Let u = 1 + sin(x)cos(x) = 1 + (1/2)sin(2x)
and du = cos(2x) dx → dx = du/cos(2x)
So ∫ cos(2x)/(1+sin(x)cos(x)) dx
= ∫ 1/u du
= ln|u| + C
= ln| 1 + sin(x)cos(x) | + C
or = ln| sin(2x) + 2 | + Cand du = cos(2x) dxwhy?1/2sin2x DX1/2怎么消失了呢？du = d[1 + (1/2)sin(2x)] = 0 + 1/2 * cos(2x) * (2x)' = 1/2 * cos(2x) * 2 = cos(2x)Do you understand? If not,plz ask me again.