A

Let's solve the integrals one by one:

1. For the integral:
$$ \int \sin 60^\circ \, dx $$
We know that: $$ \sin 60^\circ = \frac{\sqrt{3}}{2} $$
Thus, the integral becomes:
$$ \int \sin 60^\circ \, dx = \int \frac{\sqrt{3}}{2} \, dx $$
This is a constant multiplication of the integral of 1, which yields:
$$ = \frac{\sqrt{3}}{2} x + C $$

2. For the integral:
$$ \int x^{-\frac{3}{2}} \, dx $$
Using the formula for integrating powers:
$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $$
Setting $$ n = -\frac{3}{2} $$, we have:
$$ n + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} $$
So:
$$ = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + C = -2x^{-\frac{1}{2}} + C = -\frac{2}{\sqrt{x}} + C $$

Combining the results:
The results of the integrals are:
1. $$ \frac{\sqrt{3}}{2} x + C $$
2. $$ -\frac{2}{\sqrt{x}} + C $$

Final Answer:
Therefore, the integral results are correctly obtained.