∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx就这么三道,
问题描述：
∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx
就这么三道,
答
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)^3-sect+C
3、∫(secx)^2/(4+tan^2) dx
=∫1/(4+(tanx)^2) d(tanx)
=1/2arctan[(tanx)/2]+C
答
∫ 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
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∫ tan³t•sect dt
= ∫ tan²t d(sect)
= ∫ (sec²t - 1) d(sect)
= (1/3)sec³t - sect + C
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∫ sec²x/(4 + tan²x) dx
= ∫ d(tanx)/(4 + tan²x)
= (1/2)arctan[(tanx)/2] + C