Question

Let $I={\int }_0^{\pi /2}\frac{dx}{1+\sin x},$ then $I={\int }_0^{\pi /2}\frac{x^2\cos x}{{(1+\sin x)}^2}dx$

• A. $-{\pi }^2$
• B. $-{\pi }^2-2\pi I$
• C. $2\pi I$
• D. $2\pi I-{\pi }^2$

Answer 1

The correct option is D $2\pi I-{\pi }^2$

${\int }_0^{\pi /2}\frac{x^2\cos x}{{(1+\sin x)}^2}dx$
$={[-\frac{x^2}{1+\sin x}]}_0^{\pi }+2{\int }_0^{\pi }\frac{x}{1+\sin x}dx=-{\pi }^2+2I_1$
$I_1={\int }_0^{\pi }\frac{x}{1+\sin x}dx={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin (\pi -x)}dx$
$=\pi {\int }_0^{\pi }\frac{dx}{1+\sin x}-I_1$
$\Rightarrow 2I_1=\pi {\int }_0^{\pi }\frac{dx}{1+\sin x}$
$\Rightarrow I_1=\pi {\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\sin x}=I\pi$
$\therefore {\int }_0^{\pi /2}\frac{x^2\cos x}{{(1+\sin x)}^2}dx=-{\pi }^2+2\pi I$