As a first step, we use the trigonometric Pythagoras to rearrange the integral a little:
# int 3 sin^3 ( 3 x ) cos^3 ( 3 x )\ d x =#
#= int 3 sin ( 3 x ) [ 1 - cos^2 ( 3 x ) ] cos^3 ( 3 x )\ d x#.
At this point, we may view #cos ( 3 x )# as an inner function, with derivative #- 3 sin ( 3 x )#, which is why the integral is the same as
#= [ - int ( 1 - t^2 ) t^3 \ d t ]_{t = cos ( 3 x )} =#
#= [ int t^5 - t^3 \ d t ]_{t = cos ( 3 x )} =#
#= [ \frac{t^6}{6} - \frac{t^4}{4} + C ]_{t = cos ( 3 x )} =#
#= \frac{cos^6 ( 3 x )}{6} - \frac{cos^4 ( 3 x )}{4} + C#.
Notice:
If you check the integral on Wolfram Alpha, then it will give you
#\frac{1}{192} [ cos ( 18 x ) - 9 cos ( 6 x ) ] + C#
as result. It turns out, however, that this describes the same function. (Make sure you understand why! ;-))
