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Trig Substitution

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int\frac{1}{\sqrt{16-x^2}}dx[/itex]


    2. Relevant equations
    [itex]csc\theta=\frac{4}{\sqrt{16-x^2}}[/itex]

    [itex]4cos\theta=x[/itex]

    [itex]-4sin\theta d\theta=dx[/itex]

    [itex]\theta=arccos(\frac{x}{4})[/itex]

    3. The attempt at a solution
    Using these facts, I concluded that the integral, after all of the substitution, was
    [itex]-arccos(\frac{x}{4})[/itex] but the actual answer is the arcsin
     
  2. jcsd
  3. Sep 15, 2012 #2

    Mark44

    Staff: Mentor

    You don't show your complete answer, but what I think you have is probably correct. There is an identity that involves inverse trig functions:
    arcsin(x) + arccos(x) = ##\pi/2##

    So arcsin(x) = -arccos(x) + ##\pi/2##

    Your answer (-arccos(x/4)) and the book's answer (arcsin(x/4)) differ by a constant.

    Also, you can always check your answer to an integration problem by differentiating it. If the result is your original integrand, then your answer is correct.
     
  4. Sep 15, 2012 #3
    Oh, okay. It certainly makes perfect sense that there would be a inverse trig identity. Thank you.
     
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