Question

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

Answer 1

\[\text{We have}, \]

\[ I = \int_0^\pi \frac{x}{1 + \cos\alpha \sin x} d x . . . . . \left( 1 \right)\]

\[ = \int_0^\pi \frac{\pi - x}{1 + \cos\alpha \sin\left( \pi - x \right)}dx\]

\[ = \int_0^\pi \frac{\pi - x}{1 + \cos\alpha \sin x}dx . . . . . \left( 2 \right)\]

\[\text{Adding} \left( 1 \right) and \left( 2 \right) \text{we get}, \]

\[2I = \int_0^\pi \frac{x + \pi - x}{1 + \cos\alpha \sin x} d x\]

\[ \Rightarrow I = \frac{\pi}{2} \int_0^\pi \frac{1}{1 + \cos\alpha\ sinx} dx\]

\[= \frac{\pi}{2} \int_0^\pi \frac{1}{1 + \cos\alpha sinx}\]
\[ = \frac{\pi}{2} \int_0^\pi \frac{1}{1 + \cos\alpha \frac{2\tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \frac{\pi}{2} \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + 2\cos\alpha \tan \frac{x}{2}}dx\]
\[ = \frac{\pi}{2} \int_0^\pi \frac{\sec^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + 2\cos\alpha \tan \frac{x}{2}}dx\]

\[\text{Putting }\tan\frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \sec^2 x dx = dt\]

\[\text{When }x \to 0; t \to 0\]

\[\text{and }x \to \pi; t \to \infty \]

\[ \therefore I = \frac{\pi}{2} \int_0^\infty \frac{2}{1 + t^2 + 2\cos\alpha t}dt\]

\[ = \frac{\pi}{2} \int_0^\infty \frac{2}{\left( t + \cos\alpha \right)^2 - \cos^2 \alpha + 1}dt\]

\[ = \pi \int_0^\infty \frac{1}{\left( t + \cos\alpha \right)^2 + \sin^2 \alpha}dt\]

\[ = \pi \left[ \frac{1}{\sin \alpha} \tan^{- 1} \left( \frac{t + \cos \alpha}{\sin \alpha} \right) \right]_0^1 \]

\[ = \frac{\pi}{sin\alpha}\left[ \tan^{- 1} \left( \infty \right) - \tan^{- 1} \left( \cot\alpha \right) \right]\]

\[ = \frac{\pi}{sin\alpha}\left[ \frac{\pi}{2} - \tan^{- 1} \left( \tan\left( \frac{\pi}{2} - \alpha \right) \right) \right]\]

\[ = \frac{\pi\alpha}{sin\alpha}\]