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Thread: Integration problem

  1. #1
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    Integration problem

    I would like to know how to find u as shown below.

    This question is an indefinite integral, and it is a whole question. No method specified.

    , where a is a constant.
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  2. #2
    MHF Contributor Matt Westwood's Avatar
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    The substitution I would make would be $\displaystyle u = \sqrt {1 - x^2}$.

    If you rewrite the integral as:

    $\displaystyle \int x \frac {\sqrt {1 - ax^2}} {\sqrt {1 - x^2}} dx$ it may look less confusing.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by 300600e View Post
    I would like to know how to find u as shown below.

    This question is an indefinite integral, and it is a whole question. No method specified.

    , where a is a constant.
    Hint
    $\displaystyle \int {x\sqrt {\frac{{1 - a{x^2}}}{{1 - {x^2}}}} dx} = \left\{ \begin{gathered}
    \frac{{1 - a{x^2}}}
    {{1 - {x^2}}} = {t^2} \Leftrightarrow {x^2} = \frac{{{t^2} - 1}}
    {{{t^2} - a}}, \hfill \\
    xdx = \frac{{\left( {1 - a} \right)t}}
    {{{{\left( {{t^2} - a} \right)}^2}}}dt \hfill \\
    \end{gathered} \right\} = \left( {1 - a} \right)\int {\frac{{{t^2}}}
    {{{{\left( {{t^2} - a} \right)}^2}}}dt}$

    $\displaystyle = \frac{{a - 1}}
    {2}\int {td\left( {\frac{1}
    {{{t^2} - a}}} \right)} = \frac{{\left( {a - 1} \right)t}}
    {{2\left( {{t^2} - a} \right)}} - \frac{{a - 1}}
    {2}\int {\frac{{dt}}
    {{{t^2} - a}}}$

    Now consider three cases:

    1) $\displaystyle a<0$;
    2) $\displaystyle a=0$;
    3) $\displaystyle a>0$.
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