A

Step 1: Solve the first integral
The first integral is:
$$\int \sin 60^\circ \, dx$$
We know that
$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$
Hence, we can rewrite the integral as:
$$\int \frac{\sqrt{3}}{2} \, dx$$
This simplifies to:
$$\frac{\sqrt{3}}{2} \int 1 \, dx = \frac{\sqrt{3}}{2} x + C_1$$

Step 2: Solve the second integral
The second integral is:
$$\int x^{-\frac{3}{2}} \, dx$$
To solve this, we can apply the power rule:
$$\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C$$ where n is not equal to -1.
Here, n = -\frac{3}{2}, so n + 1 = -\frac{3}{2} + 1 = -\frac{1}{2}.
Therefore, we have:
$$\int x^{-\frac{3}{2}} \, dx = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 = -2x^{-\frac{1}{2}} + C_2$$

Final Result
Combining the results from both integrals:
$$\int \sin 60^\circ \, dx = \frac{\sqrt{3}}{2} x + C_1$$
$$\int x^{-\frac{3}{2}} \, dx = -2x^{-\frac{1}{2}} + C_2$$
Therefore, the answers are:

• (i) $$\frac{\sqrt{3}}{2} x + C_1$$
• (ii) $$-2x^{-\frac{1}{2}} + C_2$$