Question

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

Answer 1

Consider the given integral.

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

We know that

${\sin }^{2}x+{\cos }^{2}x=1$

Therefore,

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

Let $t=\sin x$

$dt=\cos xdx$

Therefore,

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

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

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

On putting the value of $t$, we get

$I=\frac{{\sin }^{4}x}{4}-\frac{{\sin }^{6}x}{6}+C$

Hence, this is the answer.