Evaluate: ^ _ 0 x 1 + x dx.

Question

Evaluate: $\int\limits^{\pi}_{0} \frac{\text{x}}{1 + \sin \alpha \sin x} \text{dx}.$

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Answer

$\text{I} = \int\limits^{\pi}_{0} \frac{(\pi - x)}{1 + \sin \alpha \sin (\pi - x)} \text{dx}$
$\text{2I} = \pi \int\limits^{\pi}_{0} \frac{\text{dx}}{1 + \sin \alpha \sin \text{x}}$
$= 2\pi \int\limits^{\pi/2}_{0} \frac{\text{dx}}{1 + \sin \alpha \sin \text{x}}$
$\frac{\text{dx}}{1 + \sin \alpha \frac{2\tan\frac{\text{x}}{2}}{1 + \tan^{2} \frac{\text{x}}{2}}}$
$\text{I} = \pi \int\limits^{1}_{0}\frac{\text{2dt}}{1 + \text{t}^{2} + 2\text{t} \sin \alpha} \text{ }\text{ }\text{ }\text{ }\text{put} \tan \frac{\text{x}}{2} = \text{t}$
$\Rightarrow \text{I} = 2\pi \int\limits^{1}_{0} \frac{\text{dt}}{\text{(t} + \sin \alpha)^{2} + \cos^{2} \alpha}$
$= \frac{2\pi}{\cos \alpha} \bigg[\tan^{-1}\bigg(\frac{\text{t} + \sin \alpha}{\cos \alpha}\bigg) \bigg]^{1}_{0}$
$\Rightarrow \text{I} = \frac{\pi}{\cos \alpha} \bigg(\frac{\pi}{2} - \alpha\bigg)$

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