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$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is
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2757 Upvotes
Verified Answer
The correct answer is:
$\pi^2$
$\pi^2$
$$
\begin{aligned}
& \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x}+2 \int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} \\
& =0+4 \int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} I=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} \\
& I=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \Rightarrow I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-4 \pi \int \frac{x \sin x}{1+\cos ^2 x} \Rightarrow 2 I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x
\end{aligned}
$$
put $\cos x=t$ and solve it.
\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.
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