Question

Solve
${\int }_0^{\pi }\frac{x}{\left(1+\sin x\right)}dx$

Answer 1

$\begin{array}{c}{\int }_0^{\pi }\frac{x}{\left(1+\sin x\right)}dx\\ LetI={\int }_0^{\pi }\frac{\left(\pi -x\right)}{\left(1+\sin x\right)}dx..........\left(i\right)\\ Then,I={\int }_0^{\pi }\frac{\left(\pi -x\right)}{1+\sin \left(\pi -x\right)}dx={\int }_0^{\pi }\frac{\left(\pi -x\right)}{\left(1+\sin x\right)}dx........\left(ii\right)\\ Adding\left(i\right)and\left(ii\right),weget\\ 2I=\pi {\int }_0^{\pi }\frac{dx}{\left(1+\sin x\right)}\\ =\pi .{\int }_0^{\pi }\frac{1}{\left(1+\sin x\right)}\times \frac{\left(1-\sin x\right)}{\left(1-\sin x\right)}dx\\ or2I=\pi {\int }_0^{\pi }\left(\frac{1-\sin x}{{\cos }^{2}x}\right)dx\\ =\pi .\left[{\int }_0^{\pi }{\sec }^{2}xdx-{\int }_0^{\pi }\sec x\tan xdx\right]\\ =\pi \left\{{\left[\tan x\right]}_0^{\pi }-{\left[\sec x\right]}_0^{\pi }\right\}=2\pi \\ \therefore I=\pi ,i.e.,{\int }_0^{\pi }\frac{x}{\left(1-\sin x\right)}dx=\pi \end{array}$