How do you integrate cos3(3x)dx?

1 Answer
Mar 9, 2017

14sin(3x)+136sin(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:

cos3(θ)=cos(θ)cos2(θ)

Use cos2θ=12+12cos2θ to obtain:

cos(θ)(12+12cos2θ)

=12cos(θ)+12cos(θ)cos(2θ)

Finally we can use: cos(A)cos(B)=12{cos(AB)+cos(A+B)} to rearrange the last term and get:

12cos(θ)+14cos(θ2θ)+14cos(θ+2θ)

=12cos(θ)+14cos(θ)+14cos(3θ)

Of course, the cosine function has even symmetry so:

cos(θ)=cos(θ)

Which will give us the exression:

34cos(θ)+14cos(3θ)

So it will naturally follow that:

cos3(3x)dx=34cos(3x)+14cos(9x)dx

Which easily integrates to give:

14sin(3x)+136sin(9x)+C