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Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part B

  • Last Updated : 02 Feb, 2021

Evaluate of each of the following integrals (1-46):

Question 1. \int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx

Solution:

We have,

\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx = \int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx

\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx =\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx

Let

I=\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx —— 1

So,I= \int\limits_0^{\frac{π}{2}} \frac{cos(\frac{π}{2}-x)} {cos(\frac{π}{2}-x)+sin(\frac{π}{2}-x)}dx ,[\because \int\limits_0^af(x)= \int\limits_0^af(a-x)dx]

\int\limits_0^{π/2}\frac{sinx}{sinx+cosx}dx ———— 2

Hence, by adding 1 and 2 ..

2I=\int\limits_0^{π/2} \frac{cosx}{cosx+sinx}dx+ \int\limits_0^{π/2} \frac{sinx}{cosx+sinx}dx

2I=\int\limits_0^{π/2} \frac{cosx+sinx}{cosx+sinx}dx

2I=\int\limits_0^{π/2}dx

2I=\frac{π}{2}

I=\frac{π}{4}

Question 2. \int\limits_0^{\frac{π}{2}} \frac{1}{1+cotx}dx ,

Solution:

We have,

\int\limits_0^{\frac{π}{2}} \frac{1}{1+cotx}dx =\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx

Let, I=\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx —– 1

So,I= \int\limits_0^{\frac{π}{2}} \frac{sin(\frac{π}{2}-x)}{sin(\frac{π}{2}-x)+ cos(\frac{π}{2}-x)}dx —— 2

Adding 1 & 2 ——-

2I=\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx +\int\limits_0^{\frac{π}{2}} \frac{cosx}{sinx+cosx}dx

2I=\int\limits_0^{\frac{π}{2}} \frac{sinx+cosx}{sinx+cosx}dx

2I=\int\limits_0^{\frac{π}{2}} dx

2I=\frac{π}{2}

I=\frac{π}{4}

Question 3. \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx ,

Solution:

We have ,\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{\sqrt{\frac{cosx}{sinx}}} {\sqrt{\frac{cosx}{sinx}}+\sqrt{\frac{sinx}{cosx}}}dx = \frac{cosx}{cosx+sinx}

\therefore \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx

Let,\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx

So,

I=\int\limits_0^{\frac{π}{2}} \frac{cos(\frac{π}{2}-x)}{sin(\frac{π}{2}-x)+cos(\frac{π}{2}-x)}dx

I=\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx ————— 2

Adding 1 and 2 ——–

2I=\int\limits_0^{\frac{π}{2}} \frac{sinx+cosx}{sinx+cosx}dx

2I=\int\limits_0^{\frac{π}{2}} dx

I=\frac{π}{4}

Question 4. \int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx

Solution:

Let,I= \int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx —— 1

I=\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}(\frac{π}{2}-x)} {sin^{3/2}(\frac{π}{2}-x)+cos^{3/2}(\frac{π}{2}-x)}dx

I=\int\limits_0^{\frac{π}{2}} \frac{cos^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx ———- 2

Adding 1 & 2 —–

2I=\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x+cos^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx

2I=\int\limits_0^{\frac{π}{2}} dx

I=\frac{π}{4}

Question 5. \int\limits_0^{\frac{π}{2}} \frac{sin^nx}{sin^nx+cos^nx}dx

Solution:

Let,\int\limits_0^{\frac{π}{2}} \frac{sin^nx}{sin^nx+cos^nx}dx ———– (1)

So,

I=\int\limits_0^{\frac{π}{2}} \frac{sin^n(\frac{π}{2}-x)}{sin^n(\frac{π}{2}-x)+cos^n(\frac{π}{2}-x)}dx

I= \int\limits_0^{\frac{π}{2}} \frac{cos^nx}{sin^nx+cos^nx}dx

I=\frac{π}{4}

Question 6. \int\limits_0^{\frac{π}{2}} \frac{1}{1+\sqrt{tanx}}dx

Solution:

\int\limits_0^{\frac{π}{2}} \frac{1}{1+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx

I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx ———- 1

\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cos(\frac{π}{2}-x)}} {\sqrt{cos(\frac{π}{2}-x)}+\sqrt{sin(\frac{π}{2}-x)}}dx

I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx —– 2

Adding 1 & 2 ——-

2I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx + \int\limits_0^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx

2I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx

2I= \int\limits_0^{\frac{π}{2}} dx

2I=\frac{π}{2}

I=\frac{π}{4}

Question 7. \int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx

Solution:

