1. ## Path Integral

I have to compute the arc length of $r = 1 + cos(\theta)$
I found the formula $\int_0^{2\pi} \sqrt{r^2+(\frac{dr}{d\theta})^2} d\theta = \int_0^{2\pi} \sqrt{2+2cos(\theta)} d\theta$*
I don't know how to do this integral.

2. Dear vincisonfire,

one of tricks: cos(theta) = cos(2*(theta/2)) and
cos(2x) = cos^2(x)-sin^2(x).

3. I would have $*\int \sqrt{1+cos(\theta)} = \int \sqrt{1+cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})} = \int \sqrt{2cos^2(\frac{\theta}{2})}$
Back to my problem *
$2 \int_0^{2\pi} cos(\frac{\theta}{2}) d\theta= 4 [sin(\frac{\theta}{2})]_0^{2\pi} = 0$
I should have 8 no ?

4. khm...khm...
$\sqrt{x^2} \neq x$
such in x = -3

5. Integrate from 0 to Pi and multiply by 2.

$8sin(\frac{\theta}{2})|_{0}^{\pi}=8$

6. Hello, vincisonfire!

I have to compute the arc length of $r \:=\: 1 + cos(\theta)$

I found the formula: . $\int_0^{2\pi} \sqrt{r^2+(\tfrac{dr}{d\theta})^2} d\theta \;\;= \int_0^{2\pi} \sqrt{2+2\cos\theta}\, d\theta$

We're expected to know this identity: . $\cos^2\tfrac{\theta}{2} \:=\:\frac{1+\cos\theta}{2} \quad\Rightarrow\quad 1 + \cos\theta \:=\:2\cos^2\!\tfrac{\theta}{2}$

We have: . $\sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right )} \;=\;2\cos\tfrac{\theta}{2}$

Therefore: . $L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th eta\quad\hdots$ . Got it?

7. It is not true!

See above!

8. Originally Posted by Soroban

We have: . $\sqrt{2(1 + \cos\theta)} \;=\;\sqrt{2\left(2\cos^2\!\tfrac{\theta}{2}\right )} \;=\;2\cos\tfrac{\theta}{2}$

Therefore: . $L \;=\;2\int^{2\pi}_o\!\!\cos\tfrac{\theta}{2}\,d\th eta\quad\hdots$ . Got it?
Here's where the mistake is, that should be $\int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta },$ hence,

\begin{aligned}
\int_{0}^{2\pi }{\left| \cos \frac{\theta }{2} \right|\,d\theta }&=2\int_{0}^{\pi }{\left| \cos \theta \right|\,d\theta } \\
& =2\left( \int_{0}^{\pi /2}{\cos (\theta )\,d\theta }-\int_{\pi /2}^{\pi }{\cos (\theta )\,d\theta } \right) \\
& =2(1+1)=4.
\end{aligned}