# Calculating $\int_{0}^{\pi }{\frac{x\sin \left( x \right)}{a+b{{\cos }^{2}}\left( x \right)}dx}$

I am trying to find for $$a>b>0$$ : $$I=\int_{0}^{\pi }{\frac{x\sin \left( x \right)}{a+b{{\cos }^{2}}\left( x \right)}dx}$$ I don't think a substitution is an option here, besides differentiation under the sign integral makes the integral harder. Fortunately Mathematica gives a closed form for the integral in question:

$$\frac{\pi \tan^{-1}\left( \sqrt{\frac{b}{a}} \right)}{\sqrt{ab}}$$

• You can replace $x$ with $\pi - x$ – Mann Mar 30 '19 at 20:30

HINT:

Enforce the substitution $$x\mapsto \pi-x$$.

Then, note that

$$I=\int_0^\pi \frac{x\sin(x)}{a+b\cos^2(x)}\,dx=\int_0^\pi \frac{(\pi-x)\sin(x)}{a+b\cos^2(x)}\,dx$$

Can you finish now?

• that was quick and simple and i can finish the rest of the solution using the Tangent half-angle substitution? – logo Mar 30 '19 at 20:43
• @logo It'd be easier to use $u=\cos x$. – J.G. Mar 30 '19 at 21:29
• you are right, thanks – logo Mar 31 '19 at 3:32


• (+1) Hello Felix, my friend! – Mark Viola Mar 31 '19 at 3:44
• @MarkViola Thanks, Mark. Hello. – Felix Marin Mar 31 '19 at 17:41