Expert: Paul Klarreich - 3/10/2007

Question
Hello, How do I go about finding the integral of a higher power function in a trig calculus problem such as when the cos is greater than to the 2nd power here is my example.
cos^3(2theta)  ,can I still use the trig substitution
for the 1/2 angle identity like this below

cos^2(n(theta)) = 1 + cos2(n(theta)) / 2   

or do I have to try another formula?

 Thanks


Answer
Questioner:   Jeff
Category:  Advanced Math
 
Subject:  Calculus trig question
Question:  Hello, How do I go about finding the integral of a higher power function in a trig calculus problem such as when the cos is greater than to the 2nd power here is my example.
cos^3(2theta)  ,can I still use the trig substitution
for the 1/2 angle identity like this below

cos^2(n(theta)) = 1 + cos2(n(theta)) / 2   

or do I have to try another formula?

Thanks
................................
Hi, Jeff,

Here's the scheme:

If you have odd powers of cos(x), it's easy (well, ....).  Write something like this:

cos^3(x) = cos x cos^2(x) = cos x(1 - sin^2(x))

Now do a substitution of u=sin x.

What about higher odd powers?

cos^5(x) = cos x cos^4(x) =
cos x (1 - sin^2(x))^2 =
cos x (multiply out and do u = sin x again.)
.....................
But if you have even powers of cos(x), then you must start with a half-angle identity.  You might have more work to do after that, as in:
              1 + cos(2x)
cos^4(x) =  (---------------)^2
                  2

And then you must multiply out, getting  cos^2(2x) in there, which means you must use the half-angle identity again, etc...

Naturally, this works for powers of  sin x  as well.

And remember -- doing examples like this increases your ability to handle frustration, which is an important life skill.