Answer mismatch in definite integral question:$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$ 
I faced a problem while evaluating the following integral.
$$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$

MY ATTEMPT:
Let $$I=\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$
By King's Rule $$I=-\int_{0}^{2\pi}{\frac{(2\pi-x)^2\sin(x)}{8+\sin^2(x)}}dx$$
On adding the above two integrals,$$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2+4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
or, $$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2)\sin(x)}{8+\sin^2(x)}}dx+\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
The first integral follows the property $f(x)=-f(2\pi-x)$ so it vanishes.
Thus,
$$2I=\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
or,
$$I=2 \pi\int_{0}^{2\pi}{\frac{ x\sin(x)}{8+\sin^2(x)}}dx$$
Applying king's rule and adding,
$$2I=2 \pi\int_{0}^{2\pi}{\frac{ 2\pi\sin(x)}{8+\sin^2(x)}}dx$$
$$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{8+\sin^2(x)}}dx$$
$$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{9-\cos^2(x)}}dx$$
$$I=[2\pi^2(\frac{1}{6} \log(3-\cos(x))-\frac{1}{6} \log(\cos(x)+3))]_{0}^{2\pi}+C$$
Thus,$I=0$
But the answer is not 0.Where did I go wrong?
P.S King's Rule is $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
 A: Symmetry is a bit more evident if we set $x=z+\pi$. Since $\sin(z+\pi)=-\sin(z)$,
$$ I = -\int_{-\pi}^{\pi}\frac{(z^2+2\pi z+\pi^2)\sin(z)}{8+\sin(z)^2}\,dz=-2\pi\int_{-\pi}^{\pi}\frac{z\sin(z)}{8+\sin(z)^2}=-4\pi\int_{0}^{\pi}\frac{z\sin z}{8+\sin(z)^2}\,dz $$
and since $\sin(\pi-z)=\sin(z)$,
$$ I = -4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\sin z}{8+\sin(z)^2}\,dz=4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\cos' z}{9-\cos(z)^2}\,dz =-4\pi^2\int_{0}^{1}\frac{du}{9-u^2}$$
and by partial fraction decomposition:
$$ I = -\frac{2\pi^2}{3}\int_{0}^{1}\left(\frac{1}{3-u}+\frac{1}{3+u}\right)\,du=-\frac{2\pi^2}{3}\left(\log\frac{3}{2}+\log\frac{4}{3}\right)=\color{red}{-\frac{2\pi^2}{3}\log 2}.$$
A: In your second application of King's rule, you forgot the negative sign!
$$I=2\pi\int_0^{2\pi}\frac{x \sin(x)}{8+\sin^2(x)}dx$$ becomes
$$I=-2\pi\int_0^{2\pi}\frac{(2\pi-x)\sin(x)}{8+\sin^2(x)}dx$$, so adding these two gives you nothing.
A: The second time apply King's rule and add you do so incorrectly due to a plus/minus error. The $x$ term should not have cancelled out.
Putting in your missing steps:
$$I=2 \pi\int_{0}^{2\pi}{\frac{ x\sin(x)}{8+\sin^2(x)}}dx$$
Applying the rule gives:
$$I=2 \pi\int_{0}^{2\pi}{\frac{ (2\pi-x)\sin(2\pi-x)}{8+\sin^2(2\pi-x)}}dx$$
$$I=-2 \pi\int_{0}^{2\pi}{\frac{ (2\pi-x)\sin(x)}{8+\sin^2(x)}}dx$$
$$=2 \pi\int_{0}^{2\pi}{\frac{ (x-2\pi)\sin(x)}{8+\sin^2(x)}}dx$$
Then adding should give you:
$$2I=2 \pi\int_{0}^{2\pi}{\frac{ (2x-2\pi)\sin(x)}{8+\sin^2(x)}}dx$$
