∫cosx^2/1+sinxcosx dx如何求解

∫ cos²x/(1 + sinxcosx) dx
= ∫ (1 + cos2x)/2 * 1/(2 + sin2x)/2 dx
= ∫ (1 + cos2x)/(2 + sin2x) dx
= (1/2)∫ dy/(2 + siny) + (1/2)∫ cosy/(2 + siny) dy
对于第一项,代u = tan(y/2),dy = 2du/(1 + u²),siny = 2u/(1 + u²)
= (1/2)∫ 1/[2 + 2u/(1 + u²)] * 2/(1 + u²) du + (1/2)∫ d(2 + siny)/(2 + siny)
= (1/2)∫ (1 + u²)/(2 + 2u² + 2u) * 2/(1 + u²) du + (1/2)ln(2 + siny)
= (1/2)∫ du/(u² + u + 1) + (1/2)ln(2 + siny)
= (1/2)∫ du/[(u + 1/2)² + 3/4] + (1/2)ln(2 + siny)
= (1/2) * (2/√3) * arctan[(u + 1/2)/(√3/2)] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2u + 1)/√3] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2tan(y/2) + 1)/√3] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2tanx + 1)/√3] + (1/2)ln(2 + sin2x) + C