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:

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

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