int_{-pi}^pi frac{x sin ^3 x}{4-cos ^2 x} d x | vedclass

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(B) Let $I = \int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$. Since $f(x) = \frac{x \sin ^3 x}{4-\cos ^2 x}$ is an even function (as $f(-x) = \fr

Vedclass · Accepted Answer

$2 \pi\left(1-\frac{3}{4} \log 3\right)$

Vedclass · Answer

(B) Let $I = \int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$. Since $f(x) = \frac{x \sin ^3 x}{4-\cos ^2 x}$ is an even function (as $f(-x) = \frac{(-x) \sin ^3(-x)}{4-\cos ^2(-x)} = \frac{(-x)(-\sin^3 x)}{4-\cos^2 x} = f(x)$),we can write: $I = 2 \int_0^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$ ....$(i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = 2 \int_0^\pi \frac{(\pi-x) \sin ^3(\pi-x)}{4-\cos ^2(\pi-x)} d x = 2 \int_0^\pi \frac{(\pi-x) \sin ^3 x}{4-\cos ^2 x} d x$ ....(ii) Adding $(i)$ and (ii): $2I = 2 \pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \Rightarrow I = \pi \int_0^\pi \frac{(1-\cos^2 x) \sin x}{4-\cos ^2 x} d x$. Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi, t=-1$. $I = -\pi \int_1^{-1} \frac{1-t^2}{4-t^2} dt = \pi \int_{-1}^1 \frac{1-t^2}{4-t^2} dt = 2\pi \int_0^1 \frac{1-t^2}{4-t^2} dt$. $I = 2\pi \int_0^1 \frac{-(t^2-4)+3}{4-t^2} dt = 2\pi \int_0^1 (1 - \frac{3}{4-t^2}) dt$. $I = 2\pi [t - \frac{3}{2(2)} \log |\frac{2+t}{2-t}|]_0^1 = 2\pi [1 - \frac{3}{4} \log 3]$.

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

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