Question

求∫(1+sinx/1+cosx)*e^x的不定积分

Answer 1

∫e^xdx/(1+cosx)+∫e^xsinxdx/(1+cosx)
=∫e^xdtan(x/2)+∫tan(x/2)de^x
=e^x tan(x/2) -∫tan(x/2)de^x+∫tan(x/2)de^x+C
=e^x tan(x/2) +C

追问

我不太明白能说一下吗第一步就不太明白怎么变为两个相加了？

追答

∫dx/(1+cosx)=∫dx/2cos(x/2)^2=∫sec(x/2)^2d(x/2)=∫dtan(x/2)
sinx/(1+cosx)=2sin(x/2)cos(x/2)/(2cos(x/2)^2)=tan(x/2)

∫[(1+sinx)/(1+cosx)]e^xdx
=∫e^xdx/(1+cosx)+∫[sinx/(1+cosx)]e^xdx
=∫e^xdtan(x/2)+∫tan(x/2)de^x