Integrate $\int sin(x)^6 $ using integration by parts? I know how to solve it using the double angle formula for $cos(2x)$. I also know how to integrate $sin(x)^2$ by parts like this:
$$\begin{align}\int \sin(x)\sin(x)dx &= −\cos(x)\sin(x)-\int(-\cos(x))\cos(x)dx
\\&= -\cos(x)sin(x)+\int(1-\sin(x)^2)dx
\\&= -\cos(x)\sin(x)+x-\int \sin(x)^2dx 
\\2\int \sin(x)^2dx &= -\cos(x)sin(x)+x
\\\int \sin(x)^2dx &= -\frac 12 \cos(x)\sin(x)+\frac 12 x \end{align}$$
Is it possible to integrate $sin(x)^6$ by parts?
 A: Of course, you can. To integrate $\sin^{6}(x)$, use method how to integrate $\sin^{2}(x)$ again and again.
$\int \sin^{6}(x) dx$
$=\int \sin(x) \sin^{5}(x)dx$
$=-\cos(x) \sin^{5}(x)-\int (-\cos(x)) (5 \sin^{4}(x) \cos(x)) dx$
$=-\cos(x) \sin^{5}(x)+5 \int \sin^{4}(x) \cos^{2}(x) dx$
$=-\cos(x)\sin^{5}(x)+5 \int \sin^{4}(x)dx-5 \int \sin^{6}(x) dx$
From this integration, you get $\int\sin^{6}(x)dx=-\dfrac{1}{6}\cos(x)\sin^{5}(x)+\dfrac{5}{6} \int \sin^{4}(x)dx$.
Repeating this for 3 times, you finally get desired integral result.
Or simply using the fact that $\sin^{4}(x)=\left(\dfrac{1-\cos(2 x)}{2} \right)^{2}=\dfrac{\cos^{2}(2 x)-2 \cos(2 x)+1}{4}$
$=\dfrac{\dfrac{\cos(4 x)+1}{2}-2 \cos(2 x)+1}{4}=\dfrac{\cos(4 x)-4 \cos(2 x)+3}{8}$ gives result quicker than repeating iteration of integrating by parts
A: $$\sin^6x=(\sin^2x)^3$$
and we know that:
$$\cos(2x)=1-2\sin^2x\Rightarrow \sin^2x=\frac{1-\cos(2x)}{2}$$
so we have:
$$\sin^6x=\left(\frac{1-\cos(2x)}{2}\right)^3=\frac{1}{8}\left(1-3\cos2x+3\cos^22x-\cos^32x\right)$$
now we can shuffle this around a bit:
$$3\cos^22x=\frac{3(1+\cos4x)}{2}$$
$$\cos^32x=\cos2x(1-\sin^22x)=\cos2x-\sin^22x\cos2x$$
so if we bring this all together we get:
$$\int\sin^6xdx=\frac18\int1-4\cos(2x)+\frac{3(1+\cos(4x))}{2}+\sin^2(2x)\cos(2x)dx$$
and all of the terms here are easy to integrate
