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mattmannmf
Nov27-09, 01:12 PM
Using U-substitution find the indefinite integral of:

[sin(2x)/cos^4(2x)] dx

So I do know that it will have to come out to it being ln... heres what i did so far
ok so i made u= cos^4(2x)
du= -8cos^3(2x)*sin(2x)dx....(just took the derivative of u and simplified it)

so made sin(2x)dx= du/(-8cos^3(2x))...so i can substitute it into my equation.

so it came out to be:
du/(u*-8cos^3(2x))...but in using u-substitution, i should not have an x variable...

So do i have to minipulate u=cos^4(2x) to get x by its self?
I get x= .5cos^-1(4sqrt(x))

It just seems like its sooooo complicated.. dont know.

Mark44
Nov27-09, 01:23 PM
No, it definitely won't come out being ln(something).
It just seems like its sooooo complicated.. dont know.
That's because you're making it too complicated by using the wrong substitution. Instead, use u = cos(2x).

mattmannmf
Nov27-09, 01:35 PM
ok thanks...
So u=cos(2x)
-du/2= sin(2x) dx...then i substitute:

=-1/2[integral]du/u^4
where i get =-1/2[integral] u^-4du
=-1/2*-1/3u^-3+C
=1/6u^-3+C

Is that correct?...then i can just substitute what u equals into the equation (since they started in terms of x, ill leave it in terms of x)

Mark44
Nov27-09, 02:36 PM
Right. And after you undo your substitution you can check your answer. Its derivative should be [sin(2x)/cos^4(2x)].