Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$

Question

I have started doing integration by parts: $$\int_{0}^{2\pi}{\sin^8(x)}{dx} = \int_{0}^{2\pi}{\sin^7(x)}\cdot{\sin(x)dx} = \int_{0}^{2\pi}{\sin^7(x)}\cdot{d(-\cos(x))} = \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi} + \int_{0}^{2\pi}{\cos(x)}{d(\sin^7(x))} = \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi} + \int_{0}^{2\pi}{\cos^2(x) \cdot 7 \cdot \sin^6(x)}{dx} = \dots$$

Is there a better way to do it?

possible duplicate of Find the average of $\sin^{100} (x)$ in 5 minutes? — Norbert, Jun 07 '14 at 06:01

score 8 · Accepted Answer · answered Jun 07 '14 at 05:58

8

$$ \begin{align} \int_0^{2\pi}\sin^8xdx &=\int_0^{2\pi}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^8dx\\ &=\int_0^{2\pi}2^{-8}(e^{ix}-e^{-ix})^8dx\\ &=2^{-8}\int_0^{2\pi}\sum_{k=0}^8\binom{8}{k}(e^{ix})^k(-e^{-ix})^{8-k}dx\\\ &=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{8-k}\int_0^{2\pi}e^{i(2k-8)x}dx\\ &=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{k}2\pi\delta_{2k-8,0}\\ &=2^{-8}\binom{8}{4}(-1)^{4}2\pi\\ &=\frac{\binom{8}{4}}{2^7}\pi \end{align} $$

answered Jun 07 '14 at 05:58

Norbert

56,803

http://math.stackexchange.com/questions/24533/find-the-average-of-sin100-x-in-5-minutes ? – chubakueno Jun 07 '14 at 05:59
@chubakueno ok, we can close that question as duplicate – Norbert Jun 07 '14 at 06:01
I don't understand why $\int_{0}^{2\pi}{e^{i(2k-8)x}}{dx} = 2\pi\delta_{2k-8,0}$ – Sergey Filkin Jun 07 '14 at 06:22
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same where did you get the delta from? – usukidoll Jun 07 '14 at 06:23
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This last comment of @Norbert is missing an integral sign and a dx. The left side is supposed to be $\int_0^{2\pi} e^{inx},dx$. If $n$ is $0$, you just have a constant $1$ function. Otherwise you can antidifferentiate to $\frac1{in}e^{inx}$. – 2'5 9'2 Jun 07 '14 at 06:38
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@SergeyFilkin $$\int_0^{2\pi}e^{inx}dx=\begin{cases}2\pi\quad &n=0\0\quad &n\neq0\end{cases}$$ – Norbert Jun 07 '14 at 06:39
@alex.jordan thank you! – Norbert Jun 07 '14 at 06:40

score 2 · Answer 2 · answered Jun 07 '14 at 05:57

You can use the combination of $$ \sin^2 \theta = \frac{1-\cos 2\theta}{2} , \cos^2 \theta = \frac{1 + \cos 2\theta}{2} $$ or an one of an assortment of other reduction formulas found here.

I personally use the basic power reduction formulas (what I listed above) in order to practice my trig/algebra ;D

score 2 · Answer 3 · answered Jun 07 '14 at 08:28

Consider Beta function $$ \text{B}(x,y)=2\int_0^{\Large\frac\pi2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\ d\theta=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Rewrite $$ \int_0^{\large2\pi}\sin^8x\ dx=4\int_0^{\Large\frac\pi2}\sin^8 x\ dx, $$ then $$ \int_0^{\large2\pi}\sin^8x\ dx=2\cdot\frac{\Gamma\left(\dfrac92\right)\cdot\Gamma\left(\dfrac12\right)}{\Gamma(5)}=2\cdot\frac{\dfrac72\cdot\dfrac52\cdot\dfrac32\cdot\dfrac12\cdot\Gamma^2\left(\dfrac12\right)}{4!}=\large\color{blue}{\frac{35}{64}\pi}, $$ where $\Gamma(n+1)=n\cdot\Gamma(n)$ and $\Gamma\left(\dfrac12\right)=\sqrt\pi$.

score 1 · Answer 4 · answered Jun 07 '14 at 06:18

$$\int_0^{2\pi}\sin^{2n}x\ dx=\int_0^{\pi}\sin^{2n}x\ dx+\int_\pi^{2\pi}\sin^{2n}x\ dx$$

Set $y=x-\pi$ in the second integral

$$\int_0^{\pi}\sin^{2n}x\ dx=\int_0^{\frac\pi2}\sin^{2n}x\ dx+\int_{\frac\pi2}^\pi\sin^{2n}x\ dx$$

Set $z=\pi-x$ in the second integral

Finally use Reduction formula

Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$

4 Answers4