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.