 # integrate: dx/(x-B){(x -A)(B -x)}^1/2 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; [A=alpha , B=Beta]

13 years ago

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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