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.

Putting $t = 2\pi - x$ we get $$C(a, b) = \int_{0}^{2\pi}\frac{2\pi - t}{a+b\cos^2t}\,dt= 2\pi\int_{0}^{2\pi}\frac{dt}{a + b\cos^2t} - C(a, b)$$ so that $$C(a, b) = \pi\int_{0}^{2\pi}\frac{dx}{a+b\cos^2x}$$ Since the integrand satisfies equation $f(2\pi - x) = f(x)$ it follows that $$C(a, b)=2\pi\int_{0}^{\pi}\frac{dx}{a+b\cos^2x}$$ and using the same argument again we get $$C(a,b)=4\pi\int_{0}^{\pi/2}\frac{dx}{a+b\cos^2x}=8\pi\int_{0}^{\pi/2}\frac{dx}{2a+b+b\cos 2x}$$ Putting $2x=t$ we get $$C(a, b) = 4\pi\int_{0}^{\pi}\frac{dt}{2a+b + b\cos t}=\frac{4\pi^2}{\sqrt{(2a+b)^2-b^2}}=\frac{2\pi^2}{\sqrt{a^2+ab}}$$ where we have used the standard integral formula $$\int_{0}^{\pi}\frac{dx}{A+B\cos x}=\frac{\pi}{\sqrt{A^2-B^2}}, A>|B|$$ which can be proved using the substitution $$(A + B\cos x)(A - B\cos t) = A^2-B^2$$
