xtan^2的不定积分谢谢

2025-06-27 16:51:51
推荐回答(1个)
回答1:

∫ xtan²x dx
=∫ x(sec²x-1) dx
=∫ xsec²x dx-∫ x dx
=∫ x d(tanx)-(1/2)x²
=xtanx-∫ tanx dx-(1/2)x²
=xtanx-∫ sinx/cosx dx-(1/2)x²
=xtanx+∫ 1/cosx d(cosx)-(1/2)x²
=xtanx+ln|cosx|-(1/2)x²+C