Evaluate the following integrals: ^ _0 x cosecx dx

Question

Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec(\pi-\text{x})\text{ cosec}(\pi-\text{x})}\text{ dx}$ $\Bigg[\text{Using}\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0\frac{-(\pi-\text{x})\tan\text{x}}{-\sec\text{x}\text{ cosec}\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^2\text{x dx}$
$=\pi\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\Big[\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$

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