[Math] How to integrate : $\sqrt{\frac{a-x}{x-b}}$

Question

Problem :

How to integrate : $\sqrt{\frac{a-x}{x-b}}$

Unable to find the substitution for this :

$\sqrt{\frac{a-x}{x-b}}$

Please help how to proceed ………..thanks..

Answer 1

Let

$$t = \frac{a-x}{x-b}$$

Then

$$x=\frac{a+b t}{1+t}$$

and

$$dx = \frac{b-a}{(1+t)^2} dt$$

Then the integral is

$$(b-a) \int dt \frac{\sqrt{t}}{(1+t)^2}$$

Now sub $t=\tan^2{u}$ and the integral becomes

$$2 (b-a) \int du \sin^2{u} = (b-a) (u-\sin{u} \cos{u}) + C$$

Now back substitute to get the integral in terms of $x$. I get

$$\int dx\, \sqrt{\frac{a-x}{x-b}} = \sqrt{(a-x)(x-b)} - (a-b) \arctan{\sqrt{\frac{a-x}{x-b}}}+C$$