<s>[INST] How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$ I'm having a bit of trouble solving this integral:  $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.
 [/INST] Set $\sqrt{x} = \sin(t)$. We then have $x = \sin^2(t)$. Hence, $1-x = \cos^2(t)$. This gives us
\begin{align}
\int \dfrac{\sqrt{1-x}}{\sqrt{x}}dx & = \int \dfrac{\cos(t)}{\sin(t)} 2 \sin(t) \cos(t) dt = 2\int \cos^2(t) dt\\
& = \int(1+\cos(2t))dt = t + \dfrac{\sin(2t)}2 + c\\
& = \arcsin(\sqrt{x}) + \sqrt{x}\sqrt{1-x} + c
\end{align}
</s>