Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$ 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?
 A: $$
\begin{align}
\int_0^{2\pi}\sin^8xdx
&=\int_0^{2\pi}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^8dx\\
&=\int_0^{2\pi}2^{-8}(e^{ix}-e^{-ix})^8dx\\
&=2^{-8}\int_0^{2\pi}\sum_{k=0}^8\binom{8}{k}(e^{ix})^k(-e^{-ix})^{8-k}dx\\\
&=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{8-k}\int_0^{2\pi}e^{i(2k-8)x}dx\\
&=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{k}2\pi\delta_{2k-8,0}\\
&=2^{-8}\binom{8}{4}(-1)^{4}2\pi\\
&=\frac{\binom{8}{4}}{2^7}\pi
\end{align}
$$
A: 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
A: 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 $\Gamma\left(\dfrac12\right)=\sqrt\pi$.
A: $$\int_0^{2\pi}\sin^{2n}x\ dx=\int_0^{\pi}\sin^{2n}x\ dx+\int_\pi^{2\pi}\sin^{2n}x\ dx$$
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
