题目

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

提问时间：2021-01-02

答案

∫ 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