integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ provides 
$$\int_{-\pi}^\pi \frac{xdx}{a+b\cos^2x}=0$$
and with symmetry, 
$$C(a,b)=2\pi\int_0^\pi \frac{dx}{a+b\cos^2x}$$
Then we use $t=\tan(x/2)$:
$$C(a,b)=4\pi\int_0^\infty \frac1{a+b\left[\frac{t^2-1}{t^2+1}\right]^2}\frac{dt}{t^2+1}$$
$$C(a,b)=\frac{4\pi}{a+b}\int_0^\infty \frac{x^2+1}{x^4+2\frac{a-b}{a+b}x^2+1}dx$$
Then we consider 
$$N_s(k)=\int_0^\infty\frac{x^{2s}}{x^4+2kx^2+1}dx$$
Then $x\mapsto 1/x$ gives
$$N_s(k)=N_{1-s}(k)$$
and with $s=0$:
$$C(a,b)=\frac{8\pi}{a+b}\int_0^\infty \frac{dx}{x^4+2kx^2+1}\qquad k=\frac{a-b}{a+b}$$
Then with 
$$x^4+(2-c^2)x^2+1=(x^2+cx+1)(x^2-cx+1)$$
we have that 
$$\begin{align}
u(c)=N_0\left(\frac{2-c^2}2\right)=&\frac1{4c}\int_0^\infty\frac{2x+c}{x^2+cx+1}dx-\frac1{4c}\int_0^\infty\frac{2x-c}{x^2-cx+1}dx\\
&+\frac14\int_0^\infty\frac{dx}{x^2+cx+1}+\frac14\int_0^\infty\frac{dx}{x^2-cx+1}
\end{align}$$
The first two integrals vanish and we have 
$$u(c)=\frac14I(1,c,1)+\frac14I(1,-c,1)$$
where 
$$I(a,b,c)=\int_0^\infty\frac{dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\left[\frac\pi2-\arctan\frac{b}{\sqrt{4ac-b^2}}\right]$$
So 
$$4u(c)=\frac2{\sqrt{4-c^2}}\left[\frac\pi2-\arctan\frac{c}{\sqrt{4-c^2}}+\frac\pi2-\arctan\frac{-c}{\sqrt{4-c^2}}\right]$$
$$u(c)=\frac{\pi}{2\sqrt{4-c^2}}$$
$$N_0(a)=\frac\pi{2\sqrt{2+2a}}$$
And then 
$$C(a,b)=\frac{8\pi}{a+b}N_0\left(\frac{a-b}{a+b}\right)$$
$$C(a,b)=\frac{4\pi^2}{\sqrt{a^2+ab}}$$
How else can you prove this? Have fun ;)

As it turns out, we may be missing a factor of $1/2$, so we may actually have 
$$C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}$$
an issue discussed in the comment section of @Song's answer.

Note that $\cos^2(x)=\frac{1+\cos(2x)}{2}$.  Therefore, 
$$\begin{align}
C(a,b)&=2\pi\int_0^\pi \frac1{a+b\cos^2(x)}\,dx\\\\
&=2\pi \int_0^\pi \frac{2}{2a+b+b\cos(2x)}\,dx\\\\
&=2\pi \int_0^{2\pi}\frac{1}{2a+b+b\cos(x)}\,dx\tag1\\\\
&=4\pi \int_0^\pi \frac{1}{2a+b+b\cos(x)}\,dx\tag2
\end{align}$$
We can proceed by either (i) letting $z=e^{ix}$ in $(1)$ and using contour integration or (ii) letting $t=\tan(x/2)$ in $(2)$.

METHODOLOGY $1$:  CONTOUR INTEGRATION
Letting $z=e^{ix}$ in $(1)$ reveals
$$\begin{align}
C(a,b)&=2\pi \oint_{|z|=1} \frac{1}{2a+b +b\left(\frac{z+z^{-1}}{2}\right)}\,\frac1{iz}\,dz\\\\
&=\frac{4\pi}{ib}\oint_{|z|=1}\frac{1}{z^2+2(1+2a/b)z+1}\,dz\\\\
&=2\pi i \left(\frac{4\pi}{ib}\right)\text{Res}\left(\frac{1}{z^2+2(1+2a/b)z+1}, z=-(1+2a/b)+\frac2b\,\sqrt{a(a+b)}\right) \\\\
&=\frac{8\pi^2}{b}\left(\frac{1}{4\frac{\sqrt{a(a+b)}}{b}}\right)\\\\
&=\frac{2\pi^2}{\sqrt{a(a+b)}}
\end{align}$$

METHODOLOGY $2$:  Tangent Half-Angle Substitution
Letting $t=\tan(x/2)$ in $(2)$, we obtain
$$\begin{align}
C(a,b)&=4\pi \int_0^\pi \frac{1}{2a+b+b\cos(x)}\,dx\\\\
&=4\pi \int_0^\infty \frac{1}{2a+b+b\frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\
&=\frac{4\pi}{ a}\int_0^\infty \frac{1}{t^2+(1+b/a)}\,dt\\\\
&=\frac{4\pi}{a}\frac{\pi/2}{\sqrt{1+b/a}}\\\\
&=\frac{2\pi^2}{\sqrt{a(a+b)}}
\end{align}$$
