Question

Evaluate :$\underset{0}{\overset{\pi }{\int }}\frac{x}{1+\sin \alpha \sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{\pi }{\int }}\frac{x}{1+\sin \alpha \sin x}\mathrm{d}x \\ \Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{\mathrm{\pi }-x}{1+\sin \alpha \sin \left(\mathrm{\pi }-x\right)}\mathrm{d}x\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ \Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{\mathrm{\pi }}{1+\sin \alpha \sin x}dx-\underset{0}{\overset{\pi }{\int }}\frac{x}{1+\sin \alpha \sin x}\mathrm{d}x \\ \Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{\mathrm{\pi }}{1+\sin \alpha \sin x}dx-I \\ \Rightarrow 2I=\underset{0}{\overset{\pi }{\int }}\frac{\mathrm{\pi }}{1+\sin \alpha \sin x}\mathrm{d}x \\ \Rightarrow 2I=\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1}{1+\mathrm{sin\alpha }\mathrm{sinx}}\mathrm{d}x \end{gathered}$$

$$\begin{gathered} \mathrm{Substituting}\sin x=\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}},\mathrm{we}\mathrm{get} \\ 2I=\mathrm{\pi }\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}+\mathrm{sin\alpha }\times 2\tan {\displaystyle \frac{x}{2}}}\mathrm{d}x \\ I=\frac{\mathrm{\pi }}{2}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{{\displaystyle {\sec }^2\frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}+\sin \alpha \times 2\tan {\displaystyle \frac{x}{2}}}\mathrm{d}x \\ \mathrm{Let}\tan \frac{x}{2}=t,d\left(\tan \frac{x}{2}\right)=dt\Rightarrow {\sec }^2\frac{x}{2}dx=2dt \\ \mathrm{Also}, \\ \mathrm{When}\mathrm{x}\to 0,t\to \tan 0=0 \\ \mathrm{When}x\to \mathrm{\pi },t\to \tan \frac{\mathrm{\pi }}{2}=\infty \\ \therefore \mathit{}I=\frac{\mathrm{\pi }}{2}\underset{0}{\overset{\infty }{\int }}\frac{2dt}{t^2+2t\sin \alpha +1} \\ \Rightarrow I=\mathrm{\pi }\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(t+\sin \alpha \right)}^2+{\cos }^2\mathrm{\alpha }}dt \\ \Rightarrow I=\frac{\mathrm{\pi }}{\cos \alpha }{\left[{\tan }^{-1}\left(\frac{t+\sin \alpha }{\cos \alpha }\right)\right]}_0^{\infty } \\ \Rightarrow I=\frac{\mathrm{\pi }}{\cos \alpha }\left[{\tan }^{-1}\infty -{\tan }^{-1}\left(\tan \alpha \right)\right] \\ \Rightarrow I=\frac{\mathrm{\pi }}{\cos \alpha }\left(\frac{\mathrm{\pi }}{2}-\alpha \right) \end{gathered}$$