Integrate the function w.r.t. $x$: $\sin mx$

$-\frac{1}{m} \cos mx + C$

The correct answer is Option (2) → $-\frac{1}{m} \cos mx + C$

We know that derivative of $mx$ is $m$. Thus, we make the substitution $mx = t$ so that $m \, dx = dt$.

Therefore, $\int \sin mx \, dx = \frac{1}{m} \int \sin t \, dt = -\frac{1}{m} \cos t + C = -\frac{1}{m} \cos mx + C \text{}$