<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] 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
$$\frac{\theta}{2}=\frac{\alpha}{2}-\frac{\pi}{4}=\arcsin\sqrt{x}-\frac{\pi}{4}$$
</s>