Evaluate $\int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}}$

Question

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.

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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?

Answer 1

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.

Answer 2

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 $$ \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} = \int \frac{\sqrt{(-a)x}dx}{\sqrt{x+(-a)}} = 2(-a)^\frac32\int \sinh^2 t\, dt$$

Answer 3

Let $x=\frac1t$ and $b =\frac1a$

\begin{align} \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} =& -\int \frac{dt}{t^2\sqrt{t-b}} =-\int \frac{1}{b\sqrt{t-b}}d\left(\frac{t-b}t\right)\\ \overset{ibp}=&-\frac{\sqrt{t-b}}{bt}+\frac1{b}\int \frac{d(\sqrt{t-b})}{(t-b)+b}\\ =&\ \begin{cases} -\frac{\sqrt{t-b}}{bt}+\frac{1}{b^{3/2}}\tan^{-1}\frac{t-b}{\sqrt b} & b>0\\ -\frac{\sqrt{t-b}}{bt}-\frac{1}{2(-b)^{3/2}}\ln\frac{\sqrt{t-b}-\sqrt{-b}}{\sqrt{t-b}+\sqrt{-b}}& b<0\\ \end{cases} \end{align}

Answer 4

Letting $b=\frac1a$ and $y^2=\frac{1}{x}-b$, then $$ \begin{aligned} \int \frac{d x}{\sqrt{\frac{1}{x}-\frac{1}{a}}}& =-2 \int \frac{d y}{\left(y^2+b\right)^2} \\ & =\int \frac{1}{y} d\left(\frac{1}{y^2+b}\right) \\ & =\frac{1}{y\left(y^2+b\right)}+\int \frac{1}{y^2\left(y^2+b\right)} d y \\ & =\frac{1}{y\left(y^2+b\right)}+\frac{1}{b} \int\left(\frac{1}{y^2}-\frac{1}{y^2+b}\right) d y \\ & =-\frac{y}{b\left(y^2+b\right)}+\frac{1}{b} \int \frac{1}{y^2+b} d y \end{aligned} $$ Now we can conclude that $$ I=\left\{\begin{array}{l} -a x \sqrt{\frac{1}{x}-\frac{1}{a}}+a^{\frac{3}{2}} \tan ^{-1}\left(\sqrt{\frac{a}{x}-1}\right) \text { if } a>0 \\ -a x \sqrt{\frac{1}{x}-\frac{1}{a}}+(-b)^{\frac{3}{2}} \ln \left|\frac{\sqrt{\frac{1}{x}-\frac{1}{a}}+\sqrt{-\frac{1}{a}}}{\sqrt{\frac{1}{x}-\frac{1}{a}}-\sqrt{-\frac{1}{a}}}\right| \text { if } a<0 \end{array}\right. $$