Question

The solution of ${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$, is

• A. $\frac{{\pi }^2}{4}$
• B. ${\pi }^2$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

${\pi }^2$

$I=2{\int }_{-\pi }^{\pi }\frac{x}{1+{\cos }^{2}x}dx+2{\int }_{-\pi }^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$
$I=0+2{\int }_{-\pi }^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx=2\cdot 2{\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$
$I=4{\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$ ..(1)

$I=4{\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}dx$ ...(2)

Adding (1) and (2), we get

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

$I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

Assume $\cos x=t\Rightarrow -\sin xdx=dt$

$I=4\pi {\int }_0^1\frac{1}{1+t^2}dt$

$I=4\pi {\left[{\tan }^{-1}t\right]}_0^1$

$I={\pi }^2$