如何半年里系统自学高数

2024-11-20 10:26:03
推荐回答(1个)
回答1:

∫ 1/(sin^3(x)*cos^5(x)) dx
= ∫ 1/[cos^2(x)*(sin^3(x)*cos^3(x))] dx
= ∫ 1/(tan^3(x)*cos^6(x)) d(tanx)
= ∫ sec^6(x)/tan^3(x) d(tanx)
= ∫ [tan^2(x) + 1]^3/tan^3(x) d(tanx)
= ∫ [tan^6(x) + 3tan^4(x) + 3tan^2(x) + 1]/tan^3(x) d(tanx)
= ∫ [tan^3(x) + 3tan(x) + 3/tanx + 1/tan^3(x)] d(tanx)
= (1/4)tan^4(x) + (3/2)tan^2(x) + 3ln|tanx| - 1/[2tan^2(x)] + C