Question

Find the integrals (primitives):
$\rm \displaystyle\int \sin^{3}x.\cos^{3}x \ dx$

Answer 1

We know that

${cos}^{2} x = 1 - {sin}^{2} x$
$\int \sin^{3}x.\cos^{3}x \ dx \\ \\\int \sin^{3}x.\cos^{2}x \: cos \: x\ dx \\ \\ \\\int \sin^{3}x.(1 - sin^{2}x) \: cos \: x\ dx$
let
$sin \: x = t \\ \\ cos \: x \: dx = dt$
$\int t^{3}.(1 - t^{2}) dt \\ \\ \int t^{3}dt \: - \int t^{5} \: dt = \frac{ {t}^{4} }{4} - \frac{ {t}^{6} }{6} + C \\ \\ redo \: substitution \\ \\ \int \sin^{3}x.\cos^{3}x \ dx = \frac{ {sin}^{4}x }{4} - \frac{ {sin}^{6}x }{6} + C$
Hope it helps you.