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

Dear tejas

 lim_{x→0   o}∫^x  t²/(x-sinx)√(a+t)  dt =1

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

 lim_x→0  [ _o∫^x  t²/√(a+t)  dt] /(x-sinx)   = 1

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

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

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

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

now apply limit

 2/√a  =1

 a =4

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