1. ## 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.

$\bg_white u = \int x\sqrt{\frac{1-ax^2}{1-x^2}}dx$ , where a is a constant.

2. 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.

3. Originally Posted by 300600e
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.

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

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

Now consider three cases:

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