Let,I=\int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx

Letx = sin\theta ,dx=acos\theta d\theta

Now, x=0 ,\theta = 0 , thenx=a , \theta = π/2

\int\limits_0^{π/2} \frac{acos\theta }{asin\theta+acos\theta}d\theta

I=\int\limits_0^{π/2} \frac{cos\theta }{sin\theta+cos\theta}d\theta ——————– 1

So,

I=\int\limits_0^{π/2} \frac{cos(\frac{π}{2}-\theta )} {sin(\frac{π}{2}-\theta )+cos(\frac{π}{2}-\theta )}d\theta

I=\int\limits_0^{π/2} \frac{sin\theta }{sin\theta+cos\theta}d\theta ————- 2

Adding (1) and (2) —————-

2I=\int\limits_0^{π/2} \frac{cos\theta +sin\theta}{sin\theta+cos\theta}d\theta

2I=\int\limits_0^{π/2} d\theta

I=\frac{π}{4}

Question 8. \int\limits_0^{\infty} \frac{logx}{1+x^2}dx

Solution:

Let,x= tan \theta, dx=sec^2\theta d\theta

thenx=0, \theta=0 ; x=\infty , \theta=\frac{π}{2}

\therefore I=\int\limits_0^{\infty} \frac{logx}{1+x^2}dx

I=\int\limits_0^{\frac{π}{2}} \frac{logtan\theta sec^2\theta d\theta}{1+tan^2\theta}dx

I=\int\limits_0^{\frac{π}{2}} {logtan\theta d\theta} ———— (1)

I=\int\limits_0^{\frac{π}{2}} {logtan(\frac{π}{2}-\theta) d\theta}

I=\int\limits_0^{\frac{π}{2}} {logcot\theta d\theta} ——————- (2)

Adding (1) & (2) ————

2I=\int\limits_0^{\frac{π}{2}} {(logtan\theta +logcot\theta )d\theta}

2I=\int\limits_0^{\frac{π}{2}} log1x dx=\int\limits_0^{\frac{π}{2}} 0x dx=0

Question 9.  \int\limits_0^{1} \frac{log(1+x)}{1+x^2}dx

Solution:

Let,x= tan \theta, dx=sec^2\theta d\theta

x=0, \theta=0 ; x=1 , \theta=\frac{π}{4}

\therefore \int\limits_0^{1} \frac{log(1+x)}{1+x^2}dx

I=\int\limits_0^{\frac{π}{4}} log(1+tan\theta) d \theta

I=\int\limits_0^{\frac{π}{4}} log(1+tan(\frac{π}{4}-\theta)) d \theta

I=\int\limits_0^{\frac{π}{4}} log(1+\frac{1-tan\theta}{1+tan\theta}) d \theta

I=\int\limits_0^{\frac{π}{4}} log(\frac{2}{1+tan\theta}) d \theta

I=\int\limits_0^{\frac{π}{4}} \{ log2-log(1+tan\theta)\} d \theta

I=\frac{π}{8}log2

Question 10. \int\limits_0^{\infty} \frac{x}{(1+x)(1+x^2)}dx

Solution:

I=\int\limits_0^{\infty} \frac{x}{(1+x)(1+x^2)}dx

Let,

\frac{x}{(1+x)(1+x^2)} = \frac{A}{1+x}+\frac{Bx+C}{1+x}

x=A(1+x^2)+(Bx+C)(1+x)

Equating coefficients, we get

A+B=0; A=-B

B+C=1;-2A=1

A+C=0;A=-C

\therefore A=-\frac{1}{2},B=\frac{1}{2},C=\frac{1}{2}

So,

I=\int\limits_0^{\infty} \frac{-\frac{1}{2}}{(1+x)}+\frac{1}{2}\frac{1+x}{(1+x^2)}dx

I=\int\limits_0^{\infty} -\frac{1}{2(1+x)}dx +\frac{1}{2} \int\limits_0^{\infty} \frac{x}{(1+x^2)}dx +\frac{1}{2} \int\limits_0^{\infty} \frac{1}{(1+x^2)}dx

=0+0+\frac{π}{4}+0-0-0

I=\frac{π}{4}

Question 11. \int\limits_0^{π} \frac{xtanx}{secx cosecx}dx

Solution:

I=\int\limits_0^{π} \frac{xtanx}{secx cosecx}dx

I=\int\limits_0^{π} \frac{x\frac{sinx}{cosx}}{\frac{1}{cosx} \frac{1}{sinx}}dx

I=\int\limits_0^{π} xsin^2xdx ————- (1)

