Xtanx平方的不定积分

Question

如题所述

Answer 1

∫ x (tanx)^2 dx
=∫ x [(secx)^2 -1 ] dx
=-(1/2)x^2 + ∫ x dtanx
=-(1/2)x^2 + x.tanx - ∫ tanx dx
=-(1/2)x^2 + x.tanx - ∫ (sinx/cosx) dx
=-(1/2)x^2 + x.tanx + ∫ dcosx/cosx
=-(1/2)x^2 + x.tanx + ln|cosx| + C追问

答案错了