Find $\int \sin px \cos qx \text{ d}x$, where $p$ and $q$ are constants such that $p \ne q$ and $p \ne -q$. Given that $m \ne 0$, find $\int x \sin mx \text{ d}x$. Using the result in part (b), for all positive integers $m$, evaluate $\int_0^{\pi} x \sin mx \text{ d}x$, giving your answers in the form $\frac{k}{m}\pi$ where the possible value(s) of $k$ are to be determined.

1. Express the product of sine and cosine as a sum using trigonometric factor formulas.

$\sin px \cos qx = \frac{1}{2} \left( \sin(px + qx) + \sin(px - qx) \right) = \frac{1}{2} \left( \sin(p+q)x + \sin(p-q)x \right)$

2. Integrate term-by-term and add the constant of integration.

$\int \sin px \cos qx \text{ d}x = \frac{1}{2} \int \left( \sin(p+q)x + \sin(p-q)x \right) \text{ d}x$
$= -\frac{\cos(p+q)x}{2(p+q)} - \frac{\cos(p-q)x}{2(p-q)} + c$

1. Identify parts for integration by parts.

Let $u = x$ and $\frac{\text{d}v}{\text{d}x} = \sin mx$.
Then $\frac{\text{d}u}{\text{d}x} = 1$ and $v = -\frac{1}{m}\cos mx$.

2. Apply the integration by parts formula.

$\int x \sin mx \text{ d}x = x \left( -\frac{1}{m}\cos mx \right) - \int \left( -\frac{1}{m}\cos mx \right) (1) \text{ d}x$
$= -\frac{x}{m}\cos mx + \frac{1}{m}\int \cos mx \text{ d}x$

3. Evaluate the remaining integral and add the constant of integration.

$= -\frac{x}{m}\cos mx + \frac{1}{m^2}\sin mx + c$

1. Determine the antiderivative for part (b) using integration by parts.

$\int x \sin mx \text{ d}x = x\left(-\frac{1}{m}\cos mx\right) - \int 1 \cdot \left(-\frac{1}{m}\cos mx\right) \text{ d}x$
$= -\frac{x}{m}\cos mx + \frac{1}{m^2}\sin mx + C$

2. Substitute the limits into the antiderivative and evaluate.

$\int_0^{\pi} x \sin mx \text{ d}x = \left[ -\frac{x}{m}\cos mx + \frac{1}{m^2}\sin mx \right]_0^{\pi}$
$= \left( -\frac{\pi}{m}\cos m\pi + \frac{1}{m^2}\sin m\pi \right) - (0 + 0)$
Since $m$ is a positive integer, $\sin m\pi = 0$ and $\cos m\pi = (-1)^m$.
$= -\frac{\pi}{m}(-1)^m$

3. Determine the possible values of $k$ based on the parity of $m$.

Rewrite the evaluated integral to match the required form $\frac{k}{m}\pi$:
$-\frac{\pi}{m}(-1)^m = \frac{-(-1)^m}{m}\pi$
For odd integer $m$, $(-1)^m = -1 \implies k = -(-1) = 1$.
For even integer $m$, $(-1)^m = 1 \implies k = -(1) = -1$.
Thus, the possible values for $k$ are $1$ and $-1$.