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..
 A: 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
$$\displaystyle I = 2\pi\left[-\frac{x}{6}\cdot \ln \left(\frac{3+\cos x}{3-\cos x}\right)_{0}^{2\pi}+\frac{2}{6}\underbrace{\int_{0}^{2\pi}\ln \left(\frac{3+\cos x}{3-\cos x}\right)}_{0}\right] = -\frac{2\pi^2}{3}\cdot \ln\left(2\right)$$
A: The integration by parts formula is 
$\int_0^{2\pi}u\mathrm{d}v=[uv]_0^{2\pi}-\int_0^{2\pi}v\mathrm{d}u $
Note that when going from $\mathrm{d}v$ to $v$ we are computing the antiderivative, not a definite integral. So $v$ is a function with derivative $\mathrm{d}v$. Try applying this in your case with $u=x^2$, $v=\frac{\sin{x}}{8+\sin^2{x}}$
A: 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 $$-\frac{2 \pi^2}{3}\log(2)$$
