Integration Problem

  • Thread starter Kenji Liew
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Homework Statement



[tex]\int[/tex]e[tex]^{x}[/tex][tex]\sqrt{1+e^{2 x}}[/tex]dx


Homework Equations


Cosh x = [ e[tex]^{x}[/tex]+ e[tex]^{-x}[/tex] ] / 2



The Attempt at a Solution


[tex]\int[/tex]e[tex]^{x}[/tex][tex]\sqrt{1+e^{2x}}[/tex]dx
= [tex]\int[/tex]e[tex]^{x}[/tex][tex]\sqrt{2e^{x}Cosh x}[/tex]dx

I have no idea on how to proceed already. I cannot obtained the solution given by Mathematica. I use substitution method by let u = e[tex]^{x}[/tex] and then use trigonometric substitution. But also no idea on how to proceed. Hoping anyone outside there can try to help me in solving this question.Thanks.
 
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Answers and Replies

  • #2
vela
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You don't want to use a trig substitution. After you use u=ex, try the substitution v=u+1. (Or if you want to do the problem in one shot, use u=ex+1.)
 
  • #3
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You don't want to use a trig substitution. After you use u=ex, try the substitution v=u+1. (Or if you want to do the problem in one shot, use u=ex+1.)

I am very sorry.I have a typo in the question and just done the correction.
 
  • #4
vela
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What trig substitution did you use and what did you get after you applied it?
 
  • #5
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What trig substitution did you use and what did you get after you applied it?

I have attached the reply in PDF
 

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  • #6
vela
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Mathematica gives you kind of complicated form of the integral of sec3 θ. You can evaluate the integral by hand using integration by parts. You should get

[tex]\int \sec^3\theta\,d\theta = \frac{1}{2}(\log \lvert \sec\theta + \tan\theta \rvert + \sec\theta \tan\theta)+c[/tex]

You could also simplify the result from Mathematica to that. Note that the variable is θ. You need to undo the substitutions to get it back in terms of x. Once you do that, it's pretty straightforward to show it equals the second answer you got from Mathematica.
 
  • #7
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Mathematica gives you kind of complicated form of the integral of sec3 θ. You can evaluate the integral by hand using integration by parts. You should get

[tex]\int \sec^3\theta\,d\theta = \frac{1}{2}(\log \lvert \sec\theta + \tan\theta \rvert + \sec\theta \tan\theta)+c[/tex]

You could also simplify the result from Mathematica to that. Note that the variable is θ. You need to undo the substitutions to get it back in terms of x. Once you do that, it's pretty straightforward to show it equals the second answer you got from Mathematica.

Thanks for your help,Vela !!! :smile::smile:
 
  • #8
dextercioby
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Making the right substitution [itex] e^x = u [/itex] you get the integral [itex] \int \sqrt{1+u^2} \, du [/itex] which you can solve by making the substitution [itex] u=\sinh p [/itex] and a use of the double angle formula for hyperbolic cosine

[tex] \cosh^2 p = \frac{1}{2}\left(1+ \cosh 2p\right) [/tex]
 
  • #9
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Making the right substitution [itex] e^x = u [/itex] you get the integral [itex] \int \sqrt{1+u^2} \, du [/itex] which you can solve by making the substitution [itex] u=\sinh p [/itex] and a use of the double angle formula for hyperbolic cosine

[tex] \cosh^2 p = \frac{1}{2}\left(1+ \cosh 2p\right) [/tex]

Thanks as well, dextercioby for the another method !
 

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