The directions state: evaluate the indefinite integrals by using the given substitutions to reduces the integrals to standard form. There were two problems i seem to not get at all.
1. (integral) (9r^2dr/(sqrt(1-r^3)) u = 1-r^3 du = -3r^2
the part that threw me off was -3r^2 is in 9r^2. All of the other cases, i had to add something namely the reciprocal to the left of the integral and do not know what to do in this case.
At any rate, the answer came to be: -6(1-r^3)^(1/2) + c
2. (integral) (sqrt(x))(sin(x^(3/2)-1)^2dx u = x^(3/2)-1 du = (3/2)(x^(1/2))
2/3 (integral) (sin u)^2
2/3 * (-1/3)(sin u)^3 *(cos u) = (-2/9)(sin (x^(3/2)-1))^3 * cos (x^(3/2)-1).
The answer came out to be (1/3)(x^(3/2)-1)-(1/6)(sin(2x^(3/2)-2)) + c