Evaluate an integral using half-angle identities? Please help! :)

evaluate the integral from 2 to 12 of
(sin 7x)^2 * (cos 7x)^2 dx

I know you should use the half angle identities where (sin x)^2 = 1/2 (1- cos 2x)
and (cos x)^2 = 1/2 (1+cos 2x).

I have tried this many many times and cannot get the right answer! It is for online homework and I don't know the answer.... Please help! The homework is due 2 hours!

Points for best answer!! :)

Yin comes to the rescue!

Do the indefinite integral first

∫ sin²(7x)cos²(7x) dx

sin²(7x) = [1 - cos(14x)]/2
cos²(7x) = [1 + cos(14x)]/2

∫ sin²(7x)cos²(7x) dx = ¼∫ 1 - cos²(14x) dx = ¼ ∫ sin²(14x) dx = ⅛ ∫ 1 - cos(28x) dx = ⅛[x - sin(28x)/28] + C

Where sin²(14x) = [1 - cos(28x)]/2

Now do the definite integral

12
∫ sin²(7x)cos²(7x) dx = ⅛[x - sin(28x)/28] (from 2 to 12) ≈ 1.247 three decimal places
2

Yin

1/2(1- cos14x) *1/2 (1+cos 14x)dx which simplifies to 1/4(1-(cos14x)^2)dx
use the half-angle identity again to get 1/4(1-1/2 (1+cos 28x))dx = 1/4-1/8cos 28x dx
find the anti-derivative which is 1/4 x -1/224sin28x
your answer should be 1/4 *12 -1/224sin28(12) - 1/4 *2 -1/224sin28(2)

(sin7x)^2 * (cos7x)^2 = x
1/2(1- cos 2x) * 1/2(1+ cos 2x) = x
1/4(1-cos^2x) = x
sin^2x/4 - x = 0
(sin x - 2) * (sin x + 2) =0

sin x = 2 and sin x = -2
(impossible) (impossible)

Final Answer is "No Solution".