How do you find the integral of ##sin^2(3x)dx##?

##=1/2x - 1/12sin6x + C##

##int \ sin^2 (3x) \ dx##

small book keeping gesture is to make the sub ##u = 3x, du = 3 dx##

##1/3int \ sin^2 (u) \ du##

then we use the cosine double angle formulae

##cos 2A = 1 - 2 sin^2 A##

so ## sin^2 A = (1 - cos 2A)/2##

##=1/6int \ 1 - cos 2u \ du##

##=1/6( u - 1/2sin 2u ) + C##

##=1/6( 3x - 1/2sin( 2*3x) ) + C##

##=1/2x - 1/12sin6x + C##