A

Step 1: Evaluate the first integral:
$$\int_{0}^{1} 9x^8 \, dx$$
Use the power rule for integration:
$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$
Therefore:
$$\int_{0}^{1} 9x^8 \, dx = 9 \left[ \frac{x^{9}}{9} \right]_{0}^{1} = 9 \left( \frac{1^{9}}{9} - \frac{0^{9}}{9} \right) = 9 \cdot \frac{1}{9} = 1$$

Step 2: Evaluate the second integral:
$$\int_{0}^{\frac{\pi}{2}} \cos x \, dx$$
The integral of cos is sin:
$$\int \cos x \, dx = \sin x + C$$
Therefore:
$$\int_{0}^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin \left( \frac{\pi}{2} \right) - \sin(0) = 1 - 0 = 1$$

Step 3: Combine both results:
$$1 + 1 = 2$$
Final Answer: Therefore, the value of the expression is 2.