$\displaystyle\int_{0}^{2\pi}\frac{1}{a+ \cos\theta}\,d\theta$

Question

So I am looking at the integral $$ \int_0^{2\pi} \frac{1}{a + \cos\theta} \mathrm{d} \theta$$ for $a > 1$. Evaluating the integral with complex analysis gives us $\frac{2 \pi}{\sqrt{a^2 - 1}} $ but I have failed to reproduce the result using standard substitution.

Here’s what I did: I found the antiderivative first, which is $$\frac{2}{\sqrt{a^2 - 1}} \arctan \left(\sqrt{\frac{a - 1}{a + 1}} \tan\left(\frac{\theta}{2}\right) \right)$$ but substituting in the bounds would give us 0.

I used the standard substitution $t = \tan\left(\frac{\theta}{2}\right) $ to obtain the following result.

EDIT: Plotting the graph shows us that the area is definitely positive, so there’s no way the answer is $0$.

EDIT2: Is it because $2\pi$ corresponds to the second “cycle” so we should $\arctan 0$ to be $\pi$ for our upper bound instead?

Answer 1

The theory about substitution actually states that when you want to use the substitution $t=g(\theta)$:

Suppose $f(\theta)=f^*(t)$ and $g’(\theta)=g^*(t)$, then $$\int^a_b f(\theta)d\theta=\int^{g(b)}_{g(a)} f^*(t)g^*(t)dt$$ if $g$ is differentiable and injective on $[a,b]$.

This statement is not concise but it is rigorous.

You should be able to see your mistake in your treatment.

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But this doesn’t mean that it is impossible to use the substitution $t=\tan(\theta/2)$.

Simply break your integral into two parts: $$\int^{\pi^-}_0 \frac{d\theta}{a+\cos\theta} +\int^{2\pi}_{\pi^+} \frac{d\theta}{a+\cos\theta} $$ and then, you are allowed to do the substitution.

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Since you have found the antiderivative, things get easier.

Let $F(\theta)$ be the antiderivative you found.

By fundamental theorem of calculus, your integral equals $$\begin{align} &~~~~F(\pi^-)-F(0)+F(2\pi)-F(\pi^+) \\ &=\frac{2}{\sqrt{a^2-1}}\frac{\pi}2-0+0-\frac{2}{\sqrt{a^2-1}}\frac{-\pi}2 \\ &=\color{red}{\frac{2\pi}{\sqrt{a^2-1}}} \end{align} $$

as desired, by noting $\operatorname{arctan}(\infty)=\frac\pi2$ and $\operatorname{arctan}(-\infty)=-\frac\pi2$.

Answer 2

If you plot the function you will see a maximum $\frac{1}{a-1}$ at $\pi$, so we can consider the bound as $[\pi, 0]$ and double the resulted value:

$$2\times[\frac {2}{\sqrt{ a^2-1}}\tan^{-1}(\sqrt{\frac{a-1}{a+1}}\tan \frac{\theta}{2})]^{\pi}_0=\frac{2\pi}{\sqrt {a^2-1}}$$