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Homework Help: Trigonometric Substitution

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] [tex]\frac{e^{3x}dx}{\sqrt{1-e^{2x}}}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Alright so I am able to do other similar problems fine, I think it is the "e" that is throwing me off as well as the fact that the "x" is in the exponent. I started the problem as so...



    Am I off to the right start and if so can I have some helpful hints as what to do next? Thanks guys.

    By the way, those symbols next to Sin & Cos are Theta, I am still learning how to use this forum, sorry.
  2. jcsd
  3. Oct 9, 2008 #2
    I think you should have sin(theta) = e^x, cos(theta) =sqrt(1-e^2x). Remember that for any n, (e^nx) =(e^x)^n
  4. Oct 10, 2008 #3
    Re-write it.


    [tex]e^x=\sin xdx[/tex]

    Take the natural log, then it's derivative.
  5. Oct 10, 2008 #4
    Alright so taking your advice I set sin(theta) to e^x dx, and cos(theta) to sqrt(1-e^2x). That then gave me sin^3(theta)/cos((theta). I set u=sin(theta)d(theta) and du=cos(theta) which gave me u^3du^-1. That then became u^2/2 which is sin^2/2. My final answer U then got from that was e^2x/2 + c. Does that sound correct, if not please advise me and I am sorry for misunderstanding your help.
  6. Oct 10, 2008 #5
    Looks wrong.
  7. Oct 10, 2008 #6
    You cannot have a du in the denominator. Think about what this would means in terms of the Riemann sum and you'll see that it doesn't make any sense. Instead, consider an expansion of sin(x)³. That is

    [tex] \frac{sin^3(x)}{cos(x)} = sin(x) \frac{1-cos^2(x)}{cos(x)} = tan(x) - sin(x)cos(x) [/tex] You can integrate this.
  8. Oct 10, 2008 #7
    I am trying to set up a triangle for visual aid, is the following correct...

    tan= (e^x)/(sqrt(1-e^2x)

    Thanks guys, greatly appreciated.
  9. Oct 10, 2008 #8
    Yes, those are correct, though you're missing a bracket in tan
  10. Oct 10, 2008 #9
    Thanks, i'll try to take it from here but I don't think there is too much left to do.
  11. Oct 10, 2008 #10
    I didn't read all the way back,and that was my mistake, but did you remember to substitute for dx in the integral as well?

    If [itex] e^x = \sin\theta [/itex] then [itex] e^x dx = \cos\theta d\theta [/itex] which implies that [itex] dx = \mathrm{cot}\theta d\theta [/itex]. That should actually make the integral quite easy
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