lim x→0 [from 0 to x∫t 2 /(x-sinx)√(a+t)]=1

limx→0[from 0 to x∫t2/(x-sinx)√(a+t)]=1


1 Answers

Badiuddin askIITians.ismu Expert
148 Points
14 years ago

Dear tejas

 limx→0   ox  t2/(x-sinx)√(a+t)  dt =1

here we can take out (x-sinx) from the intigration .

 limx→0  [ ox  t2/√(a+t)  dt] /(x-sinx)   = 1

noe its a 0/0 form so use L hospital rule

limx→0  [ x2/√(a+x) ]  /(1-cosx)   = 1

limx→0  [ x2/√(a+x) ]  /(2sin2x/2)   = 1

limx→0  [ 2(x/2)2/√(a+x) ]  /(sin2x/2)   = 1

now apply limit

 2/√a  =1

 a =4

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