dx-0π2log 2 dxLet 2x=t2dx=dtwhen,x=0t=0x=0  t=π2I2=120πlog sint dt-π2log 22I2=220π2log sint dt-π2log 22I2=I2-π2log 2I2=-π2log 2From iii, I=π0π2log sinx dx=π I2I=π-π2log 2I=-π2 log 22

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Question 15:

0πx sin x1+sin x dx

Answer:

Let  I =0πx sinx1+ sinxdx        ... (i)     =0ππ-xsinπ-x1+sinπ-x dx       =0ππ-x sinx1+ sinxdx    ...  (ii)Adding (i) and (ii) we get  2I=0πx+π-x sinx1+ sinxdx      =0ππ sinx1+ sinxdx   =π0π1+sinx-11+sinxdx    =π0πdx-π0π11+sinxdx    =π0πdx-π0π1-sinx1+sinx1-sinxdx    =π0πdx-π0π1-sinx1-sin2xdx    =π0πdx-π0π1-sinxcos2xdx    =π0πdx-π0πsec2x-secx tanxdx    =πx0π-πtanx-secx0π    =π2-π0+1-0+1    =π2-2πHence I=ππ2-1

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Question 16:

0πx1+cos α sin x dx, 0<α<π

Answer:

We have,  I =0πx1+cosα sinxdx           .....1         =0ππ-x1+cosα sinπ-xdx          =0ππ-x1+cosα sinxdx              .....2Adding 1 and 2 we get,2I=0πx+π-x1+cosα sinxdxI=π20π11+cosα sinx dx

= π20π11+cosα sinx=π20π11+cosα 2tanx21+tan2x2dx=π20π1+tan2x21+tan2x2+2cosα tan x2dx=π20πsec2x21+tan2x2+2cosα tan x2dx
Putting tanx2=t12sec2x dx=dtWhen x0; t0and xπ; tI=π2021+t2+2cosα tdt=π202t+cosα2-cos2α+1dt=π01t+cosα2+sin2αdt=π1sin αtan-1t+cos αsin α01=πsinαtan-1-tan-1cotα=πsinαπ2-tan-1tanπ2-α=παsinα

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Question 17:

0πx cos2 x dx

Answer:

Let I=0πx cos2x dx     ... (i)              =0ππ-x cos2π-x dx       =0ππ-x cos2x  dx       ... (ii)Adding (i) and (ii) we get2I=0πx+π-x cos2x  dx   =0ππ cos2x  dx   =π0π1+cos2x2   dx   =π20π1+cos2x   dx    =π2x+sin2x20π    =π2π-0    Hence I=π24

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Question 18:

Evaluate the following integrals:

π6π311+cot32xdx

Answer:


Let I = π6π311+cot32xdx                             .....(1)
Then,

I=π6π311+cot32π3+π6-xdx                            abfxdx=abfa+b-xdx=π6π311+cot32π2-xdx=π6π311+tan32xdx=π6π3cot32xcot32x+1dx                                          .....2

Adding (1) and (2), we get

2I=π6π31+cot32x1+cot32xdx2I=π6π3dx2I=xπ6π32I=π3-π6=π6I=π12

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Question 19:

Evaluate the following integrals:

0π2tan7xtan7x+cot7xdx

Answer:


Let I = 0π2tan7xtan7x+cot7xdx                                  .....(1)
Then,
I=0π2tan7π2-xtan7π2-x+cot7π2-xdx                      0afxdx=0afa-xdx=0π2cot7xcot7x+tan7xdx                                                .....2

Adding (1) and (2), we get

2I=0π2tan7x+cot7xtan7x+cot7xdx2I=0π2dx2I=x0π22I=π2-0=π2I=π4

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Question 20:

Evaluate the following integrals:

2810-xx+10-xdx

Answer:


Let I = 2810-xx+10-xdx                            .....(1)
Then,
I=2810-2+8-x2+8-x+10-2+8-xdx                            abfxdx=abfa+b-xdx=28x10-x+xdx                                                        .....2

