Evalute _0^ x x 1+ ^2 x d x - Vidyadip

Question

Evalute $\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$

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Answer

Let $I=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$...(i)
Using property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$:
$\therefore$ $I=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x$
= $\int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x$...(ii)
Adding (i) and (ii):
$\therefore$ $2 I=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x$
$\Longrightarrow 2 I=\pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t$ $\Longrightarrow-\sin x d x=d t$.
Limits:$\therefore$ $x=0 \rightarrow t=1, x=\pi \rightarrow t=-1.2 I$
=$\pi \int_1^{-1} \frac{-d t}{1+t^2}=\pi \int_{-1}^1 \frac{d t}{1+t^2}=\pi\left[\tan ^{-1} t\right]_{-1}^1$
$\therefore$ $2 I=\pi\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=\pi\left[\frac{\pi}{2}\right]=\frac{\pi^2}{2} \Longrightarrow I =\frac{\pi^2}{4}$

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