Question

The value of $\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$ is

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

Answer 1

The correct option is B ${\pi }^2$

$I=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$
$=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x}{1+{\cos }^{2}x}dx+2\underset{-\pi }{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx\cdots \left(i\right)$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$=0+4\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx$
$=4\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}$
$=4\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x)\sin x}{1+{\cos }^{2}x}dx$
$=4\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx-4\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx$
$\Rightarrow 2I=4\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$
$\Rightarrow I=2\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$
put $\cos x=t\Rightarrow -\sin xdx=dt$
When $x=0,t=1;$ and $x=\pi ,t=-1.$
$\Rightarrow I=2\pi \underset{1}{\overset{-1}{\int }}\frac{-dt}{1+t^2}$
$\Rightarrow I=2\pi \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}$
$=4\pi {\left[{\tan }^{-1}t\right]}_0^1$
$=4\pi \cdot \frac{\pi }{4}={\pi }^2$