Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\mathrm{\pi }}\frac{x}{1+\cos \alpha \sin x}\mathit{d}x\mathit{}\mathit{}.....\left(1\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }-x}{1+\cos \alpha \sin \left(\mathrm{\pi }-x\right)}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }-x}{1+\cos \alpha \sin x}dx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get}, \\ 2I={\int }_0^{\mathrm{\pi }}\frac{x+\mathrm{\pi }-x}{1+\cos \alpha \sin x}\mathit{d}x \\ \Rightarrow I=\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{1}{1+\cos \alpha \mathrm{sinx}}dx \end{gathered}$$

$$\begin{gathered} =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{1}{1+\cos \alpha \mathrm{sinx}} \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{1}{1+\cos \alpha {\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}dx \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}+2\cos \mathrm{\alpha }\tan {\displaystyle \frac{x}{2}}}dx \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}+2\cos \mathrm{\alpha }\tan {\displaystyle \frac{x}{2}}}\mathrm{d}\mathrm{x} \end{gathered}$$
$$\begin{gathered} \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^{2}xdx=dt \\ \mathrm{When}x\to 0;t\to 0 \\ \mathrm{and}x\to \mathrm{\pi };t\to \infty \\ \therefore I=\frac{\mathrm{\pi }}{2}{\int }_0^{\infty }\frac{{\displaystyle 2}}{1+t^2+2\cos \mathrm{\alpha }t}dt \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\infty }\frac{{\displaystyle 2}}{{\left(\mathrm{t}+\cos \mathrm{\alpha }\right)}^2-{\cos }^2\alpha +1}dt \\ =\mathrm{\pi }{\int }_0^{\infty }\frac{{\displaystyle 1}}{{\left(\mathrm{t}+\cos \mathrm{\alpha }\right)}^2+{\sin }^2\alpha }dt \\ =\mathrm{\pi }{\left[\frac{1}{\sin \alpha }{\tan }^{-1}\left(\frac{t+\cos \mathrm{\alpha }}{\sin \mathrm{\alpha }}\right)\right]}_0^1 \\ =\frac{\mathrm{\pi }}{\mathrm{sin\alpha }}\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(\cot \alpha \right)\right] \\ =\frac{\mathrm{\pi }}{\mathrm{sin\alpha }}\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{2}-\alpha \right)\right)\right] \\ =\frac{\mathrm{\pi \alpha }}{\mathrm{sin\alpha }} \end{gathered}$$