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

Answer 1

Better late than never.
I have a pile of scrap paper here with many tries at this, mostly trying to use integration by parts.
what a mess!

Then I tried this:
(sin(5x)cos(5x))^2 = (1/4)[sin(10x)]^2

we know cos 2A = 1 - 2sin^2 A

so cos20x - 1 - 2sin^2 (10x) , then
2sin^2 (10x) = 1 - cos (20x)
(1/4)[sin^2 (10x)] = 1/8 - (1/8)cos (20x)

then [integral](sin(5x)cos(5x))^2 dx
= [integral](1/8 - (1/8)cos (20x)) dx
= x/8 - (1/160)sin(20x)