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Math Help - Integration by Substitution

  1. #1
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    Integration by Substitution

    Hi

    Integral, I = -4 * dx/(3+2x-x^2)^0.5
    There is an integral sign between 4 and dx.

    I = -4 * dx/((4-(x-1)^2)^0.5 <--- completing the square

    Let x-1 = 2sin(#), where # is theta

    The result is 4arcsin((x-1)/2) + c, where c is an arbitrary constant

    Please explain how this works.

    Thanks
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  2. #2
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    <br />
-4 * \int{\frac{dx}{\sqrt{((4-(x-1)^2))}}} = -4 * \int{\frac{dx}{\sqrt{((2^2-(x-1)^2))}}}<br />

    Then, since
    <br />
\int{\frac{du}{\sqrt{a^2-u^2}}} = arcsin \frac{u}{a} + C,<br />

    just let u = x - 1 and then we have

    <br />
-4 * \int{\frac{dx}{\sqrt{((2^2-(x-1)^2))}}} = -4 * arcsin \frac{x-1}{2} + C<br />
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  3. #3
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    Thanks!
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  4. #4
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    The method in the notes is "let (x-1) = 2sin(theta)" for the above.

    How does this work?
    Can you show this using Pythagoras theorem?
    Why is it arcsin rather than sin?

    Thanks again
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  5. #5
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    Re: integration

    I don't see how letting

    <br />
(x-1) = 2 * sin(\theta)<br />

    is useful in this context.
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