I'm having a lot of trouble on figuring out how to evaluate this integral using the substitution $u=\sqrt{\frac{1-x}{1+x}}$. $$\int{\frac{1}{(1-x)^{2} \sqrt{1-x^{2}}} d x}$$ Do I have to normalise the denominator of the integral (i.e. multiply the numerator and denominator by $\sqrt{1-x^2}$)? I've done that but I fail to see how to progress from there.
Any help is greatly appreciated!
On
Hint:
$\frac{1}{(1-x)^2 \sqrt{1-x^2}}=\frac{1}{(1-x)^2 \sqrt{(1-x)(1+x)}}=\frac{1}{(1-x)^2 \sqrt{(1-x)} \sqrt{(1+x)}}=\frac{\sqrt{(1-x)}}{(1-x)^2 (1-x) \sqrt{(1+x)}}=\frac{1}{(1-x)^3}\sqrt{\frac{1-x}{1+x}}$
On
Hint : Try the easier substitution $x = cos(y)$.
To answer the question however if you take $u = \sqrt{\frac{1-x}{1+x}}$ we have $u^{2} = \frac{1-x}{1+x}$ or otherwise said $x = \frac{1-u^{2}}{1+u^{2}}$ you have $dx - \frac{1}{\sqrt{\frac{1-x}{1+x}}(1+x)^{2}} = du$ which leads to $-u \hspace{0.1cm}\frac{4}{(1+u^{2})^{2}}du = dx$; So the integral would become
$$\int \frac{4u}{(1+u^{2})^{2}(1- (\frac{1-u^{2}}{1+u^{2}})^{2})\sqrt{1-(\frac{1-u^{2}}{1+u^{2}})^{2}}}$$
On
Hint. Just do it. You might find that writing $$\frac{1-x}{1+x}$$ as $$\frac{2}{1+x}-1$$ helps. Then $$x=\frac{2}{1+y}-1,$$ which gives $$\mathrm dy=-\frac{2}{(1+y)^2}\mathrm dx.$$ I'll leave the rest to you, but for checks you should end up with the integral $$-\frac14\int y^{-5/2}+y^{-3/2}\,\mathrm dy.$$
$$u=\sqrt{\frac{1-x}{1+x}}\implies x=\frac{1-u^2}{1+u^2}\implies dx=-\frac{4 u}{\left(u^2+1\right)^2}\,du$$ Replace and simplify to get $$\int{\frac{dx}{(1-x)^{2} \sqrt{1-x^{2}}} }=-\int \frac{u^2+1}{2 u^4}\,du$$