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.