I=\int\limits_0^{π} (π-x)sin^2(π-x)dx

I=\int\limits_0^{π} (π-x)sin^2xdx —————— (2)

Adding (1) & (2) —————-

2I=\int\limits_0^{π} (π-x)sin^2xdx =π\int\limits_0^{π} \frac{1-cos2x}{2}dx =\frac{π}{2}[π-0-0+0]=\frac{π^2}{2}

\therefore \int\limits_0^{π} \frac{xtanx}{secx cosecx}dx= \frac{π^2}{4}

Question 12. \int\limits_0^{π}{xsinxcos^4x}dx

Solution:

Let ,I=\int\limits_0^{π}{xsinxcos^4x}dx ————– 1

So,

I=\int\limits_0^{π}{(π-x)sin(π-x)cos^4(π-x)}dx

I=\int\limits_0^{π}{(π-x) sinx.cos^4x}dx -1

2I=π\int\limits_0^{π}{sin(x)cos^4(x)}dx

Let,t=cos(x)dx; dt=-sinxdx

As, x=0, t=1 ; x=π , t=-1

Hence,

2I=π\int\limits_{-1}^{+1}t^4dt = π[\frac{1}{5}+\frac{1}{5}]

I=\frac{π}{5}

Question 13. \int\limits_0^{π}{xsin^3xdx}

Solution:

Let,I=\int\limits_0^{π} {(π-x)sin^3(π-x)dx}

=\int\limits_0^{π}πsin^3dx -\int\limits_0^{π}xsin^3dx

\therefore I=\int\limits_0^ππsin^3dx - I

2I=π\int\limits_0^π\frac{3sinx-sin3x}{4}dx

2I=\frac{π}{4}\int\limits_0^π(3sinx-sin3x)dx

I=2π/3

Question 14. \int\limits_0^{π} {xlogsinx}dx

Solution:

we have,\int\limits_0^{π} {xlogsinx}dx

=\int\limits_0^{π} {(π-x)logsin(π-x)}dx

I=π\int\limits_0^π logsin(x)dx-\int\limits_0^π xlogsinxdx

2I=π\int\limits_0^π log sinx dx

Since, f(x) = f(-x) , f(x) is an even function.

\therefore 2I=2π\int\limits_0^\frac{π}{2} log sinx dx

I=π\int\limits_0^\frac{π}{2} log sinx dx ————– 1

I=π\int\limits_0^\frac{π}{2} log sin(\frac{π}{2}-x) dx =π\int\limits_0^\frac{π}{2} log cosx dx

I=π\int\limits_0^\frac{π}{2} log cosx dx —————- 2

Adding 1 and 2 —————-

2I=π\int\limits_0^\frac{π}{2} log sinx dx +π\int\limits_0^\frac{π}{2} log cosxdx

2I=π\int\limits_0^\frac{π}{2} log sinx dx +π\int\limits_0^\frac{π}{2} log cosxdx

2I=π\int\limits_0^\frac{π}{2} log (sinx +cosx) dx = π\int\limits_0^\frac{π}{2} log sinx cosx dx

2I=π\int\limits_0^\frac{π}{2} log \frac{2sinxcosx}{2} dx =π\int\limits_0^\frac{π}{2} log \frac{sin2x}{2} dx =π\int\limits_0^\frac{π}{2} log sin2x dx +π\int\limits_0^\frac{π}{2} log 2dx

Now, letI=\int\limits_0^\frac{π}{2} log sin2x dx

Putting 2x=t, we get

2I=I-π\frac{π}{2}log2

I=- \frac{π^2}{2}log2

Question 15. \int\limits_0^{π} \frac{xsinx}{1+sinx}dx

Solution:

Let,I=\int\limits_0^{π} \frac{xsinx}{1+sinx}dx

=\int\limits_0^{π} \frac{(π-x)sin(π-x)}{1+sin(π-x)}dx

I=\int\limits_0^π \frac{πsinx}{1+sinx}dx -\int\limits_0^π \frac{xsinx}{1+sinx}dx

I=\int\limits_0^π \frac{πsinx}{1+sinx}dx -I

2I=\int\limits_0^π \frac{πsinx}{1+sinx}dx

2I=π\int\limits_0^π \frac{sinx}{1+sinx}\frac{(1-sinx)}{(1-sinx)}dx

2I=π\int\limits_0^π \frac{sinx-sin^2x}{1+sin^2x}dx

2I=π\int\limits_0^π \frac{sinx-sin^2x}{cos^2x}dx

2I=π\int\limits_0^π {tanxsecx}{-tan^2x}dx

2I=π\int\limits_0^π [{tanxsecx}{-(sec^2x-1)}]dx

2I=π\int\limits_0^π [{tanxsecx}{-sec^2x+1)}]dx

I=\frac{π}{2}(π-2)

