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

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/05/30 10:43:21

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

∫ 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

∫cosx cos3x dx ∫tan^3t sect dt ∫(sec^2x)/4+tan^2 dx ∫sec^4x dx ∫sec^2x tan^2x dx ∫(tan^(5)(x)*sec^(4)(x))dx ∫ ( tan^2 x + tan^4 x )dx 求不定积分∫(tan^2x+tan^4x)dx ∫ (1 + cos ^ 2 + tan ^2 ) sec^2 dx 求积分∫（sec^2x/2+tan^2x）dx ∫tan^2xdx=∫(sec^2x-1)dx 求不定积分 ∫ tan^2 x dx ∫ sec^2 x dx ②∫▒tan^3⁡x sec⁡〖x dx〗请解释高数例题：1、∫tan ^2 x sec xdx 2、∫1/x^2+4 dx 3、∫tanx dsec^(n-2)