<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] Note that you only need to integrate $\cos^2(2x)$ and $\cos^3(2x)$ now. 
$2\cos^2(2x)=\cos(4x)+1$
$\cos(6x)=4\cos^3(2x)-3\cos(2x)$
(double and triple angle formulas for cosine).
Can you proceed now?
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