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Math Help - Integration by Substitution - Confusion

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    Integration by Substitution - Confusion

    So the integral question is:

    Int(1/((1-x^2)^(-3/2)))

    I thought I would substitute in x=cos(u) dx=-sin(u)du ...

    Wolfram Alpha uses x=sin(u)

    But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

    What went wrong?
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    Quote Originally Posted by mike_302 View Post
    So the integral question is:

    Int(1/((1-x^2)^(-3/2)))

    I thought I would substitute in x=cos(u) dx=-sin(u)du ...

    Wolfram Alpha uses x=sin(u)

    But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

    What went wrong?
    change both antiderivatives back into an expression in terms of x, see what happens ...
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    Quote Originally Posted by mike_302 View Post
    So the integral question is:

    Int(1/((1-x^2)^(-3/2)))

    I thought I would substitute in x=cos(u) dx=-sin(u)du ...

    Wolfram Alpha uses x=sin(u)

    But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

    What went wrong?
    You must mean 1/(1-x^2)^{3/2}= (1- x^2)^{-3/2} but not both "1/" and the negative exponent!

    Either x= cos(u) or x= sin(u) will work because 1- cos^2(u)= sin^2(u) and 1- sin^2(u)= cos^2(x) and you get rid of that 1/2 power.

    Yes, using x= cos(u) you get the integral as cot(u) and with x= sin(u) you get u= tan(x).

    No, those are not the same function but since those are the results of different integrals, why would you expect them to be?

    You haven't finished the problem yet!

    If x= cos(u) then u= cos^{-1}(x) so, with the integral equal to tan(u), you get tan(cos^{-1}(x)). If x= sin(u), then x= sin^{-1}(u) and you get cot(sin^{-1}(x)).

    Now, tan= \frac{sin}{cos} so tan(cos^{-1}(x))= \frac{sin(cos^{-1}(x))}{cos(cos^{-1}(x))}. Of course, cos(cos^{-1}(x))= x while sin(cos^{-1}(x))= \sqrt{1- cos^2(cos^{-1}(x)}= \sqrt{1- x^2}. Your integral is \frac{\sqrt{1- x^2}}{x}.

    cot= \frac{cos}{sin} so cot(sin^{-1}(x))= \frac{cos(sin^{-1}(x))}{sin(sin^{-1}(x))}. Now sin(sin^{-1}(x))= x and cos(sin^{-1}(x))= \sqrt{1- x^2} so the integral is \frac{\sqrt{1- x^2}}{x}, just as before.
    Last edited by HallsofIvy; January 23rd 2010 at 09:51 AM.
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