Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 1

intsin^2x cos^3x dx=intsin^2x cos^2x cos xdx
Use Pythagorean trigonometric identity: sin^2x+cos^2x=1=>cos^2x=1-sin^2x
int sin^2x(1-sin^2x)cos xdx=
intsin^2x cos xdx-intsin^4x cosxdx
Make substitution: t=sin x=> dt=cos x dx
int t^2dt-intt^4 dt=t^3/3-t^5/5+C=
Return substitution to get the solution.
(sin^3x)/3-(sin^5x)/5+C