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प्रश्न
Evaluate: `int_0^1 1/sqrt(1 - x^2) "d"x`
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उत्तर
`int_0^1 1/sqrt(1 - x^2) "d"x = [sin^-1x]_0^1`
= sin–1(1) – sin–1(0)
= `pi/2 - 0`
= `pi/2`
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