Question

Solve ${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$

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

Answer 1

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

$I={\int }_{-\pi }^{\pi }\frac{2x\left(1+\sin x\right)}{1+{\cos }^{2}x}dx$ ---(i)

$={\int }_{-\pi }^{\pi }\frac{-2x\left(1-\sin x\right)}{1+{\cos }^{2}x}dx$ ($\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$)

$={\int }_{-\pi }^{\pi }\frac{2x\left(-1+\sin x\right)}{1+{\cos }^{2}x}dx$ ---(ii)

adding (i) and (ii)

$\Rightarrow 2I={\int }_{-\pi }^{\pi }\frac{2x\left(1+\sin x-1+\sin x\right)}{1+{\cos }^{2}x}dx$

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

$\Rightarrow I=4{\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$ ---(iii) (since integrand is even function)

$\Rightarrow I=4{\int }_0^{\pi }\frac{\left(\pi -x\right)\sin \left(\pi -x\right)}{1+{\cos }^2\left(\pi -x\right)}dx$ ($\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$)

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

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

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

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

putting $cosx=t$

$\Rightarrow \sin xdx=-dt$

$\Rightarrow I=2\pi {\int }_1^{-1}\frac{-dt}{1+t^2}$

$\Rightarrow I=2\pi {\int }_{-1}^1\frac{dt}{1+t^2}$ ($\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx$)

$\Rightarrow I=4\pi {\int }_0^1\frac{dt}{1+t^2}$ (since integrand is even function)

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

$\Rightarrow I=4\pi \left[\frac{\pi }{4}\right]$

$\Rightarrow I={\pi }^2$