How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
 A: Perhaps the easiest, though it requires some knowledge of the complex exponential function, is to substitute $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ in the integral and expand.
A: Let $x=\cos A+i\sin A$
Then we have,
$\displaystyle x-\frac{1}{x}=2i\sin A$
$\displaystyle (x-\frac{1}{x})^{6}=-2^6\sin^6 A$
Then we have by expanding,
$x^6+\frac{1}{x^6}-6(x^4+\frac{1}{x^4})+15(x^2+\frac{1}{x^2})-20=2\cos6 A-6\cos{4}A+15\cos2 A-20$
So we have,
$\sin^6 A=\frac{-1}{2^6}(2\cos 6 A-12\cos{4}A+30\cos2A-20)$
I have used the fact that $x^k=\cos kA+i\sin kA\Rightarrow x^k+\frac{1}{x^k}=2\cos kA$.(for $k\in Z$)
From here on I think it will be easy.
A: Note that $\displaystyle \int_{0}^{\pi} \sin^n(x) dx = 2 \int_0^{\pi/2} \sin^n(x) dx$. Let $I_n = \displaystyle \int_{0}^{\pi/2} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\pi/2} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\pi/2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos\left(\pi/2\right) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$I_n = (n-1) \left(\displaystyle \int_{0}^{\pi/2} \sin^{n-2}(x) dx - \int_{0}^{\pi/2} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$.
Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \dfrac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} I_1$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\dfrac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} \dfrac{\pi}{2} = \dfrac{(2k)!}{4^k (k!)^2} \dfrac{\pi}2$$
Hence,
$$\int_0^{\pi} \sin^n(x) dx = \begin{cases} \dfrac{2^{2k+1} (k!)^2}{(2k+1)!} & \text{if $n$ is odd, i.e., $n=2k+1$}\\
\dfrac{(2k)!}{4^k (k!)^2} \pi & \text{if $n$ is even, i.e., $n=2k$}
\end{cases}$$
A: Two ways exist. 
One is to reduce with the formulae $\sin^2(x) = (1 - \cos(2x))/2$ and
$\cos^2(x) = (1 + \cos(2x))/2$.  You can then use the fact that $\sin^3(x) = (1 - \cos^2(x)\sin(x)$ to u-sub your problem away.
Another way is to integrate by parts, $u = \sin^5(x)$, $dv = \sin(x)$, and to use the Pythogorean identity of sin and cos to get a reduction formula.  Then apply the formula repeatedly.
A: As $\sin3x=3\sin x-4\sin^3x$
$$\sin^6\alpha=(\sin^3\alpha)^2=\left(\frac{3\sin\alpha-\sin3\alpha}4\right)^2$$
$$=\frac{9\sin^2\alpha+\sin^23\alpha-3(2\sin\alpha\sin2\alpha)}{16}$$
$$=\frac{9(1-\cos2\alpha)+1-\cos6\alpha-6(\cos\alpha-\cos3\alpha)}{32}$$
using  $2\sin A\sin B=\cos(A-B)-\cos(A+B)$ and $\cos3x=1-2\sin^2x$
Now, we find $\int_o^\pi\cos mx dx=\frac{\sin mx}m\big |_0^\pi=\frac{\sin  m\pi}m$ which becomes $0$ for any integer $m$
So, the definite integral for all the cosine ratios in the given integral will vanish leaving behind  $\frac{9+1}{32}=\frac5{16}$ to be integrated resulting in $$\frac5{16}\cdot \pi$$ 

Alternatively,
$$I=\int_0^\pi\sin^6\alpha d\alpha=\int_0^{\frac\pi2}\sin^6\alpha d\alpha+\int_{\frac\pi2}^\pi\sin^6\alpha d\alpha$$
Now,putting $\beta=\alpha-\frac\pi2$ $$\int_{\frac\pi2}^\pi\sin^6\alpha d\alpha=\int_0^{\frac\pi2}\cos^6\beta d\beta=\int_0^{\frac\pi2}\cos^6\alpha d\alpha$$
So, $$I=\int_0^\frac\pi2(\sin^6\alpha+ \cos^6\alpha )d\alpha$$
Now,
 $$\sin^6\alpha+ \cos^6\alpha=(\sin^2\alpha+\cos^2\alpha)^3-3(\sin^2\alpha\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)$$
$$=1-\frac34(\sin^22\alpha)=1-\frac{3(1-\cos4\alpha)}8=\frac{5+3\cos4\alpha}8$$
A: $$\int_0^a f(x)\;dx=2\int_0^\dfrac a2 f(x)\;dx\; if \;f(a-x)=f(x)\;$$
and$\int_0^a f(x)\;dx=0\;\;if \;f(a-x)=-f(x)$
here $f(x)=\sin ^6x\implies f(\pi-x)=\sin ^6(\pi-x)\implies \sin^6x\implies f(x)$
so we can write it as first identity:
$$I=\int_0^\dfrac \pi2 \sin^6 x\;dx$$
from identity:$\int_0^af(x)\;dx=\int_0^af{(a-x)}\;dx$
$$I=\int_0^\dfrac \pi2 \cos^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 \sin^6 x+\cos ^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^2 x})^3+({\cos ^2 x})^3\;dx$$
$$2I=\int_0^\dfrac \pi2 (\sin^2 x+\cos ^2 x)({\sin^4 x}+{\cos ^4 x}-\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^4 x}+{\cos ^4 x}+2\sin^2 x\cos^2 x-3\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({{\sin^2 x}+{\cos ^2 x}}^2)-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34{(2\sin x\cos x)}^2\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\sin^2 2x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\dfrac{(1-\cos 4x)}2\;dx$$
$$2I=\left[x-\dfrac 38(x-\dfrac{\sin 4x}4)\right]_0^\dfrac \pi2 \implies \left[\dfrac {5x}{8}+\dfrac{3\sin 4x}{32}\right]_0^\dfrac \pi2 \implies \left[\dfrac {5\pi}{16}\right]\implies  I=\left[\dfrac {5\pi}{32}\right]$$
proof of identity:$$\int_0^{2a} f(x)\;dx=2\int_0^a f(x)\;dx\; if \;f(a-x)=f(x)\;$$
this identity can be prove using this identity:$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(2a-x)\;dx$$
when $f(2a-x)=f(x)$ this become
$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(x)\;dx$$
$$\int _0^{2a} f(x)=2\int_0^a f(x)\;dx$$
