Question

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

Answer 1

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

$=\int {\sin }^{3}x{\cos }^{2}x\cos xdx$

$=\int {\sin }^{3}x(1-{\sin }^{2}x)\cos xdx(\because {\sin }^{2}x+{\cos }^{2}x=1)$

$=\int ({\sin }^{3}x-{\sin }^{5}x)\cos xdx$

$=\int {\sin }^{3}x\cos xdx-\int {\sin }^{5}x\cos xdx$

Let $\sin x=t\Rightarrow \cos xdx=dt$

$\therefore \int {\sin }^{3}x{\cos }^{3}xdx=\int t^{3}dt-\int t^{5}dt$

$\Rightarrow =\frac{t^4}{4}-\frac{t^6}{6}+C$

As $t=\sin x$

$\therefore \int {\sin }^{3}x{\cos }^{3}xdx=\frac{{\sin }^{4}x}{4}-\frac{{\sin }^{6}x}{6}+C$