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
I hope I didn't understand you correctly, lest you're asking a rather weird thing: in do you "integrate just the exponent" or "both the exponent and the function x"? Absurd, isn't it?
The expression is one single bundle, and it means the trigonometric function sine x raised to the second power: you can not "separate them! Now, if you want to know how to do it, here's an idea with some basic trigonometry:
Now just gather together stuff, solve the easy integral of and you're done.