Question

Prove the following:

`int_0^(pi/2) sin^3 xdx = 2/3`

Answer 1

`= 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`