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 ------ 1So,
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 ------------ 2Hence, 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 ----- 1So,
I= \int\limits_0^{\frac{π}{2}} \frac{sin(\frac{π}{2}-x)}{sin(\frac{π}{2}-x)+ cos(\frac{π}{2}-x)}dx ------ 2Adding 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 --------------- 2Adding 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 ---------- 2Adding 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 ----- 2Adding 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 , 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 -------------------- 1So,
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 ------------- 2Adding (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 -------------- 1So,
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 ---------------- 2Adding 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 =\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sinx}dx ------- 2Adding 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} I= \frac{π }{8}[\fracπ 4+\fracπ 4] dt
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 ------------ 2Adding 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 [\because \int\limits_0^{2a} f(x)dx= 2\int\limits_0^{a} f(x)dx, f(2a-x)=f(x)] 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 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 =2\int\limits_{0}^{a}f(x)dx 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
Certainly. I'll provide a summary, 10 practice questions, and FAQs for Chapter 20: Definite Integrals, Exercise 20.4 Part B of RD Sharma's Class 12 mathematics textbook. Please note that this is based on general knowledge of the topic, as I don't have direct access to the specific textbook.
Summary:
Exercise 20.4 Part B of Chapter 20 likely focuses on more advanced techniques for evaluating definite integrals. This section probably covers:
1. Integration by parts for definite integrals
2. Definite integrals involving trigonometric functions
3. Evaluation of improper integrals
4. Application of substitution method in definite integrals
5. Definite integrals with logarithmic and exponential functions
10 Practice Questions:
1. Evaluate ∫[0 to π/2] x sin x dx using integration by parts.
2. Calculate ∫[0 to 1] x ln x dx.
3. Find the value of ∫[0 to ∞] e^(-x) dx.
4. Compute ∫[0 to π/4] sec^3 x dx.
5. Evaluate ∫[1 to e] (ln x)^2 dx.
6. Calculate ∫[0 to π/2] sin^3 x cos^2 x dx.
7. Find the value of ∫[0 to 1] x^n e^x dx, where n is a positive integer.
8. Evaluate ∫[0 to 1] √(1-x^2) dx.
9. Compute ∫[0 to π/2] sin x / (1 + cos x) dx.
10. Calculate ∫[1 to 2] (ln x) / x dx.
FAQs:
1. Q: What is integration by parts and when is it used?
A: Integration by parts is a technique used when integrating the product of two functions. It's particularly useful for integrals involving algebraic and transcendental functions.
2. Q: How do you evaluate improper integrals?
A: Improper integrals are evaluated by taking the limit of a definite integral as one or both of its limits approach infinity or a point where the integrand is undefined.
3. Q: What are some common trigonometric integrals to remember?
A: Some important ones include ∫ sin x dx = -cos x + C, ∫ cos x dx = sin x + C, and ∫ tan x dx = -ln|cos x| + C.
4. Q: How does the substitution method work for definite integrals?
A: When using substitution, you change both the integrand and the limits of integration. The new limits are found by applying the substitution to the original limits.
5. Q: What is the relationship between definite and indefinite integrals?
A: The Fundamental Theorem of Calculus connects them: if F(x) is an antiderivative of f(x), then ∫[a to b] f(x) dx = F(b) - F(a).
Would you like me to elaborate on any of these points or provide solutions to any of the practice questions?
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
Summary
Exercise 20.4 Part B of Chapter 20 likely focuses on more advanced techniques for evaluating definite integrals. This section probably covers:
- Integration by parts for definite integrals
- Definite integrals involving trigonometric functions
- Evaluation of improper integrals
- Application of substitution method in definite integrals
- Definite integrals with logarithmic and exponential functions