Question

The integral $\int {\sin }^{3}x{\cos }^{3}xdx$ is equal to

• A. $\frac{1}{32}\left[-\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x\right]+c$
• B. $\frac{1}{32}\left[-\frac{3}{2}\sin 2x+\frac{1}{6}\sin 6x\right]+c$
• C. $-\frac{1}{32}\left[-\frac{3}{2}\cos 2x+\frac{1}{6}\sin 6x\right]+c$
• D. None of the above

Answer 1

Answer: (A)

$\int {\sin }^{3}x{\cos }^{3}xdx$ $=\frac{1}{8}\int {(2\sin x\cos x)}^{3}dx$ $=\frac{1}{8}\int {\sin }^{3}2xdx$ $=\frac{1}{8}\int \left[\frac{3\sin 2x-\sin 6x}{4}\right]dx$ $[\because \sin 3x=3\sin x-4{\sin }^{3}x$ $\Rightarrow$ ${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}]$ $=\frac{1}{32}[3\int \sin 2xdx-\int \sin 6xdx]$ $=\frac{1}{32}\left[-\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x\right]+c$