The value of int_{-pi}^pi frac{2 y(1+sin y)}{ | vedclass

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Question

(A) Let $I = \int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$. Split the integral into two parts: $I = \int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d

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$\pi^2$

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(A) Let $I = \int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$. Split the integral into two parts: $I = \int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y + \int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y$. The first part $\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y = 0$ because the integrand is an odd function. For the second part,since $y \sin y$ is an even function,we have: $I = 2 \int_0^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y = 4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y$. Using the property $\int_0^a f(y) dy = \int_0^a f(a-y) dy$: $I = 4 \int_0^\pi \frac{(\pi-y) \sin y}{1+\cos ^2 y} d y = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y - 4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y$. $I = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y - I \implies 2I = 4\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y \implies I = 2\pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y$. Let $t = \cos y$,then $dt = -\sin y dy$. When $y=0, t=1$; when $y=\pi, t=-1$. $I = 2\pi \int_1^{-1} \frac{-dt}{1+t^2} = 2\pi \int_{-1}^1 \frac{dt}{1+t^2} = 2\pi [	an^{-1} t]_{-1}^1$. $I = 2\pi [	an^{-1}(1) - 	an^{-1}(-1)] = 2\pi [\frac{\pi}{4} - (-\frac{\pi}{4})] = 2\pi [\frac{\pi}{2}] = \pi^2$.

The value of $\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ is :

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