Evaluate $\int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}}$ 
Evaluate the following integral:
  $$\displaystyle \int\dfrac{dx}{\sqrt{\dfrac{1}{x}-\dfrac{1}{a}}}$$
  Where $a$ is an arbitrary constant.

How do I solve this? 
EDIT: I would appreciate it if you do consider the case when $a<0$, but this is an integral that I encountered in a physics problem. Considering $a>0$ will suffice. 

I tried the substitution $$x=a\cos \theta$$
And I ended up with:
$$\displaystyle a^{3/2}\int\dfrac{\sqrt {\cos\theta}.\sin\theta.d\theta}{\sqrt{1-\cos\theta}}$$
How do I simplify this further?
 A: Your substitution seems to make the problem more complicated, since the square root terms still remain. 
Instead, we first simply $\displaystyle \frac{1}{\sqrt{\frac{1}{x} - \frac{1}{a}}}$ by writing the two fractions on the denominator as a single fraction, and then flipping the fraction, to obtain $\displaystyle \frac{\sqrt {ax}}{\sqrt {a-x}}$ (assuming $a>0$). From here, making the substitution $x=a \sin^2 \theta$ yields $$\int \frac{a \sin \theta}{\sqrt a \cos \theta} 2a \sin \theta \cos \theta d\theta = \int 2a^{\frac{3}{2}} \sin^2 \theta d \theta$$
which can now be easily solved.
For $a<0$, simply let $b=-a$. We will obtain $\displaystyle \frac{\sqrt b}{\sqrt {b+x}}$ in the integrand, from which the substitution $x=b \tan^2 \theta$ will work.
A: It can be rewritten (for $0<x<a$) as
$$ \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} = \int \frac{\sqrt{ax}dx}{\sqrt{a-x}}$$
We can use substitution $x = a\sin^2\theta$, $0<\theta<\frac{\pi}{2}$ to get
$$ \int \frac{\sqrt{ax}dx}{\sqrt{a-x}} = 2a^\frac32 \int \frac{\sqrt{\sin^2\theta}\sin\theta\cos\theta \,d\theta}{\sqrt{1-\sin^2\theta}} = 2a^\frac32\int \sin^2\theta\, d\theta$$
For $x<a<0$ we use subsstitution $x=a\cosh^2t$, $t>0$ and we get
$$ \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} = \int \frac{\sqrt{(-a)(-x)}dx}{\sqrt{(-x)-(-a)}} = -2(-a)^\frac32\int \cosh^2t\, dt$$
Finally, for $a<0<x$ we use $x=-a\sinh^2t$, $t>0$ to get
$$ \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} = \int \frac{\sqrt{(-a)x}dx}{\sqrt{x+(-a)}} = 2(-a)^\frac32\int \sinh^2 t\, dt$$
