<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] Just a guideline:
Use geometric series to get
$$
I=\frac{1}{8}\sum_{n=0}^{\infty}\frac{(-1)^n}{8^n}\underbrace{\int_{0}^{2 \pi}x^2 \sin^{2n+1}(x)}_{I_n}
$$
Show that the inner integral is equal to (use integration by parts and find a recurion relation):
$$
I_n=2\pi ^2\frac{\sqrt{\pi }\Gamma (n+1)}{\Gamma(n+\frac{3}{2})}
$$
Use the Series expansion of $\text{arcsinh}(x)$  together with $\Gamma$-duplication formulas
To obtain
$$
I=-\frac{4\pi^2}{3} \text{arcsinh}\left(\frac{1}{2 \sqrt{2}}\right)
$$
which is after some algebra equal to $$-\frac{2 \pi^2}{3}\log(2)$$
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