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प्रश्न
Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2) "d"x`
उत्तर
Let I = `int_0^1 1/sqrt(3 + 2x - x^2) "d"x`
= `int_0^1 1/sqrt(4 - 1 + 2x - x^2) "d"x`
= `int_0^1 1/sqrt(4 - (x^2 - 2x + 1)) "d"x`
= `int_0^1 1/sqrt((2)^2 - (x - 1)^2) "d"x`
= `[sin^-1 ((x - 1)/2)]_0^1`
= `sin^-1 (0) - sin^-1 (1/2)`
= `0 - (- pi/6)`
∴ I = `pi/6`
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