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

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Answer 1 · 2017-03-09T23:48:48

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

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#