Answer is: option3

\( \frac{\pi}{12} (\sin \frac{\pi}{2} + 2 \sin \frac{2\pi}{3} + 2 \sin \frac{5\pi}{6} + \sin \pi) \)

Solution:

Given:

1. Interval: \( \left[ \frac{\pi}{2}, \pi \right] \)
2. Subdivisions: \( n = 3 \)

So,

\[ h = \frac{\pi - \frac{\pi}{2}}{3} = \frac{\pi}{6} \]

Subinterval points:

1. \( x_0 = \frac{\pi}{2} \)
2. \( x_1 = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3} \)
3. \( x_2 = \frac{\pi}{2} + 2 \cdot \frac{\pi}{6} = \frac{5\pi}{6} \)
4. \( x_3 = \pi \)

Trapezoidal rule formula:

\[ \int_a^b f(x) \, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3) \right] \]

With:

1. \( h = \frac{\pi}{6} \)
2. \( f(x) = \sin x \)

So the approximation becomes:

\[ \frac{\pi}{12} \left[ \sin\left(\frac{\pi}{2}\right) + 2 \sin\left(\frac{2\pi}{3}\right) + 2 \sin\left(\frac{5\pi}{6}\right) + \sin(\pi) \right] \]

Option (C).