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प्रश्न
Prove the following:
`int_0^(pi/2) sin^3 xdx = 2/3`
उत्तर
`= int_0^(pi/2) sin^3 x dx`
`= 1/4 int_0^(pi/2) (3 sinx - sin 3x) dx` `....[∵ sin 3x = 3 sin x - 4 sin 3x]`
`= 1/4 [-3 cos x + (cos 3x)/3]_0^(pi/2)`
`= 1/4 [- 3 cos pi/2 + 1/3 cos (3pi)/2] - 1/4 [- 3 cos 0 + (cos0)/3]`
`= 1/4 [0 + 0 + 3 - 1/3]`
`= 1/4 (8/3)`
`= 2/3`
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