`int_(pi//3)^(pi//2)sqrt(cos\ x)\ sin^3x\ dx`

Write `sin^3x ` `= sin^2x\ sin\ x` ` = (1 - cos^2x)(sin\ x)`

So, taking the indefinite case first:

`sqrt(cos\ x) sin^3x`

`=cos^(1//2)x(1-cos^2x)(sin x)`

`=(cos^(1//2)x-cos^(5//2)x)(sin\ x)`

Put `u = cos\ x` then `du = -sin\ x\ dx`

So

`intsqrt(cos\ x)\ sin^3x\ dx =int(cos^(1//2)x-cos^(5//2)x)(sin\ x)dx`

`=-int(u^(1//2)-u^(5//2))du`

`=-(2u^(3//2))/3+(2u^(7//2))/7+K`

`=-(2\ cos^(3//2) x)/3+(2\ cos^(7//2)x)/7+K`

So we have for the definite case:

`int_(pi//3)^(pi//2)sqrt(cos\ x)\ sin^3x\ dx =[-(2 cos^(3//2)x)/3+(2 cos^(7//2)x)/7]_(pi//3)^(pi//2)`

`=[(0+0)-(-0.23570+0.02525)]`

`=0.2104`

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