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प्रश्न
Evaluate the definite integral:
`int_(pi/6)^(pi/3) (sin x + cosx)/sqrt(sin 2x) dx`
उत्तर
Let `I = int_(pi/6)^(pi/3) (sin x + cos x)/sqrt(sin 2x)`dx
`= int_(pi/6)^(pi/3) (sin x + cos x)/sqrt(1 - (1 - sin 2x))`dx
`= int_(pi/6)^(pi/3) (sin x + cos x)/sqrt(1 - (sin x - cos x)^2)`dx
Put sin x - cos x = t
(cos x + sin x) dx = dt
When `x = pi/6, t = sin pi/6 - cos pi/6`
`= 1/2 - sqrt3/2`
`= (sqrt3 - 1)/2`
When x = `pi/3, t = sin = pi/3 - cos pi/3`
`= sqrt3/2 - 1/2`
`(sqrt3 - 1)/2`
∴ `I = int_(1/2 - sqrt3/2)^(sqrt3/2-1/2) dt/sqrt(1-t^2) = [sin^-1 t]_(1/2-sqrt3/2)^(sqrt3/2-1/2)`
`= sin^-1(sqrt3/2 - 1/2) - sin^-1 (1/2 - sqrt3/2)`
`= sin^-1 (sqrt3/2 - 1/2) + sin^-1 (sqrt3/2 - 1/2)`
`= 2 sin^-1 1/2 (sqrt3 - 1)`
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