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