Arc length of y=2ln(sin(x/2)) from π/3 to π

[iRaid]

Homework Statement

Find the length of the curve $y=2ln(sin\frac{1}{2}x)$, $\frac{\pi}{3}\leq x\leq\pi$

Homework Equations

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.

 

[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.

 

[iRaid]

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.

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

 

[iRaid]

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

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

 

[Infrared]

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