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

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, $I=\int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx$
Let $x = sin\theta$ , $dx=acos\theta d\theta$
Now, x=0 , $\theta = 0$ , then $x=a , \theta = π/2$
$\int\limits_0^{π/2} \frac{acos\theta }{asin\theta+acos\theta}d\theta$
$I=\int\limits_0^{π/2} \frac{cos\theta }{sin\theta+cos\theta}d\theta$ ——————– 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, $x= tan \theta, dx=sec^2\theta d\theta$
then $x=0, \theta=0 ; x=\infty , \theta=\frac{π}{2}$
$\therefore I=\int\limits_0^{\infty} \frac{logx}{1+x^2}dx$
$I=\int\limits_0^{\frac{π}{2}} \frac{logtan\theta sec^2\theta d\theta}{1+tan^2\theta}dx$
$I=\int\limits_0^{\frac{π}{2}} {logtan\theta d\theta}$ ———— (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, let $I=\int\limits_0^\frac{π}{2} log sin2x dx$
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 s $sinx= \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$
$2I = π\int\limits_0^π \frac{sec^2xdx}{1+tan^2\frac{x}{2} 2cos\alpha tan\frac{x}{2}}$
$2I = π\int\limits_0^π \frac{sec^2xdx} {1-cos^2\alpha+ (cos\alpha tan\frac{x}{2})^2}$
$tan\frac{x}{2} = t; \frac{1}{2} sec^2 \frac{x}{2}dx=dt$
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:

$I=\int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta$
Let $f(\theta)= log(\frac{a-sin\theta}{a+sin\theta})$
$f(-\theta)= log(\frac{a-sin(-\theta)}{a+sin(-\theta)}) =-log(\frac{a-sin\theta}{a+sin\theta})=-f(\theta)$
$\therefore f(\theta)= log(\frac{a-sin\theta}{a+sin\theta})$ 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:

$I=\int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx$
$I=\int\limits_{0}^{π} (\frac{(π-x)}{1+cos\alpha sin(π-x)}) dx$
$I=\int\limits_{0}^{π} (\frac{π-x}{1+cos\alpha sinx}) dx$
$2I=π\int\limits_{0}^{π} (\frac{1}{1+cos\alpha sinx}) dx$
$2I=π\int\limits_{0}^{π} (\frac{1+tan^2(\frac{x}2)} {( 1+tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}}) dx$
$2I=π\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx$
$I=\frac{π}2\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx$
Put $tan(\frac{x}2)=t$ then $sec^2(\frac{x}2)dx=2dt$
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. If $f(2a-x)=-f(x)$ , 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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Last Updated : 02 Feb, 2021

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

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

Question 1.

Question 2. ,

Question 3. ,

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Question 29.

Question 30.

Question 31.

Question 32.

Question 33.

Question 34.

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that

Question 41. If, prove that

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

(i)

(ii)

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

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

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

Question 46. Prove that:

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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