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Math Help - what gets done to f(t) in this integral?

  1. #1
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    what gets done to f(t) in this integral?

    \arcsin{(t)}=\int\frac{1}{\sqrt{1-t^2}}dt

    Does this mean that...

    \arcsin{(f(t))}=\int\frac{1}{\sqrt{1-f(t)^2}}dt ?

    Or what happens to f(t) when this integral is evaluated? Does it just stay the same? That would be nice if it did, but I imagine something complicated happens to it.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Suppose to start from...

    \int \frac{dt}{\sqrt{1-t^{2}}}= \sin^{-1} t + c (1)

    ... and that we want to insert under the integral sign x=f(t) instead of t. It will be...

    \frac{dx}{dt} = f^{'} (t) \rightarrow dx= f^{'}(t)\cdot dt (2)

    ... so that we obtain...

     \int \frac{dx}{\sqrt {1-x^{2}}}= \int \frac{f^{'}(t)}{\sqrt{1-f^{2}(t)}}\cdot dt = \sin^{-1} \{f(t)\} + c (3)



    Merry Christmas from Italy

    \chi \sigma
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  3. #3
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    Think of it in terms of something simpler. You know that \int x dx= \frac{1}{2}x^2+ C. But \int f(x) dx is NOT equal to \frac{1}{2} f^2(x)+ C for every function f! That would make integration trivial. What you are doing is essentially a substitution. And if you let u= f(x), then you also have to let du= f'(x)dx.
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  4. #4
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    Quote Originally Posted by chisigma View Post
    Suppose to start from...

    \int \frac{dt}{\sqrt{1-t^{2}}}= \sin^{-1} t + c (1)

    ... and that we want to insert under the integral sign x=f(t) instead of t. It will be...

    \frac{dx}{dt} = f^{'} (t) \rightarrow dx= f^{'}(t)\cdot dt (2)

    ... so that we obtain...

     \int \frac{dx}{\sqrt {1-x^{2}}}= \int \frac{f^{'}(t)}{\sqrt{1-f^{2}(t)}}\cdot dt = \sin^{-1} \{f(t)\} + c (3)


    \chi \sigma

    Does the f^{2}(t) in equation 3 mean "f of t squared" or "second derivative of f(t)"?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Is...

    f(t)\cdot f(t) = f^{2} (t) (1)

    ... and...

    \frac{d^{2}f}{dt^{2}} = f^{(2)} (t) (2)



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