Integrate $\int \frac{4}{5+3\cos(2x)}\,d x$ How do you integrate $$ \int \frac{4}{5+3\cos(2x)}\,dx $$ ?
I tried with substitution method ($u = 2x$, $u = \cos(2x)$, ...) without success.
Hints are accepted :)
 A: 
I thought it might be instructive to present a way forward that relies on Euler's Formula along with straightforward partial fraction expansion.  To that end, we proceed.


Another way forward is to use Euler's Formula to write $\cos(2x)=\frac{e^{i2x}+e^{-i2x}}{2}$.  Then, we have
$$\begin{align}
\int \frac{4}{5+3\cos(2x)}\,dx&=\int \frac{8e^{i2x}}{3e^{i4x}+10e^{i2x}+3}\,dx\\\\
&=\int\left(\frac{3e^{i2x}}{3e^{i2x}+1}-\frac{e^{i2x}}{e^{i2x}+3}\right)\,dx\\\\
&=\frac1{2i}\log\left(\frac{3e^{i2x}+1}{e^{i2x}+3}\right)+C\\\\
&=\frac12 \left(\arctan\left(\frac{3\sin2x)}{1+3\cos(2x)}\right)-\arctan\left(\frac{\sin(2x)}{3+\cos(2x)}\right)\right)+C\\\\
&=\frac12\arctan\left(\frac{4\sin(2x)}{3+5\cos(2x)}\right)+C 
\end{align}$$
A: Let's see a more general form
$$I=\int \frac{1}{a+b\cos x}\,\mathrm dx$$
let $t=\tan\left(\dfrac{x}{2}\right)$ we get
$$I=\int \frac{2}{(a+b)+t^2(a-b)}\,\mathrm dt$$
If $a^2 ＞ b^2$ , then we have
$$I=\frac{2}{a-b}\int \frac{1}{\left(\sqrt{\dfrac{a+b}{a-b}}\right)^2+t^2}\,\mathrm dt$$
use the known formula
$$\int \frac{1}{x^2+a^2}\,\mathrm dx=\frac{1}{a}\arctan\frac{x}{a}+C$$
we will get the answer of $I$. Hope you can take it from here.
