<s>[INST] 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.
 [/INST] HINT:
$\displaystyle32\sin^6x=2(4\sin^3x)^2$
Using $\sin3A$ formula, $\displaystyle32\sin^6x=2(3\sin x-\sin3x)^2=3(2\sin^2x)+2\sin^23x-3(2\sin x\sin3x)$
Now apply $\displaystyle\cos2B=1-2\sin^2B\implies2\sin^2B=1-\cos2B$
and $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B)$
</s>