求积分：∫dx/sin2x+2sinx

1/[sin2x+2sinx]
=1/[2sinxcosx+2sinx]
=1/[2sinx(1+cosx)]（上下都乘以sinx）
=sinx/[2sinx*sinx*(1+cosx)]

所以
∫dx/sin2x+2sinx
=1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx
=-1/2∫1/[(1-(cosx)^2)(1+cosx)]dcosx(凑微分法，记cosx=t)
=-1/2∫1/[(1-t^2)(1+t)]dt
=-1/2{-1/4*ln(t-1)-1/2*1/(1+t)+1/4*ln(1+t)}+C
=1/8*(ln(cosx-1)+ln(cosx-1)*cosx+2-ln(1+cosx)-ln(1+cosx)*cosx)/(1+cosx)+C

∫dx/sin2x+2sinx
=∫dx/2sinx(cosx+1)
=∫dx/8sin(x/2)cos(x/2){cox(x/2)}^2
=1/4∫1/sin(x/2)cos(x/2)dtan(x/2)
=1/4∫(cos(x/2)/sin(x/2)+sin(x/2)/cos(x/2)dtan(x/2)
=1/4∫1/tan(x/2)dtan(x/2)+1/4∫tan(x/2)dtan(x/2)
=1/4ln绝对值tan(x/2)+1/8{tan(x/2)}^2+C

let t = tan x/2
x = 2 tan-1 (t)
dx = 2/1 + t^2 dt

因为 t = tan x/2, sin x = 2t/1+ t^2, cos x = 1-t^2/1+ t^2
把积分拆开
= ∫dx/2sinxcosx+2sinx
= 1/2∫dx/sin x cos x + sin x
代入
= 1/2 ∫dx/ (2t/1+t^2) * ( 1-t^2/1+t^2) + (2t/1+ t^2)
dx 换成 dt，
= 1/2 ∫1 / (2t/1+t