Adding (1) and (2), we have

2I=2810-x+xx+10-xdx2I=28dx2I=x282I=8-2=6I=3

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Question 21:

Evaluate the following integrals:

0πxsinxcos2xdx                      [NCERT EXEMPLAR]

Answer:


Let I = 0πxsinxcos2xdx              .....(1)
Then,
I=0ππ-xsinπ-xcos2π-xdx                  0afxdx=0afa-xdx=0ππ-xsinxcos2xdx                    .....2

Adding (1) and (2), we have

2I=0ππ-x+xsinxcos2xdx2I=π0πsinxcos2xdx2I=-π0πcos2x-sinxdx2I=-π×cos3x30π                                       fxnf'xdx=fxn+1n+1+C2I=-π3cos3π-cos20
2I=-π3-1-1=2π3I=π3

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Question 22:

0π/2x sin x cos xsin4 x+cos4 x dx 

Answer:

Let  I=0π2x sinx cosxsin4x+cos4xdx              ...(i)        =0π2π2-x sinπ2-x cosπ2-xsin4π2-x+cos4π2-xdx        =0π2π2-xcosx sinxcos4x+sin4x dx        =0π2π2-xsinx cosxsin4x+cos4x dx           ... (ii)Adding (i) and (ii) we get2I =0π2x+π2-xsinx cosxsin4x+cos4xdx    =π20π2sinx cosxsin4x+cos4xdxLet sin2x=t, Then 2 sinx cosx dx=dtWhen x=0, t=0, x=π2, t=1Therefore2I =π401dtt2+1-t2     =π801dtt-122+14    =π8×2tan-12t-101    =π4π4+π4Hence I=π216 



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Question 23:

-π/2π/2sin3 x dx

Answer:

Let  I=-π2π2sin3x dx       =-π2π2 sinx1-cos2xdx       =-π2π2sinx dx--π2π2sinx cos2x dx       =-cosx-π2π2+cos3x3-π2π2       =0 +0 =0

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Question 24:

-π/2π/2sin4 x dx

Answer:

Let I= -π2π2sin4x dx       =-π2π2sin2x2 dx       =-π2π21-cos2x22 dx      =14-π2π21-2cos2x+cos22x  dx     =14-π2π2dx-12-π2π2cos2x dx+18-π2π21+cos4x dx     =14-π2π2dx-12-π2π2cos2x dx+18-π2π2dx+18-π2π2cos4x dx     =38-π2π2dx-12-π2π2cos2x dx+18-π2π2cos4x dx    =38x-π2π2-14sin2x-π2π2+132sin4x-π2π2   =38π2+π2-140-0+1320-0Hence I=3π8

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Question 25:

(i) -11log2-x2+x dx

(ii) -ππ1-x2 sin x cos2x dx

Answer:

(ii) -ππ1-x2 sinx cos2x dx


Let I=-ππ1-x2 sinx cos2x dxWe know,-aafxdx=0                          ,if f-x=-fx20afxdx              ,if f-x=fxHere,fx=1-x2 sinx cos2xf-x=1--x2 sin-x cos2-x        =1-x2 sin-x cos2x        =-1-x2 sinx cos2x        =-fxThus,-ππ1-x2 sinx cos2x dx=0


Hence, -ππ1-x2 sinx cos2x dx = 0.

