Question

Find the integral of the function: ${\sin }^{3}x{\cos }^{3}x$

Answer 1

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

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

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

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

Let $\cos x=t$
Differentiating w.r.t. $x$

$\Rightarrow \sin x=\frac{dt}{dx}\Rightarrow \sin xdx=-dt$

Thus,
$\int (1-{\cos }^{2}x){\cos }^{3}x.\sin xdx$

$=-\int (1-t^2)t^{3}dt$

$=\int (t^5-t^3)dt$

$=\int t^{5}dt-\int t^{3}dt$

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

$=\frac{{\cos }^{6}x}{6}-\frac{{\cos }^4}{4}+C$ $[\because t=\cos x]$

Where $C$ is constant of integration.