Answer
$$\int\sin\theta\sin2\theta\sin3\theta d\theta=\frac{\cos6\theta}{24}-\frac{\cos4\theta}{16}-\frac{\cos2\theta}{8}+C$$
Work Step by Step
$$A=\int\sin\theta\sin2\theta\sin3\theta d\theta$$
We use the identity: $$\sin mx\sin nx=\frac{1}{2}[\cos(m-n)x-\cos(m+n)x]$$
to analyze $\sin\theta\sin2\theta$ first:
$$\sin\theta\sin2\theta=\frac{1}{2}[\cos(-\theta)-\cos3\theta]$$ $$\sin\theta\sin2\theta=\frac{1}{2}[\cos\theta-\cos3\theta]$$
That means $$A=\frac{1}{2}\int(\cos\theta-\cos3\theta)\sin3\theta d\theta$$ $$A=\frac{1}{2}\Big(\int\sin3\theta\cos\theta d\theta-\int\sin3\theta\cos3\theta d\theta\Big)$$
Now use the identity: $$\sin mx\cos nx=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]$$
$$A=\frac{1}{2}\Big(\frac{1}{2}\int(\sin2\theta+\sin4\theta)d\theta-\frac{1}{2}\int(\sin0+\sin6\theta)d\theta\Big)$$ $$A=\frac{1}{4}\Big(\int(\sin2\theta+\sin4\theta)d\theta-\int\sin6\theta d\theta\Big)$$ $$A=\frac{1}{4}\Big(-\frac{\cos2\theta}{2}-\frac{\cos4\theta}{4}+\frac{\cos6\theta}{6}\Big)+C$$ $$A=\frac{\cos6\theta}{24}-\frac{\cos4\theta}{16}-\frac{\cos2\theta}{8}+C$$
