I have written (x-a)(b-x) as

xb-x² -ab +ax

= -(x²-2x(a+b)/2 +ab)

= [(a+b)/2]² - [(x-(a+b)/2]²

So it becomes the form dt/root a² -x²

=sin^-1 x/a form. Is it right?

Juana,For such type of integrals involving (x-a)&(b-x),the standard thing to do is to put x=acos²k+bsin²k to make it easier.

Here,dx=(b-a)sin2kdk and (x-a)=(b-a)sin²k,(b-x)=(b-a)cos²k.

So,the integral becomes

=2k+c.

K can be evaluated using the above eqns. from which we get k=

You made a mistake here.

-[x²-(a+b)x+ab]=(a+b/2)²-ab-[x-(a+b/2)]².

Plz check it,U forgot the ab term.