Evaluating indefinite integrals Evaluate the following indefinite integral.
$ \int { { \sin }^{ 6 } } x\quad dx $
My try :
$ \int { ({ \sin^2x } } )^{ 3 }dx\\ \int { (\frac { 1 }{ 2 }  } (1-\cos2x))^{ 3 }dx\\ \int { \frac { 1 }{ 8 }  } (1-\cos2x)^{ 3 }dx\\ \frac { 1 }{ 8 } \int { (1-\cos2x)^{ 3 } } dx\\ \frac { 1 }{ 8 } \int { 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x\quad dx $
Then i got stuck.

The general rule for $\sin^m\cos^n$:
If $m$ is odd: $m=2k+1$, $\sin^m\cos^n=\sin^{2k+1}\cos^n=\cos^n(1-\cos^2)^k\sin$.
If $n$ is odd: same trick using this time $\cos^2 = 1-\sin^2$
If $m$ and $n$ are both even: use the double angle formulas
$\sin^2 x={1\over 2}(1-\cos 2x),\ \cos^2 x={1\over 2}(1+\cos 2x)$.
Repeat while required.
