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Integration with variable substitution

  1. Dec 21, 2016 #1

    Rectifier

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    Hello, I am having trouble with solving the problem below

    The problem

    Find all primitive functions to ## f(x) = \frac{1}{\sqrt{a+x^2}} ##.
    (Translated to English)

    The attempt
    I am starting with substituting ## t= \sqrt{a+x^2} \Rightarrow x = \sqrt{t^2 - a} ## in $$ \int \frac{1}{\sqrt{a+x^2}} \ dx \rightarrow \int \frac{1}{t} \ dx $$
    ## \frac{dx}{dt} = D \left( \sqrt{t^2 - a} \right) = \frac{1}{2 \sqrt{t^2 - a}} \cdot 2t = \frac{t}{\sqrt{t^2 - a}} ##
    .
    ## dx = \frac{t}{\sqrt{t^2 - a}} \cdot dt ##

    Together we get:
    $$ \int \frac{1}{t} \frac{t}{\sqrt{t^2 - a}} \cdot dt = \int \frac{1}{\sqrt{t^2 - a}} \cdot dt $$

    This is where I get stuck since this integral is not much easier to solve that the original.
     
  2. jcsd
  3. Dec 21, 2016 #2

    PeroK

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    With these sorts of integrals, you should be thinking about a trig substitution.
     
  4. Dec 21, 2016 #3

    Rectifier

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    But the answer to this problem is ## \ln|x + \sqrt{x^2+a}| + C ## and not a trig function.
     
  5. Dec 21, 2016 #4

    PeroK

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    Ha! You might be better off in this case not knowing the answer. That is actually a trig function in disguise!
     
  6. Dec 21, 2016 #5

    Rectifier

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    Unfortunately, that does not make my life easier :,(
     
  7. Dec 21, 2016 #6

    PeroK

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    Let's assume ##a## is positive. And, let ##x = \sqrt{a} f(u)##, where ##f## is some unknown trig function, then we need to look at:

    ##x^2 + a = a(f^2(u) + 1)## and ##dx = \sqrt{a} f'(u)du##

    You're looking for a trig function where ##f^2(u) + 1## is an identity. The obvious one is ##f(u) = \tan(u)##. You can try that and you should (eventually) get the answer. But, perhaps there's a better one to try?
     
  8. Dec 21, 2016 #7

    ehild

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    Think of the hyperbolic functions, and the most basic relationship between cosh and sinh.
    img5.gif
    Use the substitution √a sinh(u)=x
     
    Last edited: Dec 21, 2016
  9. Dec 21, 2016 #8

    Ray Vickson

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    If ##a > 0## you can write the integrand as ##1/\sqrt{x^2+b^2}##. Now the identity ##b^2 \cosh^2(t) = b^2 \sinh^2(t) + b^2## comes to mind. If ##a < 0## you can write the integrand as ##1/\sqrt{x^2 - b^2}##, and the identity ##b^2 \sinh^2(t) = b^2 \cosh^2(t) - b^2## cones to mind.
     
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