Integral $\int\limits_{\sqrt{2}}^{2}\frac1{t^3\sqrt{t^2-1}}$ This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$
Because of the $\sqrt{t^2-1} = \sqrt{t^2-1^2}$ in the integrand, I set $t = 1\sec(\theta) = \sec(\theta)$, and $\theta = \sec^{-1}(t)$. Thus $\text{d}t = \sec(\theta)\tan(\theta)\text{ d}\theta$ and
$$\begin{align}\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}} \text{ d}t&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{3}(\theta)\sqrt{\sec^{2}(\theta)-1}}\sec(\theta)\tan(\theta)\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{2}(\theta)}\text{ d}\theta \\
&= \int\limits_{\pi/4}^{\pi/3}\cos^{2}(\theta)\text{ d}\theta \\
&= \dfrac{1}{2}\int\limits_{\pi/4}^{\pi/3}[1+\cos(2\theta)]\text{ d}\theta \\
&= \dfrac{1}{2}\left\{\dfrac{\pi}{3}-\dfrac{\pi}{4}+\dfrac{1}{2}\left[\sin\left(\dfrac{2\pi}{3}\right) - \sin\left(\dfrac{\pi}{2}\right)\right]\right\}\\
&= \dfrac{1}{2}\left[\dfrac{\pi}{12}+\dfrac{1}{2}\left(\dfrac{\sqrt{3}}{2}-1\right)\right] \\
&= \dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}-2}{2}\right)\text{.}
\end{align}$$
A quick side question: without access to a calculator and assuming I have the values of $\sin(x)$ memorized for $x = 0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$, what is the easiest way to find $\sin\left(\dfrac{2\pi}{3}\right)$?

Your working looks alright.$$\sin \theta= \sin (\pi - \theta)$$
Is another identity. 