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Question 26:

-π/4π/4sin2 x dx

Answer:

Let I=-π4π4sin2x  dxHere fx = sin2xf-x=sin2-x=sin2x =fxHence sin2x is an even functionTherefore,I=20π4sin2x  dx      =20π41-cos2x2dx      =0π41-cos2x dx        =x-sin2x20π4      =π4-12

Page No 19.96:

Question 27:

0πlog1-cos x dx

Answer:

Let, I=0πlog1-cosx dx       =0πlog2sin2x2 dx       =0πlog2 dx+ 20πlog sinx2  dx Let, t =x2 in the secong integral. then dt= 12dxWhen x0 ; t0 and xπ ; tπ2I=log2x0π+40π2log sint dt =πlog2+4×-π2log2              Where, 0π2log sint dt=-π2log2 =-π log2

Page No 19.96:

Question 28:

-π/2π/2log2-sin x2+sin x dx

Answer:

Let I=-π2π2log2-sinx2+sinxdxHere, fx=log2-sinx2+sinxf-x=log2-sin-x2+sin-x=log2+sinx2-sinx=-log2-sinx2+sinx=-fxHence fx is an odd function I =0

Page No 19.96:

Question 29:

Evaluate the following integrals:

-ππ2x1+sinx1+cos2xdx

Answer:


Let I = -ππ2x1+sinx1+cos2xdx
Then,
I=-ππ2x1+cos2xdx+-ππ2xsinx1+cos2xdx=I1+I2

Consider fx=2x1+cos2x.
Now,
f-x=2-x1+cos2π-x=-2x1+-cosx2=-2x1+cos2x=-fx
I1=-ππ2x1+cos2xdx=0                       -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx

Again, consider gx=2xsinx1+cos2x.

g-x=2-xsin-x1+cos2-x=2xsinx1+cos2x=gx                 sin-x=-sinx and cos-x=cosx
I2=-ππ2xsinx1+cos2xdx=2×20πxsinx1+cos2xdx                       -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx=40πxsinx1+cos2xdx                  .....1  
Then,
I2=40ππ-xsinπ-x1+cos2π-xdx=40ππ-xsinx1+cos2xdx                .....2                 0afxdx=0afa-xdx

Adding (1) and (2), we get

2I2=40ππsinx1+cos2xdx2I2=4π0πsinx1+cos2xdx

Put cosx = z

-sinxdx=dz

When x0, z1

When xπ, z-1

2I2=-4π1-1dz1+z22I2=-4π×tan-1z1-12I2=-4πtan-1-1-tan-112I2=-4π-π4-π4=2π2I2=π2

I=I1+I2=0+π2=π2

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Question 30:

Evaluate the following integrals:

-aaloga-sinθa+sinθdθ

Answer:


Let I = -aaloga-sinθa+sinθdθ

Consider fθ=loga-sinθa+sinθ.

f-θ=loga-sin-θa+sin-θ          =loga+sinθa-sinθ                               sin-x=-sinx          =loga-sinθa+sinθ-1          =-loga-sinθa+sinθ                           logab=bloga          =-fθ

f-θ=-fθI=-aaloga-sinθa+sinθdθ=0                    -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx

Page No 19.96:

Question 31:

Evaluate the following integrals:

-223x3+2x+1x2+x+1dx

Answer:


Let I = -223x3+2x+1x2+x+1dx

=-223x3x2+x+1dx+-222x+1x2+x+1dx=I1+I2

Consider fx=-223x3x2+x+1dx.

f-x=-223-x3-x2+-x+1dx=-22-3x3x2+x+1dx=--223x3x2+x+1dx=-fx

I1=-223x3x2+x+1dx=0                      -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx

Now, consider gx=-222x+1x2+x+1dx.

g-x=-222-x+1-x2+-x+1dx=-222x+1x2+x+1dx=gx

I2=-222x+1x2+x+1dx                      =2022x+1x2+x+1dx                                    -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx=2022x+1x2+x+1dx                                       x=x,x0-x,x<0=2×logx2+x+102                                       f'xfxdx=logfx+C=2×log7-log1=2×log7-0=2log7         

I=I1+I2=0+2log7=2log7

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Question 32:

Evaluate the following integrals:

-3π2-π2sin23π+x+π+x3dx

Answer:


