\int _{\frac{\pi }{2}}^{\pi }\left(-x\:cos\:x\right)dx\:

We are tasked with evaluating the integral:

\[
I = \int_{\frac{\pi}{2}}^{\pi} \left(-x \cos x \right) dx
\]

### Step 1: Apply Integration by Parts
To solve this, we use integration by parts, where the formula is:

\[
\int u \, dv = uv - \int v \, du
\]

Let:
- \( u = -x \), so \( du = -dx \)
- \( dv = \cos x \, dx \), so \( v = \sin x \)

Now apply the integration by parts formula:

\[
I = uv \bigg|_{\frac{\pi}{2}}^{\pi} - \int_{\frac{\pi}{2}}^{\pi} v \, du
\]

Substitute \( u \), \( v \), \( du \), and \( dv \):

\[
I = \left( -x \sin x \right) \bigg|_{\frac{\pi}{2}}^{\pi} - \int_{\frac{\pi}{2}}^{\pi} (-\sin x) dx
\]

### Step 2: Evaluate the Boundary Terms
We now compute the boundary values for the term \( -x \sin x \):

\[
-x \sin x \bigg|_{\frac{\pi}{2}}^{\pi} = \left( -\pi \sin \pi \right) - \left( -\frac{\pi}{2} \sin \frac{\pi}{2} \right)
\]

Since \( \sin \pi = 0 \) and \( \sin \frac{\pi}{2} = 1 \), this simplifies to:

\[
= 0 - \left( -\frac{\pi}{2} \cdot 1 \right) = \frac{\pi}{2}
\]

### Step 3: Evaluate the Remaining Integral
Now we compute the remaining integral:

\[
\int_{\frac{\pi}{2}}^{\pi} (-\sin x) dx = \left[ \cos x \right]_{\frac{\pi}{2}}^{\pi}
\]

Evaluate the boundaries:

\[
\cos \pi - \cos \frac{\pi}{2} = -1 - 0 = -1
\]

### Step 4: Combine the Results
Now, combine the results from both steps:

\[
I = \frac{\pi}{2} - (-1) = \frac{\pi}{2} + 1
\]

Thus, the value of the integral is:

\[
I = \frac{\pi}{2} + 1
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you determine the functions for integration by parts?
2. Can this method be applied to any integral of the form \( x \cos x \)?
3. What if we integrate from 0 to \( \pi/2 \) instead?
4. How would the result change if \( \cos x \) were replaced by \( \sin x \)?
5. What are some other methods to solve integrals involving products of trigonometric and polynomial functions?

### Tip:
When choosing \( u \) and \( dv \) in integration by parts, it's often helpful to choose \( u \) as a polynomial function (if present) since differentiating it simplifies the expression.

This detailed solution demonstrates how to evaluate the integral ∫ (-x cos x) dx from π/2 to π using integration by parts. Learn step-by-step how to apply integration techniques, compute trigonometric boundaries, and combine results for a clear and concise solution, ideal for advanced calculus students.

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