# 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 $z=0$?

• yes ${}{}{}{}{}$ Apr 1, 2014 at 20:37
• It's really messy :-( I'm not sure if I'm doing it right Apr 1, 2014 at 20:39
• I'm getting $0$ for the residue. Apr 1, 2014 at 20:59
• @N3buchadnezzar do you get the same? Apr 1, 2014 at 20:59
• $\int_0^{2 \pi} \sin^{2n} x dx = 2\pi {2n \choose n} \frac{1}{2^{2n}}$ Apr 1, 2014 at 21:16

## 5 Answers

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 $$\oint_{|z|=1}\left(z-\frac1z\right)^4\frac{\mathrm dz}{z}=2\pi\mathrm i\cdot6.$$

• We should be getting $\frac{3 \pi}{4}$ Apr 1, 2014 at 21:15
• We do get $3π/4$. Can't you recover the integral you are asking the value of, from the contour integral in my post?
– Did
Apr 1, 2014 at 21:16
• @user69127, you still have to divide out by $i\cdot (2i)^4 = 16\cdot i$. Then you get the desired result. Apr 1, 2014 at 21:18
• ah yes my mistake! Apr 1, 2014 at 21:19
• @Thesinus the only singularity is at $z=0$. In $z=1$, $z+ 1/z$ is perfectly defined and equals $2$. Aug 24, 2019 at 14:39

# Hint:

$$\sin^4x=\sin^2x\cdot \sin^2x \\=\frac{1}{4}(1-\cos2x)(1-\cos2x)$$ and $$\cos^22x=\frac{1}{2}(1+\cos4x)$$

• I see. How would I use residue theorem on this? Anyways I tried using a residue formula and ended up getting $0$ for the integral. Is this correct? Apr 1, 2014 at 21:05
• No, it should be (3/4)π Apr 1, 2014 at 21:07

My approach:

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

1. $I = \int_0^{2\pi} \sin^4(\theta)\,d\theta = \int_0^{2\pi} \cos^4(\theta)\,d\theta$ (the integrands are shifts of one another...)
2. $\left(\sin^2(\theta) + \cos^2(\theta)\right)^2=\sin^4(\theta) + \cos^4(\theta) + 2\sin^2(\theta)\cos^2(\theta)$
3. $2\pi = \int_0^{2\pi} 1\,d\theta = \int_0^{2\pi} (\sin^2(\theta) + \cos^2(\theta))^2\,d\theta \stackrel{\text{by (2)}}{=} 2I + \int_0^{2\pi}2\sin^2(\theta)\cos^2(\theta)\,d\theta$
4. $\int_0^{2\pi} 2\sin^2(\theta) \cos^2(\theta)\,d\theta = \frac 1 2 \int_0^{2\pi}\sin^2(2\theta)\,d\theta = \frac 1 4 \int_0^{4\pi}\sin^2(\theta)\,d\theta = \frac \pi 2$

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

$$I=\int \sin^4xdx=\int\sin^3x\sin xdx=\int \sin^3x(-\cos x)'dx=-\sin^3x\cos x+\int3\sin^2x\cos^2dx=-\sin^3x\cos x+3\int(\frac{1}{2}\sin 2x)^2dx$$

$$\cos 4x=1-2\sin^22x\Rightarrow\sin^22x=\frac{1-\cos4x}{2}\\$$

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

• I have to use facts from complex analysis though. Apr 1, 2014 at 21:14

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.