Question 16. \int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx, 0<\alpha <π

Solution:

We have , I=\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx ———- 1

I=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sin(π-x)}dx [Tex]=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sinx}dx [/Tex]——- 2

Adding 1 and 2 —-

2I=\int\limits_0^{π} \frac{(π)}{1+cos\alpha sin(π-x)}dx

substituting ssinx= \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}

2I = π\int\limits_0^π \frac{sec^2xdx}{1+tan^2\frac{x}{2} 2cos\alpha tan\frac{x}{2}}

2I = π\int\limits_0^π \frac{sec^2xdx} {1-cos^2\alpha+ (cos\alpha tan\frac{x}{2})^2}

tan\frac{x}{2} = t; \frac{1}{2} sec^2 \frac{x}{2}dx=dt

when x=0 , t=0 ; x=π ,t=\alpha

2I = \int\limits_0^\alpha \frac{dt.dx} {(1+cos^2\alpha)+ (cos\alpha+t )^2}

I =\frac{\alphaπ}{sin\alpha}

Question 17. \int\limits_0^{π} xcos^2xdx

Solution:

Let,I=\int\limits_0^{π} xcos^2xdx

I=\int\limits_0^{π}(π-x)cos^2(π-x)dx

I=\int\limits_0^{π} cos^2(x)dx - \int\limits_0^{π}x cos^2(x)dx

2I=π \int\limits_0^{π} cos^2(x)dx

=π\int\limits_0^π(\frac{1+cos2x}{2})dx

=\frac{π}2\int\limits_0^π(1+cos2x)dx

I=\frac{π^2}{4}

Question 18. \int\limits_{π/6}^{π/3} \frac{1}{1+cot^{3/2}x}dx

Solution:

I=\int\limits_{π/6}^{π/3} \frac{1}{1+cot^{3/2}x}dx

I=\int\limits_{π/6}^{π/3} \frac{sin^{3/2}(\frac{π}{2}-x)} {sin^{3/2}(\frac{π}{2}-x)+cos^{3/2}(\frac{π}{2}-x)}dx

I=\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx

2I=\int\limits_{π/6}^{π/3} \frac{sin^{3/2}x} {sin^{3/2}x+cos^{3/2}x}dx+\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx

2I=\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x+sin^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx

I=\frac{π}{12}

Question 19. \int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx

Solution:

I=\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx

I=\int\limits_{0}^{π/2} \frac{tan^7(\frac{π}{2}-x)}{tan^7(\frac{π}{2}-x)+cot^7(\frac{π}{2}-x)}dx

I=\int\limits_{0}^{π/2} \frac{cot^7x}{tan^7x+cot^7x}dx

2I=\int\limits_{0}^{π/2} \frac{cot^7x}{tan^7x+cot^7x}dx+\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx

2I=\int\limits_{0}^{π/2} dx

2I=\frac{π}{2}

I=\frac{π}{4}

Question 20. \int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx

Solution:

I=\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx

I=\int\limits_{2}^{8} \frac{\sqrt{10-(8+2-x)}} {\sqrt{(8+2-x)} +\sqrt{10-(8+2-x)}}dx

I=\int\limits_{2}^{8} \frac{\sqrt{x}}{\sqrt{x} +\sqrt{10-x}}dx

2I=\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx+ \int\limits_{2}^{8} \frac{\sqrt{x}}{\sqrt{x} +\sqrt{10-x}}dx

2I=\int\limits_{2}^{8} dx

2I=6

I=3

Question 21. \int\limits_{0}^{π} {xsinxcos^2x}dx

Solution:

\int\limits_{0}^{π} {xsinxcos^2x}dx =\int\limits_{0}^{π} {(π-x)sin(π-x)cos^2(π-x)}dx

=\int\limits_{0}^{π} {(π-x)sinxcos^2x}dx

2\int\limits_{0}^{π} {xsinxcos^2x}dx= \int\limits_{0}^{π} {πsinxcos^2x}dx

2\int\limits_{0}^{π} {xsinxcos^2x}dx= \int\limits_{0}^{π} {πsinxcos^2x}dx

\int\limits_{0}^{π} {xsinxcos^2x}dx =\frac{π}{2} \int\limits_{0}^{π}sinxcos^2xdx

Now,

\int\limits_{0}^{π} sinxcos^2xdx

Let cosx=t

sinx dx=-dt

-\int\limits_{1}^{-1} t^2dt

\int\limits_{1}^{-1} t^2dt

\int\limits_{0}^{π} {xsinxcos^2x}dx =\frac{π}{3} [Tex]I= \frac{π }{8}[\fracπ 4+\fracπ 4] dt[/Tex]

