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How do you integrate $\int {(sin (3 x))}^{3} d x$ $int (sin(3x))^3 dx$ ?

1 Answer

Jun 22, 2015

I found: $- \frac{cos (3 x)}{3} = + \frac{{cos}^{3} (3 x)}{9} + c$ $-cos(3x)/3=+cos^3(3x)/9+c$

Explanation:

I changed the variale of integration from $x$ $x$ to $cos$ $cos$ and solved as:
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