Evaluate int3sin^3(3x)cos^3(3x)dx?

1 Answer
May 9, 2018

int 3 sin^3 ( 3 x ) cos^3 ( 3 x )\ d x = \frac{cos^6 ( 3 x )}{6} - \frac{cos^4 ( 3 x )}{4} + C

Explanation:

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! ;-))