Hard integration question.

Ok heres, the question:

Evaluate: integral cos^2(5x)*sin^2(5x) dx.

I've taken about three approaches, and always get (1/80)sin(20x) + C. My book says the answer is (1/8)x - (1/160)sin(20x) + C, and I have no idea where I'm going. I've attempted substitution, half/double angle formulas, and can't seem to get near the answer.

10 points for best explanation :) thanks!

There is an easier way than integration by parts.

You need to know
sin(2θ) = 2 sin(θ) cos(θ)
sin^2(θ) = (1 - cos(2θ))/2

cos^2(5x)*sin^2(5x)

= [cos(5x)*sin(5x)]^2

= [(1/2) sin(10x)]^2

= (1/4) sin^2(10x)

= (1/8)(1 - cos(20x))

= 1/8 - cos(20x) / 8

Their result follows.

Use double-angle trig identities twice:

integral cos^2(5x)*sin^2(5x) dx
= integral [2cos(5x)*sin(5x)/2]^2 dx
= integral [sin(10x)/2]^2 dx
= integral sin^2(10x)/4 dx
= integral [1 - (1 - 2sin^2(10x))]/8 dx
= integral [1 - cos(20x)]/8 dx
= [x - (1/20)sin(20x)]/8 + C
= (1/8)x - (1/160)sin(20x) + C

∫ cos^2(5x)sin^2(5x) dx

u = 5x
du = 5 dx
dx = (1/5) du

(1/5) ∫ cos^2(u)sin^2(u) du

Now, remember that sin(2x) = 2cos(x)sin(x). Thus, cos^2(u)sin^2(u) = (1/4)sin^2(2u):

(1/20) ∫ sin^2(2u) du

Let q = 2u
dq = 2 du
du = (1/2) dq

(1/40) ∫ sin^2(q) du

Now, sin^2(q) = (1/2)(1 - cos(2q))

(1/80)∫ [1 - cos(2q)] dq

And now this is easy:

(1/80) ∫ dq - (1/80) ∫ cos(2q) dq
(1/80)q - (1/160)sin(2q)
(1/40)u - (1/160)sin(4u)
(1/8)x - (1/160)sin(20x) + C

Done!

cos^2A = (1+cos2A)/2
sin^2A = (1-cos2A) /2

cos^2 A sin^2A = (1/4) (1-cos^2 2A)

= (1/4) INT (1-cos^2 (10x)) dx
= (1/4) ( INT dx - INT cos^2 (10x) dx

= (1/4) ( x - INT ( 1- cos(20x) ) /2 dx
= (1/4) (x -(x/2) + (1/2) INT cos (20x) dx )

= (1/4) (x -(x/2) + (1/40) sin (20x) )

= (1/8) x + (1/160) sin (20x) + C

∫ cos^2(5x)*sin^2(5x) dx =
(1/4)∫ (2cos(5x)*sin(5x))^2 dx =
(1/4)∫ sin^2 (10x) dx =
(1/8)∫ 1-cos(20x) dx =
(1/8)∫ 1dx - (1/8)∫cos(20x) dx =
(1/8)x - (1/160)sin(20x) +C

Barring someone going into any more detail, what you need to do is integration by parts.