Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Using properties of definite integrals, evaluate the following : $ \int_0^{\pi}\large\frac{x\: sin\: x}{1+cos^2x}$$dx $

Can you answer this question?

1 Answer

0 votes
  • $\int\limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
  • $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$$+c$
Step 1:
$I=\int\large\frac{x\sin x}{1+\cos^2x}$$dx$-------(1)
$\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Applying the above property we get,
$I=\int\large\frac{(\pi-x)\sin(\pi- x)}{1+\cos^2(\pi-x)}$$dx$-------(2)
But $\sin(\pi-x)=\sin x$
Step 2:
Adding equ(1) and (2) we get
$2I=\int_0^\pi\large\frac{\pi\sin x}{1+\cos^2x}$$dx$
$\;\;\;=\pi\int_0^\pi\large\frac{\sin x}{1+\cos^2x}$$dx$
Put $\cos x=t$
Differentiating w.r.t $x$ we get
$-\sin xdx=dt$
When $x=0\Rightarrow t=\cos 0=1$
When $x=\pi\Rightarrow t=\cos \pi=-1$
Step 3:
$\therefore 2I=\pi\int_1^{-1}\large\frac{-dt}{1+t^2}$
This is of the form $\int\large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$$+c$
$\therefore 2I=\pi\big[\tan^{-1}t\big]_{-1}^1$
Step 4:
Now applying the limits we get,
But $\tan^{-1}(1)=\large\frac{\pi}{4}$
$\therefore 2I=\large\frac{\pi^2}{2}$
answered Oct 2, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App