I need to proof that \begin{align} \int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}} \end{align} is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the residual theorem:

I started with $\gamma: [0,\pi] \to \mathbb{C}, t \to e^{2it}$. Since $\cos(t) = \frac{1}{2}\left(e^{it}+e^{-it}\right)$ and $\gamma'(t) = 2ie^{2it}$ we find that

\begin{align} \int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(e^{it}+e^{-it}\right)} \cdot \frac{2ie^{2it}}{2ie^{2it}}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t \end{align}

I did this with the aim to use

\begin{align} \int_{\gamma} f(z) \mathrm{d}z = \int_{a}^{b} (f\circ \gamma)(t)\gamma'(t) \mathrm{d}t, \end{align}

so we find

\begin{align} \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t = \int_{\gamma} \frac{-i}{6z+2z\sqrt{z}+2\sqrt{z}} \mathrm{d}z \end{align}

At this point I don't know how to continue. Can anyone help?

  • 3
    $\begingroup$ Take a look at this. $\endgroup$ – Pspl Mar 6 '20 at 10:39

Use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx.~~~(1)$$ $$I=\int_{0}^{\pi} \frac{dt}{3+2 \cos t}~~~~(2)$$ Using (1), we get $$I=\int_{0}^{\pi} \frac{dt}{3-2 \cos t}~~~~(3)$$ Add (2) and (3), then $$2I=\int_{0}^{\pi} \frac{6\,dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{\sec^2 t}{9\tan^2 t+5}=\frac{4}{3} \int_{0}^{\infty}\frac{du}{u^2+(\sqrt{5}/3)^2}$$ $$2I=\frac{4}{3} \frac{3}{\sqrt{5}}\tan^{-1}(3u/\sqrt{5})|_{0}^{\infty}=\frac{2\pi}{\sqrt{5}}\implies I=\frac{\pi}{\sqrt5} $$ In above we have used $\tan t =u$.

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    $\begingroup$ You have lost a factor of $2$ somewhere: if $2I=\frac{\pi}{\sqrt 5}$, then $I=\frac{\pi}{2\sqrt 5}$. $\endgroup$ – TonyK Mar 6 '20 at 11:37
  • $\begingroup$ Thanks for the alert, there was a typo. $\endgroup$ – Z Ahmed Mar 6 '20 at 11:40
  • $\begingroup$ And shouldn't $\tan^1(\sqrt 5/2)$ be $\tan^{-1}(u\sqrt 5/3)$? (That's three more typos!) $\endgroup$ – TonyK Mar 6 '20 at 11:49
  • $\begingroup$ Oh! yes, I have edited it. thanks. $\endgroup$ – Z Ahmed Mar 6 '20 at 11:52
  • $\begingroup$ Still one to go... $\endgroup$ – TonyK Mar 6 '20 at 13:27

We proved here that

$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$

set $a=3$ and $b=2$ then integrate both sides we have

$$\int_0^\pi\frac{dt}{3+2\cos t}=\int_0^\pi\frac{dt}{\sqrt{5}}+\frac{2}{\sqrt{5}}\sum_{n=1}^\infty\left(\frac{\sqrt{5}-3}{2}\right)^n\underbrace{\int_0^\pi\cos(nt)\ dt}_{0}=\frac{\pi}{\sqrt{5}}$$


Using function transformations, compress the integral by a factor of $2$ in the $x$-axis, then multiply by $2$ to get:

$$2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{3+2\cos(2t)} = 2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{4 \cos^2 t+1} = 2 \int_{0}^{\pi/2} \frac{\mathrm{\sec^2 t\ d}t}{4 + \tan^2 t+1}$$

and substituting $u = \tan t$, $\mathrm{d} u = \sec^2 t \ \mathrm{d}t$:

$$2 \int_{0}^{\infty} \frac{\mathrm{d} u}{(\sqrt5)^2 + u^2} = \lim_{a \to \infty}2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{u}{\sqrt5} \right]_0^{a} = 2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{\tan t}{\sqrt5} \right]_0^{\pi/2} $$

$$=\frac{2}{\sqrt5} \left( \tan^{-1} ( \tan \pi/2) - \tan^{-1} (\tan 0)\right)= \frac{2}{\sqrt5} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt5} $$

since $\tan^{-1} (\tan x)= x, x \in [-\pi/2, \pi/2]$.


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