Evaluate the integral $\int\limits_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$. Evaluate the integral $\displaystyle \int_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$.
Hint: put $z = e^{i\theta}$. 
Is there a way to solve this without using the Residue Theorem and $\tan(z)$? Is Cauchy's integral formula applicable? 

Another approach.
If $|b|<1$ then we can write:
$$\frac{1}{1-b\cos x}=\sum_{n=0}^{\infty} b^n\cos^n x$$
The constant term of the Fourier expansion of $\cos^n x$ is zero when $n$ odd and $\frac{1}{2^n}\binom{n}{n/2}$ when $n$ is even.
Now, $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}$ is the constant of the Fourier series expansion, and, since the series converges absolutely, you get that:
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=\sum_{n=0}^{\infty}\left(\frac{b}{2}\right)^{2n}\binom{2n}{n}$$
But when $|z|<\frac{1}{4}$, $$\sum_{n=0}^{\infty} \binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}.$$
So, with $z=b^2/4$, $$\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=2\pi \frac1{\sqrt{1-b^2}}$$
Setting $b=\frac{3}{5}$ we get:
we get:
$$\int_{0}^{2\pi}\frac{dx}{1-\frac{3}{5}\cos x}=\frac{5\pi}{2}$$
And thus your integral is $\frac{\pi}{2}$.
You get the general formula, when $|b|<a$ that:
$$\int_{-\pi}^{\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
