## University Calculus: Early Transcendentals (3rd Edition)

$$\int\sin^3\theta\cos2\theta d\theta=\frac{1}{2}\cos\theta-\frac{1}{8}\cos3\theta+\frac{1}{40}\cos5\theta+C$$
$$A=\int\sin^3\theta\cos2\theta d\theta$$ Use the identity: $$\sin^3\theta=\frac{3\sin\theta-\sin3\theta}{4}$$ we have $$A=\int\frac{3\sin\theta-\sin3\theta}{4}\cos2\theta d\theta$$ $$A=\frac{3}{4}\int\sin\theta\cos2\theta d\theta-\frac{1}{4}\int\sin3\theta\cos2\theta d\theta$$ Use the identity: $$\sin mx\cos nx=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]$$ we have $$A=\frac{3}{4}\int\frac{1}{2}[\sin(-\theta)+\sin3\theta]d\theta-\frac{1}{4}\int\frac{1}{2}[\sin\theta+\sin5\theta]d\theta$$ $$A=\frac{3}{8}\int(-\sin\theta+\sin3\theta)d\theta-\frac{1}{8}\int(\sin\theta+\sin5\theta)d\theta$$ $$A=\frac{3}{8}(\cos\theta-\frac{1}{3}\cos3\theta)-\frac{1}{8}(-\cos\theta-\frac{1}{5}\cos5\theta)+C$$ $$A=\frac{3}{8}\cos\theta-\frac{1}{8}\cos3\theta+\frac{1}{8}\cos\theta+\frac{1}{40}\cos5\theta+C$$ $$A=\frac{1}{2}\cos\theta-\frac{1}{8}\cos3\theta+\frac{1}{40}\cos5\theta+C$$