मूल्यांकन कीजिएः
`int_(1//4)^(1//2) (1)/(sqrt(x-x^(2)))dx`

Explanation

`underset(1//4)overset(1//2)int (1)/(sqrt(x-x^(2)))dx`
`=underset(1//4)overset(1//2)int (1)/(sqrt(-(x^2-x)))dx`
`=underset(1//4)overset(1//2)int (1)/(sqrt(-{(x-1/2)^(2)-(1/2)^(2)}))dx`
`=underset(1//4)overset(1//2)int (1)/(sqrt((1/2)^(2)-(1/2)^(2)})dx`
`=underset(1//4)overset(1//2)int (1)/(sqrt((1/2)^2-(x-1/2)^(2)))dx`
`=[sin^(-1){((x-1/2))/((1/2))}]_(1//4)^(1//2)=[sin^(-1)(2x-1)]_(1//4)^(1//2)`
`=sin^(-1)(0)-sin^(-1)(-1/2)=0+sin^(-1)(1/2)`
`=0+pi/6=pi/6`