Question

求不定积分∫sinxdx/(sinx+1)

Answer 1

∫ sinx/(1+sinx) dx
=∫ (1+sinx-1)/(1+sinx)
=∫1dx-∫ 1/(1+cos(x-π/2)) dx
=x-∫1/(2cos²(x/2-π/4)) dx
=x-(1/2)∫sec²(x/2-π/4) dx
=x-tan(x/2-π/4)+C
=x-[tan(x/2)-1]/[tan(x/2)+1]+C
=x-[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)]+C
=x+[cos²(x/2)-sin²(x/2)]/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]+C
=x+cosx/(1+sinx)+C
C为任意常数

Answer 2

∫sinxdx/(sinx+1)=∫(1-1/(1+sinx)dx=∫(1-(1-sinx)/cos^2x)dx
=x-∫sec^2xdx-∫1/cos^2xdcosx
=x-tanx+1/cosx+C（常数)

追问

=x-∫sec^2xdx-∫1/cos^2xdcosx
怎么到=x-tanx+1/cosx+C，第三项不对吧

追答

第三项怎么不对？