Question
Evaluate: $\int\limits_{0}^{\pi}\frac{4\text{x}\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}.$
✓
Answer
Let $\text{I} = \int\limits_{0}^{ \pi}\frac{4\text{x}\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
$\text{x}\rightarrow(\pi - \text{x})\text{gives }\text{ I} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin(\pi - \text{x})}{1 + \cos^{2}(\pi - \text{x})}\text{dx} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
$\therefore2\text{I} =4\pi\int\limits_{0}^{\pi}\frac{\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
Put cos x = t
$\therefore$ sin x dx = – dt
$\therefore\text{I} = 2 \pi\int\limits_{1}^{-1}\frac{-\text{dt}}{1 + \text{t}^{2}}\ \ \ \text{or}\ \ 2\pi\int\limits_{-1}^{1}\frac{\text{dt}}{1 + \text{t}^{2}}$
$ = 2\pi\bigg[\tan^{-1}\text{t}\bigg]_{-1}^{1} = 2\pi\bigg[\frac{\pi}{4} - \bigg( - \frac{\pi}{4}\bigg)\bigg] = \pi^{2}.$
$\text{x}\rightarrow(\pi - \text{x})\text{gives }\text{ I} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin(\pi - \text{x})}{1 + \cos^{2}(\pi - \text{x})}\text{dx} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
$\therefore2\text{I} =4\pi\int\limits_{0}^{\pi}\frac{\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
Put cos x = t
$\therefore$ sin x dx = – dt
$\therefore\text{I} = 2 \pi\int\limits_{1}^{-1}\frac{-\text{dt}}{1 + \text{t}^{2}}\ \ \ \text{or}\ \ 2\pi\int\limits_{-1}^{1}\frac{\text{dt}}{1 + \text{t}^{2}}$
$ = 2\pi\bigg[\tan^{-1}\text{t}\bigg]_{-1}^{1} = 2\pi\bigg[\frac{\pi}{4} - \bigg( - \frac{\pi}{4}\bigg)\bigg] = \pi^{2}.$
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