Question

Evaluate: $\int {\sin }^{3}x{\cos }^{3}xdx$

• A. $\frac{1}{32}\left\{\frac{3}{2}\cos 2x-\frac{1}{6}\cos 6x\right\}+C$
• B. $\frac{1}{32}\left\{-\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x\right\}+C$
• C. $\frac{1}{32}\left\{-\frac{3}{2}\cos 2x-\frac{1}{6}\cos 6x\right\}+C$
• D. None of these

Answer 1

Answer: (B)

Let $I=\int {\sin }^{3}x{\cos }^{3}xdx$. Then,
$I=\frac{1}{8}\int (2\sin x\cos x{)}^{3}dx$
$\Rightarrow I=\frac{1}{8}\int {\sin }^{3}2xdx$
$\Rightarrow I=\frac{1}{8}\int \frac{3\sin 2x-\sin 6x}{4}dx$
$\Rightarrow I=\frac{1}{32}\int (3\sin 2x-\sin 6x)dx$
$=\frac{1}{32}\left\{-\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x\right\}+C$