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 $z=0$?

My approach:
NOTE: I will use that $\int_0^{2\pi} \sin^2(\theta) = \pi$, which is a different exercise.


*

*$I = \int_0^{2\pi} \sin^4(\theta)\,d\theta = \int_0^{2\pi} \cos^4(\theta)\,d\theta$ (the integrands are shifts of one another...)

*$\left(\sin^2(\theta) + \cos^2(\theta)\right)^2=\sin^4(\theta) + \cos^4(\theta) + 2\sin^2(\theta)\cos^2(\theta)$

*$2\pi = \int_0^{2\pi} 1\,d\theta = \int_0^{2\pi} (\sin^2(\theta) + \cos^2(\theta))^2\,d\theta \stackrel{\text{by (2)}}{=} 2I + \int_0^{2\pi}2\sin^2(\theta)\cos^2(\theta)\,d\theta$

*$\int_0^{2\pi} 2\sin^2(\theta) \cos^2(\theta)\,d\theta = \frac 1 2 \int_0^{2\pi}\sin^2(2\theta)\,d\theta = \frac 1 4 \int_0^{4\pi}\sin^2(\theta)\,d\theta = \frac \pi 2$


Therefore $2\pi = 2I + \frac \pi 2$, i.e. $I = \frac {3\pi} 4$.
