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Trig Substitution

  1. Mar 13, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{dx}{\sqrt{x^2 + 16}}[/tex]

    2. Relevant equations

    [tex]x = 4\tan\theta[/tex]
    [tex]dx = 4\sec^2\theta \ d\theta[/tex]

    3. The attempt at a solution

    [tex]\int\frac{4\sec^2\theta}{\sqrt{16\tan^2\theta + 16}}\ d\theta[/tex]

    [tex]\int\frac{4\sec^2\theta}{\sqrt{16(\tan^2\theta + 1)}}\ d\theta[/tex]

    [tex]\int\frac{4\sec^2\theta}{4\sec\theta}\ d\theta[/tex]

    [tex]\int\sec\theta\ d\theta[/tex]

    How do I reduce past this step?
    Integration by parts returns me to [tex]\int\sec\theta\ d\theta[/tex]

    The answer at the back of the book is as follows: [tex]\ln(\sqrt{x^2 + 16} + x) + C[/tex]


    EDIT: notational mistakes corrected.
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2
    Nothing to see here...move on. :biggrin:
    Last edited: Mar 13, 2007
  4. Mar 13, 2007 #3
    Nope, his integral is correct.

    And the answer in the book is also correct.

    The integral of secant is not very trivial, and the best method is to multiply by '1'.
    In this case by (sec@ + tan@)/(sec@ + tan@).
    Then do a 'u' substitution if the integral isn't all that obvious yet.
    (let u = sec@ + tan@)


    P.S. Teneleven -> you notated your trig-sub wrong, yet used it correctly.
    P.P.S. sec³@ is significantly harder it integrate the first time through.
  5. Mar 13, 2007 #4
    Oh. Well, the first time I saw the problem the radical was in the numerator. :smile:
  6. Mar 14, 2007 #5

    Gib Z

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    Just incase your wondering, by multiply by one, hes hinting at Integration by parts.
  7. Mar 14, 2007 #6


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    The way I would integrate sec x, not necessarily the simplest, is to write sec x as 1/cos x, multiply both numerator and denominator by cos x:
    [tex]\int \frac{cos x dx}{cos^2 x}[/tex]
    rewrite as
    [tex]\int \frac{cos x dx}{1- sin^2 x}[/tex]
    Let u= sin(x) and use partial fractions on the remaining integral.
  8. Mar 14, 2007 #7


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    Make [itex] x=4\sinh t [/itex] and see what you get. Don't forget the integration constant.
  9. Mar 14, 2007 #8
    I solved it using ChaoticLlama's method, but without integration by parts. I'm having trouble reducing it down to the answer in the book. I end up with a "4" in the denominator and haven't figured out how to get rid of it.

    [tex]\int\sec\theta\frac{(\sec\theta + \tan\theta)}{(\sec\theta + \tan\theta)}\ d\theta[/tex]

    [tex]\int\frac{\sec^2\theta}{\sec\theta + \tan\theta}\ d\theta + \int\frac{\sec\theta\tan\theta}{\sec\theta +\tan\theta}\ d\theta[/tex]

    [tex]\ln(\sec\theta + \tan\theta) + C[/tex]

    [tex]\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C[/tex]
  10. Mar 14, 2007 #9
    You've done exactly as required.

    Just Remember the property of logarithms ln(a/b) = ln(a) - ln(b)

    Then you'll recognize that...
    [tex]\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C =
    \ln(\sqrt{x^2 + 16} + x) + D[/tex]

    (What does D equal with respect to C?)
  11. Mar 14, 2007 #10
    Thanks for the prompt reply.

    I'm looking at it and I recall the property of logarithms, but I still don't understand how the "4" is reduced to [tex]\ln(4)[/tex] and removed in the final answer.

    EDIT: So you're saying that D = C + [tex]\ln(4)[/tex]?
    Last edited: Mar 14, 2007
  12. Mar 15, 2007 #11

    Gib Z

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    Yes thats exactly what hes saying. Remember C is just any constant, when you differentiate it disappears! ln4 is also just a constant, A constant plus another constant is just another constant!!

    BTW Chaoticlama, No Integration by parts for sec x will not help in the least, Halls of Ivys method got me the answer fine though.
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