Answer
$$6 \sin \left(\frac{\theta}{12}\right)+\frac{6}{7} \sin \left(\frac{7 \theta}{12}\right)+C$$
Work Step by Step
Use the formula
$$\int \cos a x \cos b x d x=\frac{\sin (a-b) x}{2(a-b)}+\frac{\sin (a+b) x}{2(a+b)}+C, \quad a^{2} \neq b^{2}$$
with $ a=\dfrac{1}{3},\ \ \ b =\dfrac{1}{4}$, then we get
\begin{align*}
\int \cos \frac{\theta}{3} \cos \frac{\theta}{4} d \theta=6 \sin \left(\frac{\theta}{12}\right)+\frac{6}{7} \sin \left(\frac{7 \theta}{12}\right)+C
\end{align*}