Compute $\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$ Compute $$\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
Tried use substitution
$$\mu=5-3\cos x$$
But made the problem even complecated...Any help?
Thank you~
 A: $$I=\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
$\cos(-x)=\cos(x)$, hence we can write:
$$
I=\frac{1}{2}\int^{\pi}_{-\pi}\frac{1}{(5-3\cos x)^3}dx=\frac{1}{2}\int^{2\pi}_{0}\frac{1}{(5+3\cos x)^3}dx
$$
Last expression can be calculated through methods of complex analysis:
$$
I=\frac{1}{2}\int_{|z|=1}\frac{1}{\left(5+\frac{3}{2}(z+\frac{1}{z})\right)^3}\frac{dz}{iz}=-i\frac{1}{2}\int_{|z|=1}\frac{z^2}{\left(5z+\frac{3}{2}(z^2+1)\right)^3}dz\to\\
I=-\frac{4i}{27}\int_{|z|=1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}dz
$$
Roots of the denominator: $z_1=-\frac{1}{3},z_2=-3$. Obviously, within unit circle only $z_1$, then:
$$
I=2\pi i*\left(-\frac{4i}{27}\right)Res_{z=z_1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}
$$
Here we have pole of order 3, hence:
$$
Res_{z=z_1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}\to\\\frac{1}{2!}\frac{d^2}{dz^2}\frac{z^2}{(z-z_2)^3}|_{z=z_1}=\frac{1}{2!}\left(\frac{2}{(z_1-z_2)^3}-\frac{12z_1}{(z_1-z_2)^4}+\frac{12z^2_1}{(z_1-z_2)^5}\right)=\frac{1593}{16384}
$$
Thus, 
$$
I=2\pi i*\left(-\frac{4i}{27}\right)*\frac{1593}{16384}=\frac{59}{2048}\pi
$$
A: Hint:
put
$t=tan(\frac {x}{ 2})$
with
$$cos(x)=\frac{1-t^2}{1+t^2}$$
and
$$dx=\frac{2}{(1+t^2)}dt$$
the integral becomes
$\int_0^{+\infty}2\frac{(1+t^2)^2}{(5(1+t^2)-3(1-t^2))^3} dt=$
$2\int_0^{+\infty}\frac{(1+t^2)^2}{(2+8t^2)^3}dt$
which we can compute using partial fractions.
