Question

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

Answer 1

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

\[ \Rightarrow I = \int\limits_0^\pi \frac{\pi - x}{1 + \sin\alpha \sin\left( \pi - x \right)}dx \left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]

\[ \Rightarrow I = \int\limits_0^\pi \frac{\pi}{1 + \sin\alpha \sin x}dx - \int\limits_0^\pi \frac{x}{1 + \sin\alpha \sin x}dx\]

\[ \Rightarrow I = \int\limits_0^\pi \frac{\pi}{1 + \sin\alpha \sin x}dx - I\]

\[ \Rightarrow 2I = \int\limits_0^\pi \frac{\pi}{1 + \sin\alpha \sin x}dx\]

\[ \Rightarrow 2I = \pi \int\limits_0^\pi \frac{1}{1 + sin\alpha sin x}dx\]

\[\text { Substituting }\sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}}, \text { we get }\]

\[2I = \pi \int\limits_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + sin\alpha \times 2\tan\frac{x}{2}}dx\]

\[I = \frac{\pi}{2} \int\limits_0^\pi \frac{\sec^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + \sin\alpha \times 2\tan\frac{x}{2}}dx\]

\[\text { Let } \tan\frac{x}{2} = t, d\left( \tan\frac{x}{2} \right) = dt \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]

\[\text { Also }, \]

\[\text { When } x \to 0, t \to \tan0 = 0\]

\[\text { When } x \to \pi, t \to \tan\frac{\pi}{2} = \infty \]

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

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

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

\[ \Rightarrow I = \frac{\pi}{\cos\alpha}\left[ \tan^{- 1} \infty - \tan^{- 1} \left( \tan\alpha \right) \right]\]

\[ \Rightarrow I = \frac{\pi}{\cos\alpha}\left( \frac{\pi}{2} - \alpha \right)\]