Let I = -3π2-π2sin23π+x+π+x3dx

Put π+x=z

dx=dz

When x-3π2, z-π2

When x-π2, zπ2

I=-π2π2sin22π+z+z3dz=-π2π2sin2z+z3dz                                        sin2π+θ=sinθ=-π2π21-cos2z2dz+-π2π2z3dz
=12-π2π2dz-12-π2π2cos2zdz+-π2π2z3dz=12×z-π2π2-12×sin2z2-π2π2+z44-π2π2=12π2--π2-14sinπ-sin-π+14π416-π416
=12×π-140+0+14×0=π2

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Question 33:

02x2-x dx

Answer:

Let  I=02x2-xdx       =022-x2-2+xdx       =022-xxdx       =022x-xx    dx       =022x12-x32  dx       =2x3232-x525202       =43x32-25x5202        =823-825=16215

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Question 34:

01log1x-1 dx

Answer:

Let I =01log1x-1dx      ... (i)                 =01log11-x-1dx       Using 0af(x) dx = 0af(a-x) dx    I   =01logx1-x dx        ... (ii)Adding (i)  and (ii)2I=01log1-xx+logx1-x dx   =01 log1-xx×x1-x  dx    =01log1 dx      =0Hence I=0

Page No 19.96:

Question 35:

Evaluate the following integrals:

-11xcosπxdx                                  [NCERT EXEMPLAR]

Answer:


Let I = -11xcosπxdx

Consider fx=xcosπx.

f-x=-xcosπ-x=-xcosπx=xcosπx=fx

I=-11xcosπxdx=201xcosπxdx                         -aafxdx=20afxdx,if f-x=fx0,if f-x=-fx     
Now,

xcosπx=xcosπx,if 0x12-xcosπx,if 12<x1

I=2012xcosπxdx+121-xcosπxdx=2xsinπxπ012-1π012sinπxdx-2xsinπxπ121-1π121sinπxdx=212πsinπ2-0-2π×-cosπxπ012-21πsinπ-12πsinπ2+2π×-cosπxπ121=1π+2π2cosπ2-cos0+1π-2π2cosπ-cosπ2=1π-2π2+1π+2π2=2π

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Question 36:

Evaluate the following integrals:

0πx1+sin2x+cos7xdx

Answer:


Let I = 0πx1+sin2x+cos7xdx                     .....(1)
Then,

I=0ππ-x1+sin2π-x+cos7π-xdx=0ππ-x1+sin2x-cos7xdx                             .....2

Adding (1) and (2), we get

2I=0πx1+sin2x+cos7x+π-x1+sin2x-cos7xdx2I=π0π11+sin2xdx

Dividing the numerator and denominator by cos2x, we get

2I=π0πsec2xsec2x+tan2xdx2I=π0πsec2x1+2tan2xdx2I=2π0π2sec2x1+2tan2xdx                         02afxdx=20afxdx,if f2a-x=fx0,if f2a-x=-fx
Put tanx = z

Then sec2xdx=dz

When x0, z0

When xπ2, z

2I=2π0dz1+2z22I=2π×tan-12z20I=π2tan-1-tan-10I=π2×π2-0I=π222

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Question 37:

Evaluate :0πx1+sin α sin xdx

Answer:

Let I=0πx1+sinα sinxdxI=0ππ-x1+sinα sinπ- xdx                0afxdx=0afa-xdxI=0ππ1+sinα sinxdx-0πx1+sinα sinxdxI=0ππ1+sinα sinxdx-I2I=0ππ1+sinα sinxdx2I=π0π11+sinα sinxdx

Substituting sinx=2tanx21+tan2x2, we get2I=π0π1+tan2x21+tan2x2+sinα ×2tanx2dxI=π20πsec2x21+tan2x2+sinα ×2tanx2dxLet tanx2=t, dtanx2=dtsec2x2dx=2dtAlso, When x0, ttan0=0When xπ, ttanπ2= I=π202dtt2+2tsinα+1I=π01t+sinα2+cos2αdtI=πcosαtan-1t+sinαcosα0I=πcosαtan-1-tan-1tanαI=πcosαπ2-α

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Question 38:

Evaluate the following integrals:

02πsin100xcos101xdx

Answer:

Let I = 02πsin100xcos101xdx

Suppose fx=sin100xcos101x.

