Question
Evaluate: $\int\limits^{\pi}_{0} \frac{x \sin x}{1 + \cos^{2} x} \text{d}x.$
✓
Answer
$\text{I} = \int\limits^{\pi}_{0} \frac{\text{x} \sin \text{x}}{1 + \cos^{2} \text{x}} \text{dx}$
$= \int\limits^{\pi}_{0} \frac{(\pi - \text{x}) \sin \text{x}}{1 + \cos^{2} \text{x}} \text{x}$
$\Rightarrow \text{2I} = \pi \int\limits^{\pi}_{0} \frac{\sin \text{x dx}}{1 + \cos^{2}\text{x}}$
$\text{Put} \cos \text{x = t and} -\sin \text{x dx = dt}$
$= -\pi \int\limits^{-1}_{1} \frac{\text{dt}}{1 + \text{t}^{2}}$
$= \pi [ \tan^{-1} \text{t}]^{1}_{-1} = \frac{\pi^{2}}{2}$
$\Rightarrow \text{I} = \frac{\pi^{2}}{4}$
$= \int\limits^{\pi}_{0} \frac{(\pi - \text{x}) \sin \text{x}}{1 + \cos^{2} \text{x}} \text{x}$
$\Rightarrow \text{2I} = \pi \int\limits^{\pi}_{0} \frac{\sin \text{x dx}}{1 + \cos^{2}\text{x}}$
$\text{Put} \cos \text{x = t and} -\sin \text{x dx = dt}$
$= -\pi \int\limits^{-1}_{1} \frac{\text{dt}}{1 + \text{t}^{2}}$
$= \pi [ \tan^{-1} \text{t}]^{1}_{-1} = \frac{\pi^{2}}{2}$
$\Rightarrow \text{I} = \frac{\pi^{2}}{4}$
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