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प्रश्न
Evaluate:
`int_0^(pi/2) sqrt(cos x) sin^3x * dx`
उत्तर
Let I = `int_0^(pi/2) sqrt(cos x) sin^3x * dx`
= `int_0^(pi/2) sqrt(cosx)sin^2x sinx * dx`
= `int_0^(pi/2) sqrt(cosx)(1 - cos^2x)sinx*dx`
Put cos x = t
∴ – sin x · dx = dt
∴ sin x · dx = – dt
When x = 0, t = cos 0 = 1
When x = π/2, t = cos 2π = 0
`I = - int_1^0 sqrt(t)(1 - t^2)(dt)`
`I = - int_1^0 (t^(1//2) - t^(5//2)) dt`
`I = -[(2t)^(3//2)/3 - (2t)^(7//2)/7]_1^0`
`I = -[0 - (2/3 - 2/7)]`
`I = (14 - 6)/21`
`I = 8/21`
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