作业帮∫dx/(1+sinxcosx)^2
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/27 16:41:54
∫dx/(1+sinxcosx)^2
∫dx/(1+sinxcosx)^2
∫dx/(1+sinxcosx)^2
令v = tanx,dx = dv/(1 + v^2),sinx = v/√(1 + v^2),cosv = 1/√(1 + v^2)
∫ 1/(1 + sinxcosx)^2 dx
= ∫ 1/[1 + v/√(1 + v^2) * 1/√(1 + v^2)]^2 * 1/(1 + v^2) dv
= ∫ (1 + v^2)^2/[(1 + v^2) + v]^2 * 1/(1 + v^2) dv
= ∫ (1 + v^2)/(v^2 + v + 1)^2 dv
= ∫ (1 + v^2)/[(v + 1/2)^2 + 3/4]^2 dv
令v + 1/2 = (√3/2)tanz,dv = (√3/2)sec^2(z) dz
tanz = (2v + 1)/√3,sinz = (2v + 1)/[2√(v^2 + v + 1)],cosz = √3/[2√(v^2 + v + 1)]
= ∫ [1 + ((√3/2)tanz - 1/2)^2]/[(3/4)tan^2(z) + 3/4]^2 * (√3/2)sec^2(z) dz
= ∫ [1 + (3/4)tan^2(z) - (√3/2)tanz + 1/4]/[(9/16)sec^4(z)] * (√3/2)sec^2(z) dz
= ∫ 8/(3√3) * [(3/4)tan^2(z) - (√3/2)tanz + 5/4] * cos^2(z) dz
= [8/(3√3)]∫ [(3/4)sin^2(z) - (√3/2)sinzcosz + (5/4)cos^2(z)] dz
= [8/(3√3)]∫ [(3/8)(1 - cos2z) - (√3/4)sin2z + (5/8)(1 + cos2z)] dz
= [8/(3√3)]∫ [1 + (1/4)cos2z - (√3/4)sin2z] dz
= [8/(3√3)][z + (1/8)sin2z + (√3/8)cos2z] + C
= [8/(3√3)][z + (1/4)sinzcosz + (√3/8)(cos^2(z) - sin^2(z))] + C
= [8/(3√3)]{arctan[(2v + 1)/√3] + (√3/16)(2v + 1)/(v^2 + v + 1) + (√3/8) * 3/[4(v^2 + v + 1)] - (√3/8) * (2v + 1)^2/[4(v^2 + v + 1)]} + C
= [8/(3√3)]{arctan[(2v + 1)/√3] + (√3/32) * 1/(v^2 + v + 1) * [2(2v + 1) + 3 - (2v + 1)^2]} + C
= [8/(3√3)]{arctan[(2v + 1)/√3] + (√3/32) * 1/(v^2 + v + 1) * 4(1 - v^2)} + C
= [8/(3√3)]{arctan[(2tanx + 1)/√3] + (√3/8)[1 - tan^2(x)]/(1 + tan^2x + tanx)} + C
= [8/(3√3)]arctan[(2tanx + 1)/√3] + (1/3)(cos^2x - sin^2x)/(cos^2x + sin^2x + sinxcosx) + C
= [8/(3√3)]arctan[(2tanx + 1)/√3] + [2cos(2x)]/[3(2 + sin(2x))] + C