<s>[INST] Prove that $\int_{0}^{2\pi}\frac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$ Prove that $$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$$
My Attempt:
$$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\int_{0}^{2\pi}x^2\frac{\sin x}{8+\sin^2x}$$
I applied integration by parts,considering $x^2$ as first function and $\dfrac{\sin x}{8+\sin^2x}$ as second function
$$=x^2\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx-\int_{0}^{2\pi}2x\left(\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx\right)dx$$
but this calculates out to be zero because $$\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx=0$$
What has gone wrong?Please help me..
 [/INST] Let $$\displaystyle I = \int_{0}^{2\pi}\frac{x^2\cdot \sin x}{8+\sin^2 x}dx ...................(1)$$
Now Replace $x\rightarrow (2\pi-x)\;,$ We get 
$$\displaystyle I = -\int_{0}^{2\pi}\frac{\left(2\pi-x\right)^2\cdot \sin x}{8+\sin^2 x}dx...........(2)$$
Now Add these Two equations, We get
$$\displaystyle 2I = \int_{0}^{2\pi}\frac{2\pi(2\pi-2x)\cdot \sin x}{8+\sin^2 x}dx$$
So we get $$\displaystyle I = 2\pi\int_{0}^{2\pi}\frac{x\cdot \sin x}{9-\cos^2 x}dx-2\pi^2\underbrace{\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2 x}dx}_{J}$$
Now value of $J=0\;,$ Using $\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$
Now  Using Integration By parts for $I\;,$ We get
$$\displaystyle I = 2\pi\left[-\frac{x}{6}\cdot \ln \left(\frac{3+\cos x}{3-\cos x}\right)_{0}^{2\pi}+\frac{2}{6}\underbrace{\int_{0}^{2\pi}\ln \left(\frac{3+\cos x}{3-\cos x}\right)}_{0}\right] = -\frac{2\pi^2}{3}\cdot \ln\left(2\right)$$
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