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Integration using sinh^2 substitution

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{1}{\sqrt{x + x^2}} dx[/tex]
    We have been told to use the substitution [itex]x = \sinh^2{t}[/itex].


    2. Relevant equations
    [tex]\int \frac{1}{\sqrt{a^2 + x^2}}dx = \sinh^{-1}(\frac{x}{a}) + C[/tex]
    Maybe?


    3. The attempt at a solution
    I'm not really sure where to start, we haven't done any questions involving letting [itex]x = \sinh^2{t}[/itex], only [itex]x = \sinh{t}[/itex]. Substituting x into the integral doesn't seem to get me anywhere. I feel like I might have to derive [itex]x[/itex] which gives [itex]\sinh{2t}[/itex] but I don't know what to do with that. :confused:
     
    Last edited: Apr 21, 2013
  2. jcsd
  3. Apr 21, 2013 #2
    Try completing the square in the denominator and see what you end up with.
     
  4. Apr 21, 2013 #3
    Ah, I think I've got it. I ended up with [itex]2\sinh^{-1}(\sqrt{x}) + C[/itex].

    Now we're supposed to let [itex]y = 2\sinh^{-1}(\sqrt{x}) + C[/itex], find the inverse, then "use implicit differentiation to prove that the fundamental theorem of calculus holds, i.e. prove that [itex]\frac{dy}{dx} = \frac{1}{\sqrt{(x+x^2)}}[/itex]". I can find the inverse but I'm not sure how I can use implicit differentiation to prove that. Anything just to get me started would be much appreciated. :confused:
     
  5. Apr 21, 2013 #4
    I think they want you to find [itex] \frac{d}{dx}2sinh^{-1}\sqrt{x} [/itex] by saying that if [itex] y=2sinh^{-1}\sqrt{x} [/itex] then [itex] sinh\frac{y}{2}=\sqrt{x} [/itex], which you can differentiate implicitly.
     
  6. Apr 21, 2013 #5
    Oh right, that makes sense. Thanks!
     
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