Asking for help on a question about integral: $\int_0^\pi \frac{x\sin x}{3+\cos^2 x}\mathrm{d}x$

Question

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.

Answer 1

Hint...there is a useful trick you could try using here:

$$I=\int_0^{\pi}xf(\sin x)dx=\pi\int_0^{\pi}f(\sin x)dx-I$$

which is readily seen if you substitute $u=\pi-x$ into the first integral...

Answer 2

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 $a+b+c = 11$