Question

${\int }_{-\pi }^{\pi }\frac{2x(1+sinx)}{1+co{s}^{2}x}dxis$

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

Answer 1

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

$LetI={\int }_{-\pi }^{\pi }\frac{2x(1+sinx)}{1+co{s}^{2}x}dx$
$={\int }_{-\pi }^{\pi }\frac{2x}{1+co{s}^{2}x}dx+{\int }_{-\pi }^{\pi }\frac{2xsinx}{1+co{s}^{2}x}dx$
$\Rightarrow I=4{\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}dx--------\left(1\right)$
$\Rightarrow I=4{\int }_0^{\pi }\frac{(\pi -x)sin(\pi -x)}{1+co{s}^2(\pi -x)}dx-------\left(2\right)$
From (1) and (2)
$\Rightarrow I=4\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx-I$
$\Rightarrow I=2\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx$
$Putcosx=t\Rightarrow -sinxdx=dt$
$\therefore I=-2\pi {\int }_1^{-1}\frac{1}{1+t^2}dt={\pi }^2$