∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx

桃艳杏娇环溪啄岸2022-10-04 11:39:542条回答

∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx
就这么三道,

已提交,审核后显示!提交回复

共2条回复
henand 共回答了12个问题 | 采纳率91.7%
∫ cosx•cos3x dx
= (1/2)∫ [cos(x + 3x) + cos(x - 3x)] dx
= (1/2)∫ cos4x dx + (1/2)∫ cos2x dx
= (1/2)(1/4)sin4x + (1/2)(1/2)sin2x + C
= (1/8)sin4x + (1/4)sin2x + C
_________________________________
∫ tan³t•sect dt
= ∫ tan²t d(sect)
= ∫ (sec²t - 1) d(sect)
= (1/3)sec³t - sect + C
__________________________
∫ sec²x/(4 + tan²x) dx
= ∫ d(tanx)/(4 + tan²x)
= (1/2)arctan[(tanx)/2] + C
1年前
harryabc12 共回答了150个问题 | 采纳率
1、∫cosx cos3x dx
用积化和差
∫cosx cos3x dx=1/2∫(cos2x+cos4x) dx=1/4sin2x+1/8sin4x+C
2、注意(secx)'=secxtanx
∫ (tant)^3 sect dt
=∫ (tant)^2d(sect)
=∫ [(sect)^2-1]d(sect)
=1/3(sect...
1年前

相关推荐