int_{-pi}^pi frac{x sin x}{1+cos ^2 x} d x= | vedclass

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Question

(A) Let $f(x) = \frac{x \sin x}{1+\cos^2 x}$.Since $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,the function

Vedclass · Accepted Answer

$\frac{\pi^2}{2}$

Vedclass · Answer

(A) Let $f(x) = \frac{x \sin x}{1+\cos^2 x}$.Since $f(-x) = \frac{(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{(-x)(-\sin x)}{1+\cos^2 x} = f(x)$,the function is even.Using the property $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$,we have $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$.Applying $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $I = 2 \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$.Adding the two expressions for $I$:$2I = 2 \int_0^\pi \frac{\pi \sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$.Let $t = \cos x$,then $dt = -\sin x dx$. Limits change from $0$ to $\pi$ to $1$ to $-1$.$I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2} = 2\pi \int_0^1 \frac{dt}{1+t^2}$.$I = 2\pi [	an^{-1} t]_0^1 = 2\pi (\frac{\pi}{4} - 0) = \frac{\pi^2}{2}$.

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

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