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 / B^3/2.tan³β.{(A - Bsec²β)}^1/2

     Put , tan β = t

        sec²β.dβ = dt

= ∫ 2B.t.dt / B^3/2.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