1
$\begingroup$

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?

$\endgroup$
1
  • 3
    $\begingroup$ Take a look at this. $\endgroup$ – Pspl Mar 6 '20 at 10:39
6
$\begingroup$

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 $\tan t =u$.

$\endgroup$
5
  • 2
    $\begingroup$ 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
1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.