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प्रश्न
Evaluate:
`int_0^1 (1)/sqrt(3 + 2x - x^2)*dx`
उत्तर
`int_0^1 (1)/sqrt(3 + 2x - x^2)*dx`
= `int_0^1 (1)/sqrt(3 - (x^2 - 2x + 1) + 1)*dx`
= `int_0^1 (1)/sqrt((2)^2 - (x - 1)^2)*dx`
= `[sin^-1 ((x - 1)/2)]_0^1`
= `sin^-1(0) - sin^-1(-1/2)`
= `0 - sin^-1 (-sin pi/6)`
= `-sin^-1[sin(- pi/6)]`
= `-(- pi/6)`
= `pi/(6)`.
APPEARS IN
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