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.
 A: Let $$u=\frac{\sqrt{1-x}}{\sqrt{x}}$$
Then $u^2=\frac{1-x}{x}=\frac{1}{x}-1$. Hence
$$x=\frac{1}{u^2+1}$$
$$dx=-\frac{2u}{(u^2+1)^2}$$
Your integral becomes
$$-2 \int \frac{u^2}{(u^2+1)^2}du=-2 \int \frac{u^2+1}{(u^2+1)^2}du+2 \int \frac{1}{(u^2+1)^2}du$$
which can be calculated integrating by parts $\frac{1}{u^2+1}$ or via a standard trig substitution.
A: Alternative solution, making the substitution $x=t^2$ integral becomes :
$$I=\int\frac{\sqrt{1-x}}{\sqrt{x}}\;\mathrm{d}x = 2\int\sqrt{1-t^2}\;\mathrm{d}t =2J$$
But that means :
$$J=\int\sqrt{1-t^2}\;\mathrm{d}t = \int\frac{1-t^2}{\sqrt{1-t^2}}\;\mathrm{d}t = \arcsin t - \int\frac{t^2}{\sqrt{1-t^2}}\;\mathrm{d}t =  $$
 ... via per partes ...
$$= \arcsin t + t\sqrt{1-t^2}-\int\sqrt{1-t^2}\;\mathrm{d}t = \arcsin t + t\sqrt{1-t^2} -J $$
Therefore
$$I=2J =\arcsin t + t\sqrt{1-t^2} = \arcsin \sqrt{x} + \sqrt{x}\sqrt{1-x} $$
However, your original way is not bad after all, if you continued - see :
$$\frac{1}{2}\int\frac{\cos^2{\theta}}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int\frac{1-\sin^2{\theta}}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int\frac{(1-\sin{\theta})(1+\sin{\theta})}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int1-\sin{\theta}\;\mathrm{d}\theta $$
Therefore
$$I=\frac{1}{2}\theta+\frac{1}{2}\cos{\theta}=\frac{1}{2}\arcsin{(2x-1)}+\frac{1}{2}\sqrt{1-(2x-1)^2}= \frac{1}{2}\arcsin{(2x-1)}+\sqrt{x^2-x}$$
and these results are indeed equivalent, because
$$\frac{1}{2}\arcsin{(2x-1)}=\arcsin\sqrt{x}-\frac{\pi}{4}$$
multiplying by $2$ and taking sin of both sides :
$$2x-1=-\cos\left( 2\arcsin\sqrt{x}\right)=2\sin^2\arcsin\sqrt{x}-1=2x-1$$
Or let $\theta=\alpha-\pi/2$, then
$$x=\frac{1+\sin\theta}{2}=\frac{1-\cos\alpha}{2}=\sin^2\left(\frac{\alpha}{2}\right)$$
So
$$\frac{\theta}{2}=\frac{\alpha}{2}-\frac{\pi}{4}=\arcsin\sqrt{x}-\frac{\pi}{4}$$
A: 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}
A: Integrate by parts
$$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx
=\sqrt x \sqrt{1-x}-\int \frac{d(\sqrt x)}{\sqrt{1-x}}
= \sqrt{x(1-x)}-\sin^{-1}\sqrt x$$
A: 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.
