Question
Using properties of definite integrals, prove the following:
$\int\limits_0^{\pi} \frac{x \tan x}{\sec x\text{ }cosec\text{ x}} dx = \frac{\pi^{2}}{4}$
$\int\limits_0^{\pi} \frac{x \tan x}{\sec x\text{ }cosec\text{ x}} dx = \frac{\pi^{2}}{4}$
$\therefore \text{I} = \int\limits_0^{\pi} \pi \sin^{2} \text{x dx} - \int\limits_0^{\pi} x \sin^{2} \text{x dx}$
$\text{2 I} =\pi \int\limits_0^{\pi} \sin^{2} \text{x dx} = \frac{\pi}{2} = \int\limits_0^{\pi}(1 - \cos 2x) \text{dx}$
$= \frac{\pi}{2} \bigg[ x- \frac{\sin 2x}{2} \bigg]^{\pi}_{0}$
$= \frac{\pi}{2} [\pi] = \frac{\pi^2}{2}$
$\Rightarrow \text{I} = \frac{\pi^{2}}{4}$
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