Now,

f2π-x=sin1002π-xcos1012π-x=-sinx100cosx101=sin100xcos101x=fx

I=02πsin100xcos101xdx=20πsin100xcos101xdx                      02afxdx=20afxdx,if f2a-x=fx0,if f2a-x=-fx
Again,

fπ-x=sin100π-xcos101π-x=sinx100-cosx101=-sin100xcos101x=-fx

I=2×0=0                    02afxdx=20afxdx,if f2a-x=fx0,if f2a-x=-fx

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Question 39:

Evaluate the following integrals:

0π2asinx+bsinxsinx+cosxdx

Answer:


Let I =0π2asinx+bcosxsinx+cosxdx              .....(1)

Then,
I=0π2asinπ2-x+bcosπ2-xsinπ2-x+cosπ2-xdx                     0afxdx=0afa-xdx
=0π2acosx+bsinxcosx+sinxdx                     .....(2)

Adding (1) and (2), we get

2I=0π2asinx+bcosxcosx+sinx+acosx+bsinxsinx+cosxdx2I=0π2asinx+bcosx+acosx+bsinxsinx+cosxdx2I=0π2a+bsinx+a+bcosxsinx+cosxdx2I=0π2a+bsinx+cosxsinx+cosxdx
2I=0π2a+bdx2I=a+b×x0π22I=a+b×π2-02I=π2a+bI=π4a+b

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Question 40:

03/2|x cosπx| dx

Answer:

Ans

Page No 19.96:

Question 41:

01 | xsin πx | dx

Answer:


For 0<x<1, x>0 and sinπx>0xsinπx>0
01xsinπxdx=01xsinπx dxLet I=xsinπx dx=xsinπx dx-ddxxsinπx dxdx=x-cosπxπ--cosπxπdx
=-xcosπxπ+sinπxπ2

Applying the limits, we get

01xsinπxdx=-xcosπxπ+sinπxπ201=-cosππ+sinππ2-0+0
=1π+0-0=1π

Page No 19.96:

Question 42:

Evaluate : 03/2x sin πxdx

Answer:


For 0<x<1, x>0 and sinπx>0xsinπx>0For 1<x<32, x>0 and sinπx<0xsinπx<0
032xsinπxdx=01xsinπx dx-132xsinπx dxLet I=xsinπx dx=xsinπx dx-ddxxsinπx dxdx=x-cosπxπ--cosπxπdx
=-xcosπxπ+sinπxπ2

Applying the limits, we get

032xsinπxdx=-xcosπxπ+sinπxπ201--xcosπxπ+sinπxπ2132=-cosππ+sinππ2-0+0--32cos3π2π+sin3π2π2--cosππ+sinππ2
=1π+0-0-1π2-1π+0=1π+1π2+1π=2π+1π2=2π+1π2

Page No 19.96:

Question 43:

If f is an integrable function such that f(2ax) = f(x), then prove that
02afx dx=20afx dx

Answer:

Let  I=02afxdxBy Additive propertyI=0afxdx+a2afxdxConsider the integral a2afxdxLet x=2a-t, then dx=-dtWhen x=a, t=a, x=2x, t=0Hence a2afxdx=-a0f2a-tdt                         =0af2a-tdt                        =0af2a-xdx                 Changeing the variableTherefore,I=0afxdx+0af2a-xdx  =0afxdx+0afxdx          Given 0afxdx=0af2a-xdx   =20afxdx

Hence Proved

Page No 19.96:

Question 44:

If f(2ax) = −f(x), prove that 02afx dx=0.