Question 22. \int\limits_{0}^{π/2} \frac{xsinxcosx}{sin^4x+cos^4x}dx

Solution:

I=\int\limits_{0}^{π/2} \frac{xsinxcosx}{sin^4x+cos^4x}dx —————— 1

\int\limits_{0}^{π/2} \frac{(\frac{π}{2}-x)sinxcosx}{sin^4x+cos^4x}dx ———— 2

Adding 1 & 2 ————-

2I= \frac{π }{2} \int\limits_0^{π/2} \frac{sinxcosx}{cos^4x+sin^4x}dx

2I= \frac{π }{4} \int\limits_0^{π/2} \frac{2sinxcosx}{cos^4x+sin^4x}dx

Let ,t=sin^2x

2I= \frac{π }{4} \int\limits_0^1 \frac{1}{(1-t^2)+t^2}dt

2I= \frac{π }{8} \int\limits_0^1 \frac{1}{(t-\frac{1}{2})^2+\frac12^2}dt

I=\frac{π^2}{16}

Question 23. \int\limits_{-π/2}^{π/2} sin^3xdx

Solution:

Let,I=\int\limits_{-π/2}^{π/2} sin^3xdx

f(-x)=\int\limits_{-π/2}^{π/2} sin^3(-x)dx

=\int\limits_{-π/2}^{π/2} sin^3xdx

Here, f(x)=-f(x)

Hence, f(x) is odd function

Question 24. \int\limits_{-π/2}^{π/2} sin^4xdx

Solution:

We have, I=\int\limits_{-π/2}^{π/2} sin^4xdx = 2\int\limits_{-π/2}^{π/2} sin^4xdx \because sin^4x is an even function.

=2 \int\limits_{0}^{π/2} (sin^2x)^2dx

=2\int\limits_{0}^{π/2}( \frac{1-cos2x}{2})^2 dx

=\frac12 \int\limits_{0}^{π/2} ({1-cos2x})^2 dx

=\frac12[ \int\limits_{0}^{π/2}( {1+cos^22x}-2cos2x) ]dx

=\frac12[ \int\limits_{0}^{π/2} ( {1-2cos2x}+\frac{1+cos4x}{2}) ]dx

=\frac14[ \int\limits_{0}^{π/2}( {3-4cos2x}+cos4x) ]dx

\int\limits_{-π/2}^{π/2} sin^4xdx=\frac{3π }{8}

Question 25. \int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx

Solution:

we have,I=\int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx

Since,\int\limits_{-1}^{1} log(\frac{2-(-x)}{2+(-x)})dx= \int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx

\therefore this is an odd function

I=0

Question 26. \int\limits_{-π/4}^{π/4} sin^2xdx

Solution:

we have,I=\int\limits_{-π/4}^{π/4} sin^2xdx

sin2x is even function

Hence,

I=2\int\limits_{0}^{π/4} sin^2xdx =2\int\limits_{0}^{π/4} \frac{1-cos2x}{2}dx

\therefore \int\limits_{-π/4}^{π/4} sin^2xdx =\fracπ 4-\frac12

Question 27. \int\limits_{0}^{π} log(1-cosx)dx

Solution:

I=\int\limits_{0}^{π} log(1-cosx)dx

=\int\limits_{0}^{π} log(2sin^2\frac x 2)dx

=\int\limits_{0}^{π} log(2)dx +\int\limits_{0}^{π} log(sin^2\frac x 2)dx

=\int\limits_{0}^{π} log(2)dx +2\int\limits_{0}^{π} log(sin\frac x 2)dx

I=πlog2+4I1

Question 28. \int\limits_{-π/4}^{π/4} log(\frac {2-sinx}{2+sinx})dx

Solution:

we have ,I=\int\limits_{-π/4}^{π/4} log(\frac {2-sinx}{2+sinx})dx

Let,f(x)=log(\frac{2-sinx}{2+sinx})dx

Then,

f(-x)=log (\frac{2-sin(-x)}{2+sin(-x)})= -log(\frac{2-sinx}{2+sinx})=-f(x)

\therefore \int\limits_{-π /4}^{π /4} log(\frac{2-sinx}{2+sinx})dx=0

Question 29. \int\limits_{-π}^{π} \frac{2x(1+sinx)}{1+cos^2x}dx

Solution:

