_-^ 2x(1+ x)1+^2x,dx is:

Question

$\displaystyle\int_{-\pi}^\pi \dfrac{2x(1+\sin x)}{1+\cos^2x}\,dx$ is:

Accepted Answer

$\int_{-\pi}^\pi \dfrac{2x(1+\sin x)}{1+\cos^2x}\,dx = \int_{-\pi}^\pi \dfrac{2x\,dx}{1+\cos^2x} + 2\int_{-\pi}^\pi \dfrac{x\sin x}{1+\cos^2x}\,dx$

$= 0 + 4\int_0^\pi \dfrac{x\sin x}{1+\cos^2x}\,dx$ (first term is odd, integrates to 0; second is even)

Let $I = 4\int_0^\pi \dfrac{x\sin x}{1+\cos^2x}\,dx$. Using $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$:

$I = 4\int_0^\pi \dfrac{(\pi-x)\sin x}{1+\cos^2x}\,dx = 4\pi\int_0^\pi \dfrac{\sin x}{1+\cos^2x}\,dx - I$

$2I = 4\pi\int_0^\pi \dfrac{\sin x}{1+\cos^2x}\,dx$

Put $\cos x=t \Rightarrow -\sin x\,dx=dt$; limits $x=0\Rightarrow t=1$, $x=\pi\Rightarrow t=-1$

$I = 2\pi\int_{-1}^1 \dfrac{dt}{1+t^2} = 2\pi[	an^{-1}t]_{-1}^1 = 2\pi\left[\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}ight)ight] = 2\pi\cdot\dfrac{\pi}{2} = \pi^2$

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