Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\pi }\frac{xdx}{1+\sin x}$

Answer 1

Let $I={\int }_0^{\pi }\frac{xdx}{1+\sin x}$ .........(1)
$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin (\pi -x)}dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin x}dx$ ............. (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\pi }\frac{\pi }{1+\sin x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{(1-\sin x)}{1+\sin x\left)\right(1-\sin x)}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\{{\sec }^{2}x-\tan x\sec x\}dx$
$\Rightarrow 2I=\pi [\tan x-\sec x{]}_0^{\pi }$
$\Rightarrow 2I=\pi \left[2\right]\Rightarrow I=\pi$