$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is
$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is
-
$\frac{\pi^2}{4}$
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$\pi^2$
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zero
-
$\frac{\pi}{2}$
Solution
$
\begin{aligned}
& \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x}+2 \int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} \\
& =0+4 \int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} I=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} \\
& I=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \Rightarrow I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-4 \pi \int \frac{x \sin x}{1+\cos ^2 x} \Rightarrow 2 I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x
\end{aligned}
$
put $\cos x=t$ and solve it
Asked in: JEE Main 2002
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