Question

${\int }_0^{\pi }\frac{x^2\cos x}{{(1+\sin x)}^2}dx=\pi (2+\pi )$. If this is true enter 1, else enter 0.

Answer 1

Let $I={\int }_0^{\pi }\frac{x^2\cos x}{{(1+\sin x)}^2}dx$
Integrating by parts
$I={[-x^2\frac{1}{1+\sin x}]}_0^{\pi }+2\int \frac{1}{1+\sin x}xdx=-{\pi }^2+2I_1$
Where
$I_1={\int }_0^{\pi }\frac{x}{1+\sin x}dx={\int }_0^{\pi }\frac{\pi -x}{1+\sin (\pi -x)}dx$ ....( using integration property )
$2I_1=\pi {\int }_0^{\pi }\frac{1}{1+\sin x}dx=\pi {\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx$
$=\pi \int ({\sec }^{2}x-\sec x\tan x)dx=\pi {[\tan x-\sec x]}_0^{\pi }=\pi \left[2\right]=2\pi$
Therefore $I=-{\pi }^2+2\pi =\pi (2-\pi )$