\int{[sin(9x)cos(9x)]} \, dx

Question

`\int{[sin(9x)cos(9x)]} \, dx`

Answer 1

Recall the double angle identity,

sin(2x) = 2 sin(x) cos(x)

Then we can write

sin(9x) cos(9x) = 1/2 sin(2 • 9x) = 1/2 sin(18x)

Then

∫ sin(9x) cos(9x) dx

= 1/2 ∫ sin(18x) dx

= -1/2 • 1/18 cos(18x) + C

= -1/36 cos(18x) + C

though you could continue with another double angle identity,

cos(2x) = cos²(x) - sin²(x)

to rewrite the antiderivative as

= -1/36 (cos²(9x) - sin²(9x)) + C

= 1/36 (sin²(9x) - cos²(9x)) + C