Trigonometric Substitution

  • Thread starter protivakid
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  • #1
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Homework Statement



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


Homework Equations





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...

sin[tex]\vartheta[/tex]=[tex]\frac{e^{\sqrt{x}}}{\sqrt{1}}[/tex]

cos[tex]\vartheta[/tex]d[tex]\vartheta[/tex]=[tex]\frac{e^{\sqrt{x}}dx}{\sqrt{1}}[/tex]

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.
 

Answers and Replies

  • #2
73
1
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
 
  • #3
1,752
1
Re-write it.

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

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

Take the natural log, then it's derivative.
 
  • #4
17
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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.
 
  • #5
1,752
1
Looks wrong.
 
  • #6
743
1
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.
 
  • #7
17
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I am trying to set up a triangle for visual aid, is the following correct...

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

Thanks guys, greatly appreciated.
 
  • #8
743
1
Yes, those are correct, though you're missing a bracket in tan
 
  • #9
17
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Thanks, i'll try to take it from here but I don't think there is too much left to do.
 
  • #10
743
1
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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