Answer
$$\int\cos\frac{\theta}{3}\cos\frac{\theta}{4}d\theta=6\sin\frac{\theta}{12}+\frac{6}{7}\sin\frac{7\theta}{12}+C$$
Work Step by Step
$$A=\int\cos\frac{\theta}{3}\cos\frac{\theta}{4}d\theta$$
Use Formula 69c, which states that
$$\int \cos ax\cos bx dx=\frac{\sin(a-b)x}{2(a-b)}+\frac{\sin(a+b)x}{2(a+b)}+C$$
for $a=1/3$ and $b=1/4$:
$a-b=1/3-1/4=1/12$ while $a+b=1/3+1/4=7/12$
$$A=\frac{\sin\frac{1}{12}\theta}{2\times\frac{1}{12}}+\frac{\sin\frac{7}{12}\theta}{2\times\frac{7}{12}}+C$$ $$A=\frac{\sin\frac{\theta}{12}}{\frac{1}{6}}+\frac{\sin\frac{7\theta}{12}}{\frac{7}{6}}+C$$ $$A=6\sin\frac{\theta}{12}+\frac{6}{7}\sin\frac{7\theta}{12}+C$$
