What is the integral of sqrt(x^2-1)/x dx?

  • Thread starter funzone36
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  • #1
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Homework Statement



What is the integral of sqrt(x^2-1)/x dx?

Homework Equations





The Attempt at a Solution



∫ √(x^2 - 1) dx / x

let x = sec u: u = sec^-1(x) and tan u = √(x^2 - 1)

dx = sec u tan u du

now the integral becomes

∫ √sec^2(u) - 1) sec u tan u du / sec u

= ∫ tan u tan u du

= ∫ tan^2(u) du

= ∫ (sec^2(u) - 1) du

= ∫ (sec^2(u) du - ∫ du

= tan u - u + c

back substitute u = sec^-1(x) and tan u = √(x^2 - 1)

= √(x^2 - 1) - sec^-1(x) + c

However, the correct answer should be arccot(sqrt(x^2-1))+sqrt(x^2-1) + c. Can someone help me find what went wrong?
 
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Answers and Replies

  • #2
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Maple says that this integral is
sqrt(x^2-1)+arctan(1/(x^2-1))+c
 
  • #3
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Well, arctan(1/(x^2-1)) is the same as arccot(sqrt(x^2-1)). I just don't know the mistake I did when showing my steps.
 
  • #4
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funzone36, what you initially had was more or less correct. you can check by writing sec^-1(x) as arccos(1/x) and then differentiating.

The answer that you said was supposedly correct involving arccot is gotten by substituting x = csc(u) and turns out to be a slightly neater approach.
 
  • #5
Dick
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Homework Helper
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arcsec(x)=arctan(sqrt(x^2-1)). Draw a right triangle with hypotenuse x and leg 1. t=arcsec(x) is the angle between those two. What's tan(t)?
 
  • #6
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Thanks everywhere. My steps were actually correct. I learned that I should draw diagrams next time I do these problems.
 

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