We can solve this by expressing #*sin ^2 (x)# in terms of a function in the first power.
From the equation:
# cos 2x = cos ^ 2 (x) - sin^2 (x)#
and the equation
#sin^2 (x) + cos^2 (x) = 1 #
#cos ^2 (x) = 1 - sin^2 (x)#
then. we substitute to the first equation
# cos 2x = 1 - sin^2 (x) - sin ^2(x)#
simplifying,
# sin ^2 (x) = (1 - cos 2x)/ 2 #
substitute to the original problem
# int [(1-cos 2x)/2]dx#
#(1/2)int(1 - cos 2x)dx#
#(1/2)[ intdx - intcos 2xdx#
#(1/2)[x - sin 2x*intd(2x)]#
# (1/2)[x - (sin 2x) * 2*1] #
# x/2 - sin 2x + c #
