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

Ramesh V
70 Points
```										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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we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
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