Question

这题的不定积分怎么求？

Answer 1

∫(sinx+cosx)dx/[(cosx)^2(sinx-cosx)]

=∫-2(1+sin2x)dx/[(1+cos2x)(cos2x)]
=-2∫dx/[(1+cos2x)cos2x]-2∫sin2xdx/[(cos2x)(1+cos2x)]

∫sin2xdx/[(1+cos2x)cos2x]
=-∫dcos2x/cos2x+∫dcos2x/(1+cos2x)
=ln[(1+cos2x)/cos2x]+C

∫dx/[(1+cos2x)(cos2x)]
=∫dx/cos2x-∫dx/(1+cos2x)
=∫dx/cos2x-(1/2)∫dx/(cosx)^2
=(1/2)∫cos2xd(2x)/[(1+sin2x)(1-sin2x)]-(1/2)tanx
=(1/2)ln[(1+sin2x)/(1-sin2x)]-(1/2)tanx