I=\int\limits_{-π}^{π} \frac{2x(1+sinx)}{1+cos^2x}dx

I=\int\limits_{-π}^{π} \frac{2x}{1+cos^2x}dx +\int\limits_{-π}^{π} \frac{2xsinx}{1+cos^2x}dx

I=0+\int\limits_{-π}^{π} \frac{2xsinx}{1+cos^2x}dx

I=2\int\limits_0^π\frac{2xsinx}{1+cos^2x}dx

I=4\int\limits_0^π \frac{xsinx}{1+cos^2}dx

I=2π\int\limits_0^π \frac{sinx}{1+cos^2x}

Put cosx = t then -sinx dx = dt

I=-2π\int\limits_{-1}^{1} \frac{1}{1+t^2}dt

I=π^2

Question 30. \int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta ,a>0

Solution:

I=\int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta

Letf(\theta)= log(\frac{a-sin\theta}{a+sin\theta})

f(-\theta)= log(\frac{a-sin(-\theta)}{a+sin(-\theta)}) =-log(\frac{a-sin\theta}{a+sin\theta})=-f(\theta)

\therefore f(\theta)= log(\frac{a-sin\theta}{a+sin\theta}) is an odd function

\therefore \int\limits_{-a}^a log (\frac{a-sin\theta}{a+sin\theta}) d\theta =0

Question 31. \int\limits_{-2}^{2} \frac{3x^3+2| x |+1}{x^2+| x |+1}dx

Solution:

I=\int\limits_{-2}^{2} \frac{3x^3+2| x |+1}{x^2+| x |+1}dx

I=\int\limits_{-2}^{2} \frac{3x^3}{x^2+| x |+1}dx +\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx

I=0 +\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx

I=2\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx

I=2[log(4+2+1)-log(1)]

I=2log(7)

Question 32. \int\limits_{-3π/2}^{-π/2} sin^2(3π+x)+(π+x)^3)dx

Solution:

I=\int\limits_{-3π/2}^{-π/2} sin^2(3π+x)+(π+x)^3)dx

Substitute π+x=u then dx=du

I=\int\limits_{-3π/2}^{-π/2} sin^2(2π+u)+(u)^3)du

I=\int\limits_{-3π/2}^{-π/2} sin^2(u)+(u)^3)dx

I=\frac{π}{2}

Question 33. \int\limits_{0}^{2} x\sqrt{2-x} dx

Solution:

Let, I=\int\limits_{0}^{2} x\sqrt{2-x} dx

I=\int\limits_{0}^{2} ({2-x})\sqrt x dx

=\frac43 (2)^{\frac32} - \frac25x^{\frac52}

= \frac{4*2\sqrt 2}{3} - \frac{2}{5}*4\sqrt 2

=\frac{8 \sqrt 2}{3} - \frac{8\sqrt 2}{5}

=\frac{40\sqrt 2 - 24\sqrt 2} {15}

=\frac{16\sqrt 2}{15}

Question 34. \int\limits_{0}^{1} log(\frac{1}{x}-1)dx

Solution:

I=\int\limits_{0}^{1} log(\frac{1}{x}-1)dx

=\int\limits_{0}^{1} log(\frac{1-x}{x})dx

=\int\limits_{0}^{1} log(1-x)dx - \int\limits_{0}^{1} log(x)dx

Applying the property , \int\limits_0^a f(x)dx=\int\limits_0^a f(a-x)dx

Thus,I=\int\limits_0^1 log(1-(1-x))dx-\int\limits_0^1 log(x)dx

=\int\limits_0^1 log(1-1+x)dx -\int\limits_0^1log(x)dx

=0

Question 35. \int\limits_{-1}^{1} | xcosπx |dx

Solution:

I= \int\limits_{-1}^{1} | xcosπx |dx

Let,f(x)=| xcosπx |dx

f(-x)=|- xcos(-πx) |=|- xcos(πx) |=| xcosπx |=f(x)

\therefore \int\limits_{-1}^{1} | xcosπx |dx=2\int\limits_{-1}^{1}| xcosπx |dx

f(x) = | xcosπx |dx= \{xcosπx, if 0\leq x\leq \frac12 ;-xcosπx, if \frac12< x< 1

\therefore I=2\int\limits_0^1 | xcosπx |dx

I= \frac{2}{π}

Question 36. \int\limits_{0}^{π} (\frac{x}{1+sin^2x}+cos^7x) dx

Solution:

