Question

The value of $\underset{0}{\overset{\pi }{\int }}\frac{2x\sin x}{3+\cos 2x}dx$ is

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

Answer 1

The correct option is A $\frac{{\pi }^2}{4}$

$I=\underset{0}{\overset{\pi }{\int }}\frac{2x\sin x}{3+\cos 2x}dx$
$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}\frac{2(\pi -x)\sin (\pi -x)}{3+\cos \left[2\right(\pi -x\left)\right]}dx$
$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}[\frac{2\pi \sin x}{3+\cos 2x}-\frac{2x\sin x}{3+\cos 2x}]dx$
$\Rightarrow 2I=\underset{0}{\overset{\pi }{\int }}\frac{2\pi \sin x}{3+\cos 2x}dx$
$\Rightarrow I=\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{2+2{\cos }^{2}x}dx[\because \cos 2x=2{\cos }^{2}x-1]$

Let $\cos x=t\Rightarrow -\sin xdx=dt$
$\Rightarrow I=\frac{\pi }{2}\underset{1}{\overset{-1}{\int }}\frac{-dt}{1+t^2}$
$=\frac{\pi }{2}\underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}$
$=\frac{\pi }{2}{\left[{\tan }^{-1}t\right]}_{-1}^1$
$=\frac{{\pi }^2}{4}$