Answer:

Let  I=02afxdxUsing additive propertyI=0afxdx+a2afxdxConsider the integral a2afxdxLet x=2a-t, Then dx=-dtWhen x=a, t=a and x=2a, t=0Therefore,a2afxdx=-a0f2a-tdt               =0af2a-tdt               =0af2a-xdx                  changing the variableWe have  f2a-x=-fxTherefore,I=0afxdx-0afxdx =0

Page No 19.96:

Question 45:

If f is an integrable function, show that
(i) -aafx2 dx=20afx2 dx

(ii) -aax fx2 dx=0

Answer:

(i)
I=-aafx2dxHere gx=f(x2)g-x=f-x2=f(x2)=gx i.e, gx is even ThereforeI=20afx2dx        Using -aagxdx=20agxdx   when gx is even


(ii)
I=-aaxfx2dxLet gx=xfx2g-x=-xf-x2=-xfx2=-gx i.e, gx is odd ThereforeI=0        Using -aagxdx=0  when gx is odd

Page No 19.96:

Question 46:

If f (x) is a continuous function defined on [0, 2a]. Then, prove that
02afx dx=0afx+f2a-x dx

Answer:

Let I=02afxdxBy Additive propertyI=0afxdx+a2afxdxConsider the integral a2afxdxLet x=2a-t, then dx=-dtWhen x=a. t=a, x=2a, t=0Hence, a2afxdx=-a0f2a-tdt                           =0af2a-tdt=0af2a-xdx ThereforeI=0afxdx+0af2a-xdx      =0afx+f2a-x dxHence, proved.

Page No 19.96:

Question 47:

If fa+b-x=fx, then prove that

abxfxdx=a+b2abfxdx

Answer:


abxfxdx=aba+b-xfa+b-xdx                     abfxdx=abfa+b-xdxabxfxdx=aba+b-xfxdx                             fa+b-x=fxabxfxdx=aba+bfxdx-abxfxdx2abxfxdx=a+babfxdxabxfxdx=a+b2abfxdx



Page No 19.97:

Question 48:

If f(x) is a continuous function defined on [−a, a], then prove that
-aafx dx=0afx+f-x dx

Answer:

Let I=-aafxdxBy Additive propertyI=-a0fxdx+0afxdxLet x=-t, then dx = -dt,When x=-a, t=a, x=0, t=0Hence -a0fxdx=-a0f-tdt                               =0af-tdt =0af-xdx              Changing the variableTherefore,I=0af-xdx+0afxdx   =0afx+f-x dxHence, proved.

Page No 19.97:

Question 49:

Prove that: 0πxfsinxdx=π20πfsinxdx

Answer:


0πxfsinxdx=0ππ-xfsinπ-xdx                   0afxdx=0afa-xdx0πxfsinxdx=0ππ-xfsinxdx0πxfsinxdx=π0πfsinxdx-0πxfsinxdx20πxfsinxdx=π0πfsinxdx0πxfsinxdx=π20πfsinxdx

Page No 19.97:

Question 50:

Prove that: 0afx dx=0afa-xdx, and hence evaluate 01x21-xn dx

Answer:

To Prove: 0afx dx=0afa-xdx

I=0afx dxLet x=a-tdx=0-dtdx=-dtAlso, if x=0, t=aif x=a, t=0Thus,I=a0fa-t-dt =-a0fa-tdt =--0afa-tdt         abfxdx=-bafxdx =0afa-tdt =0afa-xdx                abftdt=abfxdxHence, 0afx dx=0afa-xdx


Now,
Let J=01x21-xn dx       =011-x21-1-xn dx         0afx dx=0afa-xdx       =011+x2-2x1-1+xn dx       =011+x2-2xxn dx       =01xn+xnx2-2xnxdx       =01xn+xn+2-2xn+1dx       =xn+1n+1+xn+3n+3-2xn+2n+201       =1n+1n+1+1n+3n+3-21n+2n+2       =1n+1+1n+3-2n+2


Hence, 01x21-xn dx=1n+1+1n+3-2n+2.​



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