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प्रश्न
Evaluate the definite integral:
`int_0^1 dx/sqrt(1-x^2)`
उत्तर
`int_0^1 dx/sqrt(1 - x^2)`
`= [sin^-1 x]_0^1`
`= sin^-1 (1) - sin^-1 0`
= `pi/2 - 0`
`= pi/2`
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