<s>[INST] How to integrate : $\sqrt{\frac{a-x}{x-b}}$ 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..
 [/INST] 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
$$\int dx\, \sqrt{\frac{a-x}{x-b}} = \sqrt{(a-x)(x-b)} - (a-b) \arctan{\sqrt{\frac{a-x}{x-b}}}+C$$
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