Question

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

Answer 1

\[We have, \]
\[I = \int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx . ....... . . . \left( 1 \right)\]
\[ \Rightarrow I = \int\limits_0^\pi \frac{\pi - x}{1 + \cos \alpha \sin \left( \pi - x \right)} dx ...............\left( \because \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right)\]
\[ \Rightarrow I = \int\limits_0^\pi \frac{\pi - x}{1 + \cos \alpha \sin x} dx . ....... . . . \left( 2 \right)\]
Adding (1) and (2), we get

\[2I = \int\limits_0^\pi \frac{\pi}{1 + \cos \alpha \sin x} dx \]
\[ \Rightarrow I = \frac{\pi}{2} \int\limits_0^\pi \frac{1}{1 + \cos \alpha \sin x} dx \]
\[ = \frac{\pi}{2} \int\limits_0^\pi \frac{1}{1 + \cos \alpha \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} dx \]
\[ = \frac{\pi}{2} \int\limits_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + \cos \alpha 2\tan \frac{x}{2}} dx\]
\[\text{Putting }\tan\frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]
\[\text{When }x \to 0 ; t \to 0\]
\[\text{and }x \to \pi ; t \to \infty \]
Now, integral becomes

\[I = \pi \int\limits_0^\infty \frac{dt}{1 + t^2 + 2t \cos \alpha} \]

\[ = \pi \int\limits_0^\infty \frac{dt}{\left( t + \cos \alpha \right)^2 + 1 - \cos^2 \alpha}\]

\[ = \pi \int\limits_0^\infty \frac{dt}{\left( t + \cos \alpha \right)^2 + \sin^2 \alpha}\]

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

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

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

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

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

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

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