Evaluating $\int_0^{2 \pi} \sin^4 \theta\: \mathrm{d} \theta$ Evaluate the following integral: 
$$\int_0^{2 \pi} \sin^4 \theta \:\mathrm{d} \theta$$
My approach: Parametrize and obtain $$\frac{1}{(2i)^4} \int_{|z|=1} \left (z-\frac{1}{z} \right)^4 \frac{1}{iz}\:\mathrm{d}z=\frac{1}{(2i)^4} \int_{|z|=1} \left (\frac{(z+1)(z-1)}{z} \right)^4 \frac{1}{iz}\:\mathrm{d}z$$
Can I directly use the residue theorem from here with a residue at $z=0$?

Your approach seems right. $\frac{(z-1/z)^4}{z} = z^3 - 4z + 6/z - 4/z^3 + 1/z^5$. So, by contour integration we look at the pole $z=0$ and we have that the integral is $2\cdot \pi \cdot i \cdot 6 = 12\pi \cdot i$. Then we divide out by $(2i)^4\cdot i = 16\cdot i$ and we get $3\pi /4$ as desired.
Another approach is to define $\sin(z) = \frac{e^{it} + e^{-it}}{2i}$ and expand thereby integating $(e^{4it}+e^{-4it} - 4(e^{2it}+e^{-2it}) + 6) = 2\cdot (\cos(4t) - 4\cos(2t) +3)$ from $0$ to $2\pi$. Now, the $\sin$ terms vanish and so we have an integral of $6\cdot (2\pi - 0)/16 = 3\pi/4$ as desired.
