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Math Help - 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
    Super Member Matt Westwood's Avatar
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    The substitution I would make would be u = \sqrt {1 - x^2}.

    If you rewrite the integral as:

    \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
    \int {x\sqrt {\frac{{1 - a{x^2}}}{{1 - {x^2}}}} dx}  = \left\{ \begin{gathered}<br />
  \frac{{1 - a{x^2}}}<br />
{{1 - {x^2}}} = {t^2} \Leftrightarrow {x^2} = \frac{{{t^2} - 1}}<br />
{{{t^2} - a}}, \hfill \\<br />
  xdx = \frac{{\left( {1 - a} \right)t}}<br />
{{{{\left( {{t^2} - a} \right)}^2}}}dt \hfill \\ <br />
\end{gathered}  \right\} = \left( {1 - a} \right)\int {\frac{{{t^2}}}<br />
{{{{\left( {{t^2} - a} \right)}^2}}}dt}

    = \frac{{a - 1}}<br />
{2}\int {td\left( {\frac{1}<br />
{{{t^2} - a}}} \right)}  = \frac{{\left( {a - 1} \right)t}}<br />
{{2\left( {{t^2} - a} \right)}} - \frac{{a - 1}}<br />
{2}\int {\frac{{dt}}<br />
{{{t^2} - a}}}

    Now consider three cases:

    1) a<0;
    2) a=0;
    3) a>0.
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