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 $z=0$?

$$
I=\int \sin^4xdx=\int\sin^3x\sin xdx=\int \sin^3x(-\cos x)'dx=-\sin^3x\cos x+\int3\sin^2x\cos^2dx=-\sin^3x\cos x+3\int(\frac{1}{2}\sin 2x)^2dx
$$
$$
\cos 4x=1-2\sin^22x\Rightarrow\sin^22x=\frac{1-\cos4x}{2}\\
$$
$$
I=-\sin^3x\cos x+\frac{3}{4}\int\frac{1-\cos4x}{2}dx=-\sin^3x\cos x+\frac{3}{8}\int (1-\cos4x)dx=\\=-\sin^3x\cos x+\frac{3}{8}x-\frac{3}{32}\sin4x+c\\
\Rightarrow\int_0^{2\pi} \sin^4xdx=\frac{3\pi}{4}
$$
