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 $z=0$?
 A: Hint (if one insists on using the residue theorem): $$\left(z-\frac1z\right)^4\frac1z=\left(z^4-4z^2+6-4\frac1{z^2}+\frac1{z^4}\right)\frac1z,$$ hence the $\dfrac1z$ term is $$\dfrac6z,$$
and
$$
\oint_{|z|=1}\left(z-\frac1z\right)^4\frac{\mathrm dz}{z}=2\pi\mathrm i\cdot6.
$$
A: Hint:
$$\sin^4x=\sin^2x\cdot \sin^2x
\\=\frac{1}{4}(1-\cos2x)(1-\cos2x)$$
and $$\cos^22x=\frac{1}{2}(1+\cos4x)$$
A: 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. $I = \frac {3\pi} 4$.
A: $$
I=\int \sin^4xdx=\int\sin^3x\sin xdx=\int \sin^3x(-\cos x)'dx=-\sin^3x\cos x+\int3\sin^2x\cos^2dx=-\sin^3x\cos x+3\int(\frac{1}{2}\sin 2x)^2dx
$$
$$
\cos 4x=1-2\sin^22x\Rightarrow\sin^22x=\frac{1-\cos4x}{2}\\
$$
$$
I=-\sin^3x\cos x+\frac{3}{4}\int\frac{1-\cos4x}{2}dx=-\sin^3x\cos x+\frac{3}{8}\int (1-\cos4x)dx=\\=-\sin^3x\cos x+\frac{3}{8}x-\frac{3}{32}\sin4x+c\\
\Rightarrow\int_0^{2\pi} \sin^4xdx=\frac{3\pi}{4}
$$
A: 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.
