Evaluate the integral #int cos^4(2t) dt#?

#1/2[cos^3usinu + 3intcos^2usin^2u du]#

I cannot figure out how to do the integral of #int cos^2usin^2u du#. I saw there is an identity but idk how to get that identity which is #(1-cos4x)/8#

1 Answer
Mar 7, 2017

The first part of this answer describes a new way to find this integral using identities.

The second shows how to find the identity you were confused about.

Explanation:

What I'd recommend is to immediately try (from the very beginning) to substitute a simpler expression for #cos^4(2t)#.

A good identity to know is the cosine double-angle identity in all its forms. One is:

#cos(2alpha)=2cos^2(alpha)-1#

Which when rearranged, says that:

#cos^2(alpha)=(cos(2alpha)+1)/2#

Note that this holds for any cosine where functions where the squared function's argument (here, #alpha#) is half that of the other cosine function (here, #2alpha#). So the following also holds:

#cos^2(2t)=(cos(4t)+1)/2#

Which is close to the original integrand. Squaring shows that:

#intcos^4(2t)dt=int((cos(4t)+1)/2)^2dt#

Which can be simplified and split up. This simplification will also yield a #cos^2(4t)# term, which can be rewritten by reapplying the same identity as before. See that:

#cos^2(4t)=(cos(8t)+1)/2#

This is my recommendation for the integral.


In your case, with the step you originally took, I would solve the resulting integral with the identity:

#sin(2alpha)=2sin(alpha)cos(alpha)#

Then:

#sin^2(2u)=4sin^2(u)cos^2(u)#

So in your case:

#sin^2(u)cos^2(u)=sin^2(2u)/4#

This can be rewritten using the version of the cosine double-angle formula which uses sine:

#cos(2alpha)=1-2sin^2(alpha)#

Or:

#sin^2(alpha)=(1-cos(2alpha))/2#

Then:

#sin^2(2u)=(1-cos(4u))/2#

Returning to the previous expression concerning your integrand:

#sin^2(u)cos^2(u)=((1-cos(4u))/2)/4=(1-cos(4u))/8#

Which is the identity you were looking for.