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#
I cannot figure out how to do the integral of
1 Answer
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
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,
#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)=(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.