Question
$\text{Evaluate :} \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} dx$
$= \int\limits_0^{\pi} \frac{(\pi - x) \sin (\pi - x)}{1 + \cos^{2} (\pi - x)} \text{dx} = \int\limits_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^{2} x} \text{dx}$
$\Rightarrow \text{2I} = \int\limits_0^{\pi} \frac{\sin x}{1 + \cos^{2} x} \text{dx}$
$\text{I} = \frac{\pi}{2} . 2 \int\limits_0^{\pi} \frac{\sin x dx}{1 + \cos^{2} x} = -\pi \bigg[ \tan^{-1} (\cos x)\bigg]^\frac{\pi}{2}_{0}$
$= -\pi \bigg[ -\frac{\pi}{4}\bigg] = \frac{\pi^{2}}{4}$
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$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}-3}{-3}$ and $\frac{\text{x}+3}{-1}=\frac{\text{y}-5}{8}=\frac{\text{z}-1}{4}$