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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?

snivysteel
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    MathJax hint: using \left and \right before parentheses and brackets automatically makes them the same size as whatever's inside them. – Robert Howard Oct 20 '18 at 04:48
  • @Nosrati it’s a different question? – snivysteel Oct 20 '18 at 05:01
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    When $\theta$ moves from $0 \to \pi \to 2\pi$, $\tan\frac{\theta}{2}$ first move from $0 \to +\infty$, it then jump to $-\infty$ at $\theta = \pi$ and finally move from $-\infty \to 0$... – achille hui Oct 20 '18 at 05:02
  • @achillehui does that matter because it’s not like path dependent? – snivysteel Oct 20 '18 at 05:18
  • The integral is equal to $2\int_{0}^{\pi}(a+\cos \theta) ^{-1},d\theta $ and then you can apply your anti-derivative now. Another approach is to use the slightly difficult difficult substitution $(a+\cos\theta) (a-\cos\phi) =a^2-1$ – Paramanand Singh Oct 20 '18 at 07:22

2 Answers2

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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.

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.

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 $\operatorname{arctan}(-\infty)=-\frac\pi2$.

Lee
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  • We can also just get twice the left integral by imposing $\theta\mapsto 2\pi - \theta$ in the second one. – J.G. Oct 20 '18 at 06:49
  • @J.G. That’s true. – Lee Oct 20 '18 at 06:50
  • @Szeto Thanks Sir!! – snivysteel Oct 20 '18 at 07:14
  • The "and injective" bit is something that seems to rather terribly get missed in many calculus texts, e.g. Stewart's Calculus doesn't mention this, nor do any of the other texts that I happen to have on my hands right at this particular moment in time. This is WRONG, and it will BITE you eventually if you didn't spot it first - and here it did indeed bite somebody! (And it makes sense if one thinks about it: clearly, if it is not injective then $[g(a), g(b)]$ is not the image of $[a, b]$ under transformation by $g$ and indeed it "seems a little too good to be true" to think you could – The_Sympathizer Oct 20 '18 at 08:17
  • have an arbitrarily-sophisticated $g$ merely by transforming the endpoints...!) – The_Sympathizer Oct 20 '18 at 08:17
  • @The_Sympathizer Your words echo my thoughts. I often give examples like applying $u=x^2$ on $$\int^1_{-1}f(x)dx$$ to emphasize injectivity. I didn’t see injectivity being mentioned anywhere, and I see come up with the ‘theorem’ by myself. It is very strange and unreasonable that most calculus texts missed this point. – Lee Oct 20 '18 at 08:28
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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:

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

sirous
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