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प्रश्न
Evaluate : `int_0^4 (1)/sqrt(4x - x^2)*dx`
योग
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उत्तर
`int_0^4 (1)/sqrt(4x - x^2)*dx`
= `int_0^4 (1)/sqrt(4 - (x^2 - 4x + 4))*dx`
= `int_0^4 (1)/sqrt(2^2 - (x - 2)^2)*dx`
= `[sin^-1 ((x - 2)/2)]_0^4`
= `sin^-1((4 - 2)/2)- sin^-1 ((0 - 2)/2)`
= sin–1 1 –sin–1 (– 1)
= 2 sin–1 1 ...[∵ sin–1 (– x) = – sin–1 x]
= `2(pi/2)`
= `pi`.
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