Evaluate : _0^ x x 1 + ^ 2 x dx - Vidyadip

Question

$\text{Evaluate :} \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} dx$

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Answer

$\text{I} = \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} \text{dx}$$= \int\limits_0^{\pi} \frac{(\pi - x) \sin (\pi - x)}{1 + \cos^{2} (\pi - x)} \text{dx} = \int\limits_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^{2} x} \text{dx}$
$\Rightarrow \text{2I} = \int\limits_0^{\pi} \frac{\sin x}{1 + \cos^{2} x} \text{dx}$
$\text{I} = \frac{\pi}{2} . 2 \int\limits_0^{\pi} \frac{\sin x dx}{1 + \cos^{2} x} = -\pi \bigg[ \tan^{-1} (\cos x)\bigg]^\frac{\pi}{2}_{0}$
$= -\pi \bigg[ -\frac{\pi}{4}\bigg] = \frac{\pi^{2}}{4}$

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