tan³xdx
=∫sin³x/cos³xdx
=-∫sin²x/cos³xdcosx
=-∫(1-cos²x)/cos³xdcosx
=-∫(1/cos³x)dcosx + ∫(1/cosx)dcosx
= 1/(2cos²x) + ln|cosx| + C
= ½ sec²x + ln|cosx| + C
匿名回答于2024-05-22 16:04:16
