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Math Help - Indefinite integral

  1. #1
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    Indefinite integral

    Find the following indefinite integral, identifying any rules of calculus that you use:

    \int1/\sqrt{(1-x^2)*\arcsin(x)}

    I think I need to use integration by substitution, so have rearranged to give:

    \int(1-x^2)^{\-1/2} * (sin(x))^{1/2}

    I am not quite sure what to do next. The formula I have for integration by substitution is:

    \int{f(g(x))g'(x)dx} = \int{f(u)du}, where u = g(x)

    Would I then get

    f(x) = (1-x^2)^{\-1/2}

    and

    g(x) = (sin(x))^{1/2}, so u = -cosx

    or do I need to integrate both parts first as they are both composite integrals and then use the above formula on the answers I get?

    Thanks for any help in advence
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  2. #2
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    Quote Originally Posted by cozza View Post
    Find the following indefinite integral, identifying any rules of calculus that you use:

    \int1/\sqrt{(1-x^2)*\arcsin(x)}

    \displaystyle \int \frac{1}{\sqrt{\arcsin{x}}} \cdot \frac{1}{\sqrt{1-x^2}} \, dx

    u = \arcsin{x}<br />

    \displaystyle du = \frac{1}{\sqrt{1-x^2}} \, dx

    substitute ...

    \displaystyle \int \frac{1}{\sqrt{u}} \, du<br />

    finish it
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  3. #3
    Junior Member bondesan's Avatar
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    I'll give you a huge hint. You will have to use the u-substitution, this identity:

    \displaystyle{\frac{d}{dx} arcsin(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2}}

    and the integral rewritten as: \displaystyle{\int\frac{1}{\sqrt{(1-x^2)\cdot arcsin(x)}}dx=\int (1-x^2)^{-1/2}\cdot(arcsin(x))^{-1/2}dx}

    This should work.
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  4. #4
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    Hi, thanks for the help. Can I just check that the answer I have is right:

    1/(\sqrt{1-x^2})+c
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    I think the answer is:

    2sqrt(arcsinx) + c
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  6. #6
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I think the answer is:

    2sqrt(arcsinx) + c
    Quote Originally Posted by cozza View Post
    Hi, thanks for the help. Can I just check that the answer I have is right:

    1/(\sqrt{1-x^2})+c
    On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

    1/(1-x^2) ?

    Where does the 2 come from? Have I missed something?
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  7. #7
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    Quote Originally Posted by cozza View Post
    On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

    1/(1-x^2) ?

    Where does the 2 come from? Have I missed something?
    yes, you have.


    as stated earlier ...

    \displaystyle \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C<br />

    back substitute u = \arcsin{x} ...

    2\sqrt{\arcsin{x}} + C
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