# Homework Help: Arclength (Difficult?)

1. Apr 3, 2013

### iRaid

1. The problem statement, all variables and given/known data
Find the length of the curve $y=2ln(sin\frac{1}{2}x)$, $\frac{\pi}{3}\leq x\leq\pi$

2. Relevant equations

3. The attempt at a solution
Alright so I figured out the derivative of y is cot(1/2)x so I put it into the arclength formula to get:
$$\int_{\frac{\pi}{3}}^{\pi} \sqrt{1+{cot^{2}(\frac{x}{2})}}dx$$

But I don't know how to solve that.. Looking at wolfram it seems like a really ugly integral so I don't want to put a lot of effort into solving it if I have the integral wrong.

Thanks for any help.

2. Apr 3, 2013

### Infrared

Never forget your Pythagorean identities. Divide $sin^2 \theta +cos^2 \theta = 1$ through by $sin^2 \theta$ to get $1+cot^2\theta = csc^2\theta$. Just let $\theta = \frac{x}{2}$ and the integral simplifies greatly.

3. Apr 3, 2013

### iRaid

Yeah I just kept going and realized that. So I did:
u=x/2 du=1/2 dx
$$2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{1+cot^{2}u}du=2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sqrt{csc^{2}u}du=2ln(cscu+cotu)|_{\frac{\pi}{6}}^{\frac{\pi}{2}}$$

4. Apr 3, 2013

### iRaid

Yeah I finished it and evaluated, $-2ln(2+\sqrt{3})$

Forgot that I knew what the integral of csc was :P. Thanks.

5. Apr 3, 2013

### Infrared

Don't forget to change your bounds after making a substitution.