Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+\sin x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\sin \left(\mathrm{\pi }-\mathrm{x}\right)}{1+\sin \left(\mathrm{\pi }-\mathrm{x}\right)}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\sin x}{1+\sin x}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\mathrm{\pi }}\left(x+\mathrm{\pi }-\mathrm{x}\right)\frac{\sin x}{1+\sin x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\sin x}{1+\sin x}dx \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1+\mathrm{sinx}-1}{1+\mathrm{sinx}}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1}{1+\mathrm{sinx}}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\left(1-\mathrm{sinx}\right)}{\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\left(1-\mathrm{sinx}\right)}{1-{\sin }^{2}x}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\left(1-\mathrm{sinx}\right)}{{\cos }^{2}x}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\mathrm{dx}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left({\sec }^{2}x-\sec x\tan x\right)\mathrm{dx} \\ =\mathrm{\pi }{\left[\mathrm{x}\right]}_0^{\mathrm{\pi }}-\mathrm{\pi }{\left[\mathrm{tanx}-\mathrm{secx}\right]}_0^{\mathrm{\pi }} \\ ={\mathrm{\pi }}^2-\mathrm{\pi }\left(0+1-0+1\right) \\ ={\mathrm{\pi }}^2-2\mathrm{\pi } \\ \mathrm{Hence}I=\mathrm{\pi }\left(\frac{\mathrm{\pi }}{2}-1\right) \end{gathered}$$