Using the identity cos^2(A)=1-sin^2(A)
transform
integral of sin^2(πx)cos^5(πx)dx
=integral of sin^2(πx)[1-sin^2(πx)]^2 cos(πx)dx
=integral of [sin^2(πx)-2sin^4(πx)+sin^6(πx)]cos(πx)dx
using the substitution u=sin(πx), du=πcos(πx)dx
=integral of [u^2-2u^4+u^6] (du/π)
=1/π[u^3/3-(2/5)u^5+u^7/7] + C
back substitute u=sin(πx)
=1/π[sin(πx)^3/3-(2/5)sin(πx)^5+sin(πx)^7/7]
or
=sin(πx)^3/3π-2sin(πx)^5/5π+sin(πx)^7/7π
