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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
    So far i have made
    arcsin x = u and
    du=1/(sqrt1-x^2).dx

    I am at u^2 * du

    I have spent hours on this and it appears obvious but i just don't get it! sorry
    Any pointers i would be extremely grateful. I feel its something to do with the calculus relationship between the values of arcsin and du.
    Thanks in advance
    Ken
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  2. #2
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    Quote Originally Posted by kendon
    Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
    So far i have made
    arcsin x = u and
    du=1/(sqrt1-x^2).dx

    I am at u^2 * du

    I have spent hours on this and it appears obvious but i just don't get it! sorry
    Any pointers i would be extremely grateful. I feel its something to do with the calculus relationship between the values of arcsin and du.
    Thanks in advance
    Ken
    \int \frac{(\arcsin(x))^2}{\sqrt{1-x^2}} dx

    you make the substitution \arcsin(x)=u, find du=\frac{1}{\sqrt{1-x^2}}, so the integral becomes:

    <br />
\int u^2\ du<br />
.

    Look OK to me.

    Now:

    <br />
\int u^2\ du=\frac{u^3}{3}+C=\frac{(\arcsin(x))^3}{3}+C<br />

    which can be checked by differentiating it, you should get the integrand back.

    RonL
    Last edited by CaptainBlack; August 3rd 2006 at 04:25 AM.
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  3. #3
    Eater of Worlds
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    You've got the right substitution.

    Let u=sin^{-1}(x)

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

    You have:

    \int{u^{2}}du

    Now, it's easy, ain't it?.
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  4. #4
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    thanks a lot that has been superb.
    cheers
    ken
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