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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?

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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.

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    $\begingroup$ What happen if $a<0?$ $\endgroup$ – lab bhattacharjee May 23 at 7:02
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    $\begingroup$ @labbhattacharjee If a is negative, we can replace the subtraction with addition and let $b=-a$, and now use $x=b \tan^2 \theta$. $\endgroup$ – auscrypt May 23 at 7:04
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    $\begingroup$ The assumption should find a place in the post $\endgroup$ – lab bhattacharjee May 23 at 7:14
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    $\begingroup$ @labbhattacharjee Thanks, I’ll add it in now. $\endgroup$ – auscrypt May 23 at 7:14
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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 $$ \int \frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{a}}} = \int \frac{\sqrt{(-a)x}dx}{\sqrt{x+(-a)}} = 2(-a)^\frac32\int \sinh^2 t\, dt$$

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