I=\int\limits_{0}^{π} (\frac{π-x}{1+sin^2(π-x)}+cos^7(π-x) dx

I=\int\limits_{0}^{π} (\frac{π-x}{1+sin^2x}-cos^7x) dx

2I=\int\limits_{0}^{π} (\frac{π}{1+sin^2x}) dx

2I=π\int\limits_{0}^{π} (\frac{1}{1+sin^2x}) dx

2I=π\int\limits_{0}^{π} (\frac{1}{1+tan^2x})sec^2x dx

I=π\int\limits_{0}^{π/2} (\frac{1}{1+2tan^2x})sec^2x dx [Tex][\because \int\limits_0^{2a} f(x)dx= 2\int\limits_0^{a} f(x)dx, f(2a-x)=f(x)][/Tex]

let tanx = v

dv = sec2xdx

I=π\int\limits_0^\infty \frac{1}{1+2v^2} dv

I=\frac{π^2}{2\sqrt2}

Question 37. \int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx

Solution:

I=\int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx

I=\int\limits_{0}^{π} (\frac{(π-x)}{1+cos\alpha sin(π-x)}) dx

I=\int\limits_{0}^{π} (\frac{π-x}{1+cos\alpha sinx}) dx

2I=π\int\limits_{0}^{π} (\frac{1}{1+cos\alpha sinx}) dx

2I=π\int\limits_{0}^{π} (\frac{1+tan^2(\frac{x}2)} {( 1+tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}}) dx

2I=π\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx

I=\frac{π}2\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx

Puttan(\frac{x}2)=t thensec^2(\frac{x}2)dx=2dt

x=0 ⇒ t=0 and x=π ⇒t=\infty

I=\frac{π}{2} \int\limits_0^\infty \frac{2}{t^2+2tcos\alpha +1} dt

I={π} \int\limits_0^\infty \frac{1}{(t+cos\alpha)^2+(1-cos^2\alpha)} dt

I={π} \int\limits_0^\infty \frac{1}{(t+cos\alpha)^2+(sin^2\alpha)} dt

I=\frac{π\alpha}{sin\alpha}

Question 38. \int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx

Solution:

we know,\int\limits_{0}^{2a} f(x)= \int\limits_{0}^{a}f(x)+\int\limits_{0}^{a}f(2a-x)dx

Also here,

f(x) = f(2π -x)

So,I=\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx =2\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx

I=2\int\limits_{0}^{2π} (sin^{100}(π -x)cos^{101}(π -x)) dx

I=-2\int\limits_0^π sin^{100}x cos^{101}xdx

2I=0

I=0

Question 39. \int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx

Solution:

I=\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx

then,

\int\limits_{0}^{π/2} \frac{(asin(\frac{π}2-x)+bcos(\frac{π}2-x))} {sin(\frac{π}2-x)+cos(\frac{π}2-x)} dx

I=\int\limits_{0}^{π/2} \frac{(acosx+bsinx)}{sinx+cosx} dx

2I=\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx+\int\limits_{0}^{π/2} \frac{(acosx+bsinx)}{sinx+cosx} dx

2I=(a+b)\int\limits_{0}^{π/2} \frac{(sinx+cosx)}{sinx+cosx} dx

I=\frac{(a+b)}{2}\int\limits_{0}^{π/2} 1 dx

I=\frac{(a+b)π}{4}

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that\int\limits_{0}^{2a} f(x)dx= 2\int\limits_{0}^{a} f(x)dx

Solution:

We have ,I=\int\limits_{0}^{2a} f(x)dx

Then,

I=\int\limits_{0}^{a} f(x)dx +I(1)

Let , 2a-t =x then dx=-dt

if t=a ⇒x=a

if t=2a ⇒ x=0

I(1)=\int\limits_{0}^{2a} f(x)dx= \int\limits_{a}^{0} f(2a-t)(-dt) =-\int\limits_{a}^{0} f(2a-t)dt

I(1)=\int\limits_{0}^{a}f(2a-t)dt= \int\limits_{0}^{a}f(2a-x)dx

\therefore I=\int\limits_{0}^{a}f(x)dx +\int\limits_{0}^{a}f(2a-x)dx

I=\int\limits_{0}^{a}f(x)dx +\int\limits_{0}^{a}f(x)dx [Tex]=2\int\limits_{0}^{a}f(x)dx[/Tex]

Hence Proved.

