given:   ∫ dx / (x-B)*{(x -A)(B -x)}1/2 
   put x = B.sec2 β  ; so dx = 2B.sec2β.tan β.dβ
=  ∫ 2B.sec2β.tan β.dβ / B.tan2β.{B.tan2β.(A - Bsec2β)}1/2 
=  ∫ 2B.sec2β.tan β.dβ / B3/2.tan3β.{(A - Bsec2β)}1/2 
     Put  , tan β = t
        sec2β.dβ = dt
=  ∫ 2B.t.dt / B3/2.t3.{(A - B) -(B.t2))}1/2 
    put, { (A - B) -(B.t2) } = y2   ; so   -2Bt.dt = 2y.dy
= ∫ -2y.dy / m.{(A - B) -y2)}3/2 
= -2* ∫ dy /{(A - B) -y2)}3/2 
 put , y = (A-B)1/2.sin α  ; then we have , dy = (A-B)1/2.cos α. dα
= -2* ∫ (A-B)1/2.cos α. dα /{(A - B) -(A-B).sin2 α)}3/2 
=  [-2/(A-B)]* ∫ sec2α.dα
= -2.tan α / (A-B) + C
on substitution, we have
= tan {sin-1( ( [(A-B)-(x-B)]/[A-B] )1/2 )} + C  where C is constant
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IIT Kgp - 2005 batch