`int u sqrt(1 - u^2) du` Evaluate the indefinite integral.

You need to use the following substitution  `1 - u^2 = t` , such that:

`1 - u^2 = t=> -2udu = dt=>u du = -(dt)/2`

`int u*sqrt(1 - u^2) du = -(1/2)*int sqrt t dt`

`-(1/2)*int sqrt t dt = (-1/2)*(t^(3/2))/(3/2) + c`

Replacing back `1 - u^2` for t yields:

`int u*sqrt(1 - u^2) du = (-1/3)*((1 - u^2)^(3/2)) + c`

Hence, evaluating the indefinite integral, yields` int u*sqrt(1 - u^2) du = -((1...

You need to use the following substitution  `1 - u^2 = t` , such that:

`1 - u^2 = t=> -2udu = dt=>u du = -(dt)/2`

`int u*sqrt(1 - u^2) du = -(1/2)*int sqrt t dt`

`-(1/2)*int sqrt t dt = (-1/2)*(t^(3/2))/(3/2) + c`

Replacing back `1 - u^2` for t yields:

`int u*sqrt(1 - u^2) du = (-1/3)*((1 - u^2)^(3/2)) + c`

Hence, evaluating the indefinite integral, yields` int u*sqrt(1 - u^2) du = -((1 - u^2)^(3/2))/3 + c`