what is the integral of sin^2(6x)cos^2(6x)?

Question

thanks

thanks for trying to answer my question carth... unfortunately that doesn't seem like the right answer... I took the derivative of your answer and it's not the same.

need help thanks!

Answer 1

This one requires knowing some trig identities.
∫sin^2(6x)cos^2(6x) dx

Identity: sin (2u) = 2 sin u cos u
let u = 6x:
sin 12x = 2 sin 6x cos 6x
sin^2 12x = 4 sin^2 6x cos^2 6x
1/4 sin^2 12x = sin^2 6x cos^2 6x

Back to our integral:
∫1/4 sin^2 12x dx

Identity: sin^2 u = (1 - cos (2u)) / 2
u = 12x
sin^2 12x = (1 - cos (24x)) / 2

∫1/4 (1 - cos (24x)) / 2 dx
∫1/8 - 1/4 cos (24x) / 2 dx
∫-1/8 cos (24x)
-1/8 ∫cos (24x) = 1/8 [1/24 sin 24x + c]

1/192 sin 24x + c (solution!)