# 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?

• It might be fun to do this by finding a geometric interpretation. It is reminiscent of problems where I've seen that done but I'm not adept at that just now. – Michael Hardy Jul 27 '17 at 2:26

## 4 Answers

hint

with $t=\theta-\pi$, it becomes

$$I=\int_{-\pi}^\pi \frac{dt}{5+3\cos (t)}$$ $$=2\int_0^\pi\frac {dt}{5+3\cos (t)}$$

now put $$u=\tan \left(\frac {t}{2}\right) .$$

using

$$\cos (t)=\frac {1-u^2}{1+u^2}$$ $$dt=\frac {2\,du}{1+u^2}$$

You will find $$I=2\int_0^{+\infty}\frac {du}{4+u^2} =\int_0^{+\infty}\frac {dv}{1+v^2}$$ $$=\Bigl [\arctan (v)\Bigr]_0^{+\infty}=\frac {\pi}{2}$$

where $u=2v$.

• This substitution is probably due to Euler, but Stewart's calculus textbooks, and probably others as well, wrongly attribute it to Weierstrass, who was born long after Euler died. – Michael Hardy Jul 27 '17 at 0:19
• I would call this a sketch rather than a hint. Calling a sketch of a solution a "hint" seems to by done somewhat frequently. – Michael Hardy Jul 27 '17 at 2:27

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:

$$\int_{-\pi}^{\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$


Using your hint the integral is $2i\int_{S^1}\frac{dz}{3z^2-10z+3}=\frac{i}{4}\int_{S^1}\frac{dz}{z-3}+\frac{3i}{4}\int_{S^1}\frac{dz}{3z-1}$

Now you can use Cauchy's formula on each term.

[Although I don't know why you don't like the residue theorem. This and Cauchy;s formula are pretty much the same stuff.]

• What would it look like by using the Residue Theorem? – Hwi Moon Jul 27 '17 at 0:24