logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the integral using substitution\[\int\limits_0^\frac{\large \pi}{2}\frac{\sin x}{1+\cos^2x}dx\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • Method of substitution:If $I=\int f(x)dx,Let f(x)=t\; then\; f'(x)dx=dt\; therefore\; I=\int t.dt$
  • (iii)$ \int \large \frac{\sin x}{1+x^2}dx=\tan^{-1}(x)$
Given $\int\limits_0^\frac{\large \pi}{2}\frac{\sin x}{1+\cos^2x}dx$
 
Let $\cos x=t,$ on differentiating with respect to x
 
$-\sin xdx=dt\qquad=>\sin x dx=-dt$
 
Now substituting t and dt,
 
$I=\int \limits_0^1 \large\frac{dt}{1+t^2} \;As\; x \to \pi/2,t \to 0$
 
$=-\int \limits_0^1\frac {dt}{1+t^2}$
 
This is of the form $ =\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan ^{-1}(\frac{x}{a})$
 
Here $a=1\;and\;x=t$
 
Therefore $ \int \limits_0^1\large\frac{dt}{1+t^2}=\tan^{-1}(t)$
 
On applying limits we get,
 
$=\tan^{-1}(1)$
 
$I=\large\frac{\pi}{4}$

 

answered Feb 12, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...