Yes, There's a Trick

This is the graph of $\sin(16x)$ for $0 \leq x \leq \frac{\pi}{8}$ . Notice that since both regions of different colors have the same areas, the total area must be zero. We can use this info to compute $\int\limits_0^{12\pi} \sin(16x)\,dx$ .

Express the given as $\int\limits_{0}^{12\pi} \sin(16x)\,dx + \int\limits_0^{12\pi} k\,dx$ For $\sin(16x)$ , observe that $\int\limits_{0}^{\frac{\pi}{8}} \sin(16x)\,dx = 0$ which illustrates the regions of the same areas. Since $\sin(16x)$ has a periodicity of $\frac{\pi}{8}$ , there are $12\pi \div \frac{\pi}{8} = 96$ pairs of such regions for $0 \leq x \leq 12\pi$ . In that case, $\int\limits_{0}^{12\pi} \sin(16x)\,dx = 96 \cdot \int\limits_0^{\frac{\pi}{8}} \sin(16x) = 0$ Thus, $\begin{array}{rl} \int\limits_0^{12\pi} k\,dx &= 12\pi k = 36\pi\\ & \Longrightarrow \boxed{k = 3} \end{array}$