Answer
$\frac{4\sqrt{3}}{3}$
Work Step by Step
$\int_{\pi/3}^{2\pi/3}csc^2~(\frac{1}{2}t)~dt$
Let $u = \frac{cos~(\frac{1}{2}t)}{sin~(\frac{1}{2}t)}$
$\frac{du}{dt} = \frac{-\frac{1}{2}sin^2~(\frac{1}{2}t)-\frac{1}{2}cos^2~(\frac{1}{2}t)}{sin^2~(\frac{1}{2}t)} = \frac{-\frac{1}{2}~[sin^2~(\frac{1}{2}t)+cos^2~(\frac{1}{2}t)~]}{sin^2~(\frac{1}{2}t)} = -\frac{1}{2}csc^2~(\frac{1}{2}t)$
$dt = -\frac{2~du}{csc^2(\frac{1}{2}t)}$
When $t = \frac{\pi}{3}$, then $u = \sqrt{3}$
When $t = \frac{2\pi}{3}$, then $u = \frac{\sqrt{3}}{3}$
$\int_{\sqrt{3}}^{\sqrt{3}/3} [csc^2~(\frac{1}{2}t)]\cdot [-\frac{2~du}{csc^2(\frac{1}{2}t)}]$
$= \int_{\sqrt{3}}^{\sqrt{3}/3} -2~du$
$= \int_{\sqrt{3}/3}^{\sqrt{3}} 2~du$
$= 2u~\vert_{\sqrt{3}/3}^{\sqrt{3}}$
$= 2\sqrt{3}-2{\frac{\sqrt{3}}{3}}$
$= \frac{6\sqrt{3}}{3}-{\frac{2\sqrt{3}}{3}}$
$= \frac{4\sqrt{3}}{3}$
