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

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  • $\int\limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
  • $\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
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