Question 41. Iff(2a-x)=-f(x) , prove that\int\limits_{0}^{2a} f(x)dx =0

Solution:

We have, I=\int\limits_{0}^{2a} f(x)dx = \int\limits_{0}^{2a} f(x)dx+\int\limits_{a}^{2a} f(x)dx

I=\int\limits_{0}^{2a} f(x)dx+I(1)

Let 2a-t=x then dx=-dt

t=a , x=a ; t=2a , x=0

I(1) = \int\limits_0^{2a}f(x)dx=\int\limits_a^0f(2a-t)(-dt)

= \int\limits_a^0 f(2a-t)dt

I(1)=\int\limits_0^{2a}f(2a-t)dt=\int\limits_0^{2a}f(2a-x)dx

I=\int\limits_0^a f(x)dx + \int\limits_0^a f(2a-x)dx

I=\int\limits_0^a f(x)dx - \int\limits_0^a f(x)

I=0

Question 42. If f is an integrable function, show that

(i)\int\limits_{-a}^{a} f(x^2)dx= 2\int\limits_{0}^{a} f(x^2)dx

Solution:

we have ,I=\int\limits_{-a}^{a} f(x^2)dx

clearly f(x2) is an even function .

So,\int\limits_{-a}^{a} f(t)= 2\int\limits_{0}^{a} f(t)dx

I=2\int\limits_{0}^{a} f(x^2)dx

(ii)\int\limits_{-a}^{a}x f(x^2)dx=0

Solution:

I=\int\limits_{-a}^{a}x f(x^2)dx

clearly , xf(x2) is odd function .

So,I=0

\therefore \int\limits_{-a}^{a}x f(x^2)dx=0

Question 43. If f(x) is a continuous function defined on [0,2a] . Then, prove that

\int\limits_{0}^{2a} f(x)dx =\int\limits_{0}^{a} {f(x)+f(2a-x)}dx

Solution:

We have from LHS,

I=\int\limits_{0}^{2a} f(x)dx =\int\limits_{0}^{a} f(x)dx+ \int\limits_{a}^{2a}f(x)dx

\therefore \int\limits_0^{2a}f(x)dx= - \int\limits_a^{0}f(2a-t)dt

\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(2a-x)dx

\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(2a-x)dx

substituting\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(2a-x)dx

we get,

\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(x)dx+ \int\limits_0^{a}f(2a-x)dx

\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}\{f(x)+f(2a-x)\}dx

Question 44. If f(a+b-x) = f(x), then prove that

\int\limits_{a}^{b} f(x)dx= (\frac{a+b}{2})\int\limits_{a}^{b} f(x)dx

Solution:

I=\int\limits_{a}^b xf(x)dx

I=\int\limits_{a}^{b}(a+b-x) f(a+b-x)dx

I=\int\limits_a^bf(x)dx ——————[ Given that f(a+b-x) = f(x) ]

I=\int\limits_a^b (a+b) f(x)dx-\int\limits_a^b xf(x)dx

I=\int\limits_a^b (a+b)f(x)dx - I

2I=\int\limits_a^b (a+b)f(x)dx

I=\frac{a+b}{2}\int\limits_a^bf(x)dx

Question 45. If f(x) is a continuous function defined on [-a,a], then prove that

\int\limits_{-a}^{a} f(x)dx = \int\limits_{0}^{a} f(x)+f(-x)dx

Solution:

we have ,I=\int\limits_{-a}^{a} f(x)dx = \int\limits_{-a}^{0} f(x)+ \int\limits_{0}^{a}f(-x)dx

Let, x=-t, then dx=-dt

x=-a ⇒ t=a

x=0 ⇒ t=0

\therefore \int\limits_{-a}^a f(x)dx= \int\limits_{-a}^0f(-t)(-dt) =\int\limits_{-a}^0-f(-t)dt

\int\limits_{-a}^a f(x)dx=\int\limits_{0}^a f(-t)(dt)

\int\limits_{-a}^0 f(x)dx=\int\limits_{0}^a f(-x)dx

\therefore \int\limits_{-a}^a f(x)dx=\int\limits_{0}^a f(-x)dx +\int\limits_{0}^a f(x)dx

\int\limits_{-a}^a f(x)dx = \int\limits_{0}^a \{f(-x)+f(x)\}dx

Hence, Proved.

Question 46. Prove that:\int\limits_{0}^{π} xf(sinx)dx = \frac{π}{2} \int\limits_{0}^{π}f(sinx)dx

Solution:

I= \int\limits_{0}^{π} xf(sinx)dx

I=\int\limits_{0}^{π}(π- x)f(sin(π-x))dx

I=\int\limits_0^π (π -x) f(sinx)dx

2I=\int\limits_0^π π f(sinx)dx

I=\fracπ2 \int\limits_0^π f(sinx)dx


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