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We are tasked with determining whether the given integral is convergent or divergent and to evaluate it if it is convergent. The integral is as follows:

\[
\int_{\frac{\pi}{2}}^{\pi} \frac{\cos x}{(\sin x - 1)^2} \, dx
\]

### Step 1: Investigating the behavior of the integrand

The integrand is:

\[
f(x) = \frac{\cos x}{(\sin x - 1)^2}
\]

This function may potentially cause divergence if either the numerator or the denominator tends to infinity or becomes undefined within the limits of integration.

Let's examine the denominator \((\sin x - 1)^2\). Since \(\sin x\) varies between \(-1\) and \(1\) in the interval \([\frac{\pi}{2}, \pi]\), we need to identify points where \(\sin x = 1\), as that would make the denominator zero, potentially leading to an infinite result.

- \(\sin(\frac{\pi}{2}) = 1\)
- \(\sin(\pi) = 0\)

At \(x = \frac{\pi}{2}\), the denominator becomes zero, causing the integrand to approach infinity. Therefore, we need to investigate the behavior of the integrand near this point to determine if the integral converges or diverges.

### Step 2: Behavior near \(x = \frac{\pi}{2}\)

Near \(x = \frac{\pi}{2}\), \(\sin x \approx 1\), so \(\sin x - 1 \approx 0\). To explore this in more detail, we use a Taylor expansion around \(x = \frac{\pi}{2}\) for small values of \((x - \frac{\pi}{2})\):

\[
\sin x \approx 1 + \frac{(x - \frac{\pi}{2})^2}{2}
\]

Thus,

\[
\sin x - 1 \approx \frac{(x - \frac{\pi}{2})^2}{2}
\]

Substituting this into the integrand, we approximate it near \(x = \frac{\pi}{2}\):

\[
\frac{\cos x}{(\sin x - 1)^2} \approx \frac{\cos x}{\left(\frac{(x - \frac{\pi}{2})^2}{2}\right)^2}
\]

Since \(\cos(\frac{\pi}{2}) = 0\), the integrand behaves like:

\[
\frac{\cos x}{(x - \frac{\pi}{2})^4}
\]

This implies that near \(x = \frac{\pi}{2}\), the function grows extremely rapidly, leading to a divergence at \(x = \frac{\pi}{2}\).

### Step 3: Conclusion

The integrand has a singularity at \(x = \frac{\pi}{2}\), where it tends to infinity. Because of this, the integral diverges. Thus, the given integral does **not** converge.

### Final Answer:

The integral diverges.

---

Would you like further explanation on any part of this process or more details? Here are some related questions to expand on this topic:

1. How do you identify singular points in an integral?
2. What are some techniques to test for convergence of improper integrals?
3. Can we use substitution to simplify such integrals?
4. How does the behavior of trigonometric functions affect convergence?
5. What are common types of singularities in integrals?

**Tip:** When evaluating an integral, always check for points where the function may be undefined or approach infinity to assess convergence.

Determine whether the integral \( \int_{\pi/2}^{\pi} \frac{\cos x}{(\sin x - 1)^2} dx \) is convergent or divergent. Evaluate if convergent.

Explore the convergence of the integral \( \int_{\pi/2}^{\pi} \frac{\cos x}{(\sin x - 1)^2} dx \), including analysis of the behavior near singular points and using Taylor expansion to determine divergence. Suitable for undergraduate calculus students.

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