Verified answer

∫sin(2x)cos(2x)dx = (1/2)∫sin(4x)dx = (-1/8)cos(4x) + C

If you do it another way you may get the answer

∫sin(2x)cos(2x)dx = (-1/4) [cos (2x)]^2 + c

This looks different but actually they are both correct, because

(-1/8)[cos(4x)] + C = (-1/8)[2 cos (2x)^2 - 1] + C

= (-1/4) cos (2x)]^2 + (1/8 + C)

They are the same because c = 1/8 + C

If you do it a third way you may get the answer

∫sin(2x)cos(2x)dx = (1/4) [sin(2x)]^2 + k

This is also correct because

cos(4x) = 1 - 2[sin(2x)]^2

There are (at least) 3 possible correct answers

Regards - Ian

u = sin(2x)

du = cos(2x) (2) dx

So, ∫sin(2x)cos(2x) dx = ∫u (1/2) du = (1/4)u^2 + c = (1/4)sin^2(2x) + c

sin(t) * cos(t) = (1/2) * sin(2t)

int(sin(2x) * cos(2x) * dx) =

int((1/2) * sin(4x) * dx) =

(1/2) * int(sin(4x) * dx)

u = 4x

du = 4 * dx

(1/2) * int(sin(u) * (1/4) * du) =>

(1/8) * int(sin(u) * du) =>

(1/8) * (-cos(u)) + C =>

(-1/8) * cos(4x) + C