How to Find Indefinite Integral Using U-substitution

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The discussion focuses on using U-substitution to find the indefinite integral of sin(2x)/cos^4(2x) dx. The initial attempt involved setting u = cos^4(2x), but complications arose due to the presence of x variables. A more effective substitution was suggested, using u = cos(2x), which simplified the integral to -1/2 ∫ u^-4 du. The final result was derived as 1/6 u^-3 + C, and it was confirmed that substituting back to x should yield the original integrand upon differentiation. The conversation emphasizes the importance of choosing the correct substitution for successful integration.
mattmannmf
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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... here's 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.. don't know.
 
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No, it definitely won't come out being ln(something).
mattmannmf said:
It just seems like its sooooo complicated.. don't know.
That's because you're making it too complicated by using the wrong substitution. Instead, use u = cos(2x).
 
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)
 
Right. And after you undo your substitution you can check your answer. Its derivative should be [sin(2x)/cos^4(2x)].
 
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