Calculus (3rd Edition)

$$\frac{1}{3}\sin(3 \theta)-2\tan\left(\frac{\theta}{4}\right)+c.$$
Since $(\sin(3 \theta))'= 3\cos (3 \theta)$ and $(\tan\left(\frac{\theta}{4}\right))'=\frac{1}{4}\sec ^{2}\left(\frac{\theta}{4}\right)$ , we have $$\int\left(\cos (3 \theta)-\frac{1}{2} \sec ^{2}\left(\frac{\theta}{4}\right)\right) d \theta\\ =\frac{1}{3}\sin(3 \theta)-\frac{4}{2}\tan\left(\frac{\theta}{4}\right)+c\\ =\frac{1}{3}\sin(3 \theta)-2\tan\left(\frac{\theta}{4}\right)+c.$$