Asking for help on a question about integral: $\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x$ Given the following intergral, and the fact that $a$ and $c$ are prime numbers,
$$\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x= \frac{π^a}{b\sqrt c} $$
Evaluate $a+b+c$
I've tried to solve the intergral by using intergration by part where I let $u$ be $x$ and $dv$ be  $\frac{\sin x}{3+\cos^2 x}$. Which then gave me
$$\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x = \frac{-x}{\sqrt 3}\arctan{\left(\frac{\cos x}{\sqrt 3}\right)}\Big|_0^π + \int_0^\pi \frac{1}{\sqrt 3}\arctan{\left(\frac{\cos x}{\sqrt 3}\right)}\mathrm{d}x$$
But I'm pretty much out of luck at this point since I don't know how to solve $\int_0^π v\mathrm{d}u$ in this case and I believe there is a better way to do this.

So I eventually figured out the answer to
$$I_1 = \int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x$$
By using the following property: $$\int_a^b f(x)\mathrm{d}x = \int_a^b f(a+b-x)\mathrm{d}x$$ we can proof that $$I_1 = \frac{π}{2}\int_0^π f(\sin x)\mathrm{d}x = \frac{π}{2}I_2$$ where $I_2 = \int_0^π \frac{\sin x}{3+\cos^2 x}\mathrm{d}x$. To solve for $I_2$ we substitute $u$ for $\cos x$ which gave us $$I_2=\frac{1}{\sqrt 3}\arctan \left(\frac{u}{\sqrt 3}\right) \big|_{-1}^1=\frac{π\sqrt 3}{9}$$
Therefore $$I_1=\frac{π}{2}\frac{π\sqrt 3}{9}=\frac{π^2}{6\sqrt 3}$$
And $a+b+c = 11$
