Question

If ${\int }_0^{\pi }\frac{x^2\cos x}{(1+\sin x{)}^2}dx=A\pi -{\pi }^2$ then A is

Answer 1

Integrating by parts, we have
${\int }_0^{\pi }\frac{x^2\cos x}{(1+\sin x{)}^2}dx$
$={-\frac{x^2}{1+\sin x}|}_0^{\pi }+2{\int }_0^{\pi }\frac{x}{1+\sin x}dx=-{\pi }^2+2I$
where $I={\int }_0^{\pi }\frac{x}{1+\sin x}dx={\int }_0\pi \frac{\pi -x}{1+\sin x}dx=\pi {\int }_0^{\pi }\frac{dx}{1+\sin x}-I$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{dx}{1+\sin x}=2\pi {\int }_0^{\pi /2}\frac{dx}{1+\sin x}$
$\Rightarrow I=\pi {\int }_0^{\pi /2}\frac{dx}{1+\sin x}=\pi {\int }_0^{\pi /2}\frac{dx}{1+\sin (\pi /2-x)}$
$={\int }_0^{\pi /2}\frac{dx}{1+\cos x}$
$=\frac{\pi }{2}{\int }_0^{\pi /2}{\sec }^2\left(x/2\right)dx={\pi \tan \left(x/2\right)]}_0^{\pi /2}=\pi$
Hence ${\int }_0^{\pi }\frac{x^2\cos x}{(1+\sin x{)}^2}dx=-{\pi }^2+2\pi$
Therefore $A=2$