Question

integration of
(sin3x)^3

Answer 1

(sin3x)^3
[cos(3x)+isin(3x)]^3 = cos(9x) + isin(9x) 
Expand the left side and compare real parts: 
cos(9x) = cos^3(3x) - 3cos(3x)sin^2(3x) = -3cos(3x) + 4cos^3(3x) + c 
So, 
-3 cos(3x)/12 + cos(9x)/36 
= (1/36)(-9cos(3x) -3cos(3x) + 4cos^3(3x) + c) 
= -(1/3)cos(3x) + (1/9)cos^3(3x) + C 

In addition, you can use mental substitution. 
∫ (sin(x))^3 dx 
= -∫ (sin(x))^2 dcos(x) 
= ∫ (cos(x))^2 - 1 dcos(x) 
= (1/3)(cos(x))^3 - cos(x) 

So, 
∫ (sin(3x))^3 dx 
= (1/3)∫ (sin(3x))^3 d3x 
= (1/9)(cos(3x))^3 - (1/3)cos(3x) + c