<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] You're on the right track!
Now do a couple more substitutions:
$$\cos^2 (2x) = \frac{1 + \cos(4x)}{2};$$
$$\cos^3 (2x) = (1 - \sin^2 2x)\cos 2x.$$
So from where you were:
$$I = \frac { 1 }{ 8 } \int ({ 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x)  dx$$
$$I = \frac { 1 }{ 8 } \int (1 - 3 \cos 2x+\frac{3(1+\cos 4x)}{2} - (1 - \sin^2 2x)(\cos 2x))  dx$$
$$I = \frac { 1 }{ 8 } \int (\frac{5}{2} - 2 \cos 2x + \frac{3}{2}\cos 4x + \sin^2 2x \cos 2x)  dx.$$
Can you take it from there?
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