##=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##

##=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##