I'm looking for the shortest way to integrate $\int\sqrt{\frac{x}{1-x}}dx$.
All the other integration questions I've worked through in the same book section were pretty straightforward, whereas this one seems to take me through more steps than seems reasonable based on the other questions.
Note that the domain is $x\in[0,1)$ and integrate by parts as follows
$$\int\sqrt{\frac{x}{1-x}}dx =-\int\sqrt{\frac{x}{1-x}}d(1-x) =-\sqrt{{x}(1-x)} +\int\frac1{\sqrt{1-x}}d(\sqrt x)\\ =-\sqrt{{x}(1-x)} -\cos^{-1}\sqrt x+C $$
I would like to solve the problem by the substitution $x=\sin^2 \theta.$ $$ \begin{aligned} I &=\int \sqrt{\frac{\sin ^{2} \theta}{1-\sin ^{2} \theta}}(2 \sin \theta \cos \theta d \theta) \\ &=\int 2 \sin ^{2} \theta d \theta \\ &=\int(1-\cos 2 \theta) d \theta \\ &=\theta-\frac{\sin 2 \theta}{2}+C \\ &=\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C \end{aligned} $$
The idea to compute such integral $\int f(x) d x$ is to produce a nice parametrization of the curve $y=f(x)$, say $y(t), x(t)$ so that the integral becomes $\int y(t) x'(t) dt$. Here $y^2 (1-x)=x$ or $x(1+y^2)= y^2$ using $y$ as a parameter the integral is $\int y ({y^2\over 1+y^2})' dy$, then you can use standard methods (integration by part+ usual functins)
$\int y ({y^2\over 1+y^2})' dy= y ({y^2\over 1+y^2})+ \int {y^2\over 1+y^2} dy$
and $ \int {y^2\over 1+y^2} dy= \int {1+y^2 -1\over 1+y^2} dy= y - \arctan (y) $
$$u =\sqrt{1-x}$$ $$du = \frac{-1}{2\sqrt{1-x}}$$ $$-2\int\sqrt{1-u^2}du$$ Now just make a trig sub $u=\sin(\theta)$
