integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$

By making change of variable $u=2\pi -x$, we have $$\begin{align*} C(a,b)&=\int_0^{2\pi}\frac{2\pi -u}{a+b\cos^2 u}du=\int_0^{2\pi}\frac{2\pi}{a+b\cos^2 u}du-C(a,b), \end{align*}$$ hence $\displaystyle C(a,b)=\pi\int_0^{2\pi}\frac{1}{a+b\cos^2 u}du$. On the other hand, we find that $$\begin{align*} \int_0^{2\pi}\frac{1}{a+b\cos^2 u}du&=4\int_0^{\frac\pi 2}\frac{1}{a\sin^2 u+(b+1)\cos^2 u}du \\&=4\int_0^{\frac\pi 2}\frac{1}{a\tan^2 u+b+1}\sec^2 u\ du \\&=4\int_0^\infty \frac{dv}{av^2+(b+1)}\tag{$v=\tan u$} \\&=\frac{4}{\sqrt{a^2+ab}}\left[\arctan\left(\frac{\sqrt{a}v}{\sqrt{b+1}}\right)\right]^\infty_0\\&=\frac{2\pi}{\sqrt{a^2+ab}}. \end{align*}$$ This gives $$ C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}. $$ (I guess $2\pi^2$ is the right constant since this gives the right result $2\pi^2$ when $a=1,b=0$.)