Integration by Trigonometric Substitution.

  • Thread starter azatkgz
  • Start date
  • #1
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I'm not sure about answer.It looks very strange.

Homework Statement



[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}[/tex]





The Attempt at a Solution



for u=lnx-->u'=1/x
[tex]\int \frac{du}{\sqrt{1+u^2}}[/tex]
substituting [tex]u=tan\theta[/tex]

[tex]=\int \frac{d\theta}{cos\theta}=ln|sec\theta+tan\theta|[/tex]

[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}=ln|\sqrt{-1}|[/tex]
 
Last edited:

Answers and Replies

  • #2
learningphysics
Homework Helper
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Your formula looks right but your final results isn't. Didn't you actually use u = tan(theta) ?

what limits do you get for theta?
 
  • #3
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I just put to the
[tex]ln|lnx+\sqrt{lnx-1}|[/tex]
 
  • #4
learningphysics
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I just put to the
[tex]ln|lnx+\sqrt{lnx-1}|[/tex]
Did you use u = tan(theta) or u = sec(theta) ?
 
  • #5
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Sorry i typed wrongly.I used u=tan(theta)
 
  • #6
learningphysics
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Sorry i typed wrongly.I used u=tan(theta)
ok. so this part is wrong

[tex]ln|lnx+\sqrt{lnx-1}|[/tex]

fix your substitution.
 
  • #7
Gib Z
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Express sec in terms of tan again.
 
Last edited:
  • #8
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but if I change limits
[tex]\int_{0}^{\frac{\pi}{4}}sec\theta d\theta=ln2[/tex]
 
  • #9
Gib Z
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I don't know what you just did, but continue as you were before, you had the right anti derivative: ln |tan O + sec O|, but you didn't replace the original variable back in properly for the sec O.
 
  • #10
dextercioby
Science Advisor
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Just use a "sinh" substitution when you're left only with the sqrt in the denominator and you're done.
 

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