<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] Integrate $\int \frac{\sqrt{1-x}}{\sqrt x} dx$
let
$$
\begin{equation} \tag 1
x=t^2, \,\,\,\,\, dx=2t\, dt 
\end{equation}
$$
now; $\int \frac{\sqrt{1-t^2}}{t} dt$
$$
\begin{equation} \tag{as $dt=\frac{dx}{2t}$} 
\int 2 \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 2
2 \int \sqrt{1 - t^2} dt
\end{equation}
$$
$$
\begin{equation} \tag 3
2 \left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}\sin^{-1} (t) \right]
\end{equation}
$$
put $t=\sqrt{x}$, to get answer.
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