Recall the identity
#cos^2(x)=1/2(1+cos(2x))#
Then, let's temporarily let #u=2x#
#cos^2(2x)=cos^2(u)=1/2(1+cos(2u))=1/2(1+cos(4x))#
Rewrite #intsinxcos^2(2x)dx:#
#1/2intsinx(1+cos(4x))=1/2int(sinx+sinxcos(4x))dx=1/2intsinxdx+1/2intsinxcos(4x)dx=-1/2cosx+1/2intsinxcos4xdx#
Recall the identity
#sinucosv=1/2(sin(u+v)+sin(u-v))#
Then,
#sinxcos4x=1/2(sin(5x)+sin(-3x))=1/2(sin(5x)-sin(3x))#
So, #1/2intsinxcos4xdx=1/4int(sin(5x)-sin(3x))=1/4(-1/5cos(5x)+1/3cos(3x))=-1/20cos(5x)+1/12cos(3x)#
Combining everything together,
#intsinxcos^2(2x)dx=-1/2cosx-1/20cos(5x)+1/12cos(3x)+C#
Adding in the constant of integration.