Question

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

• A. $\frac{{\sin }^{4}x}{4}+\frac{{\sin }^{6}x}{6}+C$
• B. $\frac{{\sin }^{4}x}{4}-\frac{{\sin }^{6}x}{6}+C$
• C. $-\frac{3\cos 2x}{64}+\frac{\cos 6x}{192}+C$
• D. $-\frac{3\cos 2x}{64}-\frac{4\cos 6x}{192}+C$

Answer 1

Answer: (B), (C)

$-\frac{3\cos 2x}{64}+\frac{\cos 6x}{192}+C$

$\int {\sin }^{3}x{\cos }^{3}xdx=\int \cos x(1-{\sin }^{2}x){\sin }^{3}xdx$
Let $\sin x=t$
$\Rightarrow \cos xdx=dt$
$\Rightarrow$ $\int \left(t^3\right)(1-t^2)dt$
$=\int (t^3-t^5)dt=\frac{t^4}{4}-\frac{t^6}{6}+C$
$=\frac{{\sin }^{4}x}{4}-\frac{{\sin }^{6}x}{6}+C$
OR
$\int {\sin }^{3}x{\cos }^{3}xdx$
$=\int \frac{1}{8}\left(8{\sin }^{3}x{\cos }^{3}x\right)dx=\int \frac{1}{8}(2\sin x\cos x{)}^{3}dx=\int \frac{1}{8}{\sin }^{3}2xdx$
$\{\because {\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}\}$
$=\frac{1}{8}\int \frac{3\sin 2x-\sin 6x}{4}dx=\frac{1}{32}\int (3\sin 2x-\sin 6x)dx=-\frac{3\cos 2x}{64}+\frac{\cos 6x}{192}+C$