Question

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`

Answer 1

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.

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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.