Question

${\int }_0^{\pi }\frac{x^2\cos x}{{(1+\sin x)}^2}dx$ equals

• A. $\pi (\pi -2)$
• B. ${\pi }^2(\pi +2)$
• C. $\pi (2-\pi )$
• D. None of these

Answer 1

Answer: (C)

$\pi (2-\pi )$

$I={\int }_0^{\pi }\frac{x^2\cos x}{{(1+\sin x)}^2}dx$
$={[-\frac{x^2}{1+\sin x}]}_0^{\pi }+2{\int }_0^{\pi }\frac{x}{1+\sin x}dx$
$=-{\pi }^2+2I_1$
Where $I_1={\int }_0^{\pi }\frac{x}{1+\sin x}dx={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin (\pi -x)}dx$
$=\pi {\int }_0^{\pi }\frac{x}{1+\sin x}dx-I_1$
$\Rightarrow 2I_1=\pi {\int }_0^{\pi }\frac{1}{1+\sin x}dx=2\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin x}dx$
$\Rightarrow I_1=\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin x}dx=\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin (\frac{\pi }{2}-x)}dx$
$=\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{1+\cos x}dx=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{\sec }^2\frac{x}{2}dx$
$={\left[\pi \tan \frac{x}{2}\right]}_0^{\frac{\pi }{2}}=\pi$
Therefore $I={\int }_0^{\pi }\frac{x^2\cos x}{{(1+\sin x)}^2}dx={-\pi }^2+2\pi$