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

147 Points
```										Dear tejas
limx→0   o∫x  t2/(x-sinx)√(a+t)  dt =1

here we can take out (x-sinx) from the intigration .
limx→0  [ o∫x  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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```
7 years ago
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