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Math Help - proof of derivative of arcsin(x)

  1. #1
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    proof of derivative of arcsin(x)

    Hi guys

    I have this problem I'm a little stuck on. It's a 5 part question which leads you through a proof of \frac{d}{dx}\left( \sin^{-1} x\right)

    In the previous two parts of the question, I was asked to show that

    \sin^{-1} t = 2\int^{t}_{0}\sqrt{1-x^2}\,dx - t\sqrt{1-t^2}

    and that

    \int^{t}_{0}\sqrt{1-x^2}\,dx = t\sqrt{1-t^2} + \int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\,dx

    which I did. It's the final part which I'm stuck on. It says the following

    "Use your answers to the above two parts to prove that"

    \sin^{-1} t = \int^{t}_{0}\frac{dx}{\sqrt{1-x^2}}


    I thought it would be simple enough, I proceeded as follows

    \sin^{-1} t = 2\left[ t\sqrt{1-t^2} + \int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\,dx \right] - t\sqrt{1-t^2}

    =2\int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\, dx + t\sqrt{1-t^2}

    ...and it's here I get stuck. I've looked at it for a while now and can't see how use the above information to derive the required result.

    (oh, I can already derive the result using other methods, it's this particular question which is bugging me)

    Thanks
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Hi Stonehambey.

    Quote Originally Posted by Stonehambey View Post
    \sin^{-1} t = 2\left[ t\sqrt{1-t^2} + \int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\,dx \right] - t\sqrt{1-t^2}

    =2\int^{t}_{0}\frac{x^2}{\sqrt{1-x^2}}\, dx + t\sqrt{1-t^2}

    ...and it's here I get stuck. I've looked at it for a while now and can't see how use the above information to derive the required result.
    \int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx

    =\ -\int_0^t\frac{-x^2}{\sqrt{1-x^2}}\, dx

    =\ -\int_0^t\left(\frac{1-x^2}{\sqrt{1-x^2}}-\frac1{\sqrt{1-x^2}}\right)\, dx

    =\ -\int_0^t\left(\sqrt{1-x^2}-\frac1{\sqrt{1-x^2}}\right)\, dx

    =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}-\int_0^t\sqrt{1-x^2}\,dx

    =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}-t\sqrt{1-t^2}-\int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx


    \therefore\ 2\int_0^t\frac{x^2}{\sqrt{1-x^2}}\, dx+t\sqrt{1-t^2}\ =\ \int_0^t\frac{dx}{\sqrt{1-x^2}}
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