Question

Evaluate $\stackrel{\pi }{\underset{0}{\int }}\frac{x\sin x}{1+\sin x}dx$.

• A. $0$
• B. $\pi$
• C. $\frac{{\pi }^2}{2}$
• D. $\frac{{\pi }^2}{2}-\pi$

Answer 1

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

Let $I={\int }_0^{\pi }\frac{x\sin x}{1+\sin x}dx$ .......(1)

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\pi }\frac{(\pi -x)\sin x}{1+\sin x}dx$ .......(2)

Add (1) and (2)

$2I={\int }_0^{\pi }\frac{\pi \sin x}{1+\sin x}dx=2\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+\sin x}dx$

$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}dx$

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

$=\pi {\int }_0^{\frac{\pi }{2}}(\sec x\tan x-{\tan }^{2}x)dx$

$=\pi {\int }_0^{\frac{\pi }{2}}(\sec x\tan x+1-{\sec }^{2}x)dx$

$=\pi {[\sec x-\tan x+x]}_0^{\frac{\pi }{2}}$

$=\pi {[\frac{1-\sin x}{\cos x}+x]}_0^{\frac{\pi }{2}}$

$=\pi {[\frac{\cos x}{1+\sin x}+x]}_0^{\frac{\pi }{2}}$

$=\pi [(0+\frac{\pi }{2})-(1+0)]$

$=\frac{{\pi }^2}{2}-\pi$