`int sqrt(x) sin(1 + x^(3/2)) dx` Evaluate the indefinite integral.

You need to evaluate the indefinite integral by using the substitution `1 + x^(3/2) = t` , such that:

`1 + x^(3/2) = t => (3/2)x^(3/2-1)dx = dt => sqrt x*dx = (2/3)*dt`

`int sqrt x*sin(1 + x^(3/2) )dx = (2/3) int sin t dt`

`(2/3) int sin t dt = -(2/3)*cos t + c`

Replacing back `1 + x^(3/2)` for t yields:

`int sqrt x*sin(1 + x^(3/2) )dx = -(2/3)*cos (1 + x^(3/2))+ c`

Hence, evaluating the indefinite integral, yields `int sqrt x*sin(1 + x^(3/2) )dx = -(2/3)*cos (1 + x^(3/2))+ c.`