Question

How do you integrate `int cos^3(3x)dx`?

Answer 1

`1/4sin(3x)+1/36sin(9x)+C`

Explanation:

This can be a tricky one, first we need to find a way to express the function to remove the power of 3.

To do this, we can try:

`cos^3(theta)=cos(theta)*cos^2(theta)`

Use `cos^2theta = 1/2+1/2cos2theta` to obtain:

`cos(theta)(1/2+1/2cos2theta)`

`=1/2cos(theta)+1/2cos(theta)cos(2theta)`

Finally we can use: `cos(A)cos(B) = 1/2{cos(A-B)+cos(A+B)}` to rearrange the last term and get:

`1/2cos(theta)+1/4cos(theta-2theta)+1/4cos(theta+2theta)`

=`1/2cos(theta)+1/4cos(-theta)+1/4cos(3theta)`

Of course, the cosine function has even symmetry so:

`cos(-theta)=cos(theta)`

Which will give us the exression:

`3/4cos(theta)+1/4cos(3theta)`

So it will naturally follow that:

`intcos^3(3x)dx=int3/4cos(3x)+1/4cos(9x)dx`

Which easily integrates to give:

`1/4sin(3x)+1/36sin(9x)+C`