求解不定积分一题∫((x-cosx)/(sinx+1))dx

MS在高等数学学过的方法中，对形如：∫(x/sinx)dx的积分是求不出来的

1/(1+sinx)＝1/[sin(x/2)＋cos(x/2)]^2＝(sec(x/2))^2/[1+tan(x/2)]^2
所以，
∫x/(sinx+1))dx＝∫x(sec(x/2))^2/[1+tan(x/2)]^2dx
＝2∫x/[1+tan(x/2)]^2 dtan(x/2)
＝－2∫x d[1/(1+tan(x/2))]
＝－2x/(1+tan(x/2))＋2∫ dx/(1+tan(x/2))
又－∫cosx/(sinx+1))dx＝－∫(1－tan(x/2))/(1+tan(x/2))
所以，
∫((x-cosx)/(sinx+1))dx＝－2x/(1+tan(x/2))＋∫dx＝－2x/(1+tan(x/2))＋x＋C

解题说明：利用三角函数之间的关系,可对分子分母同时乘以(1-sinx),将分母简化.
∫[(x-cosx)/(sinx+1)]dx
=∫{[(x-cosx)(1-sinx)]/[(1-sinx)(sinx+1)]}dx
=∫{[(x-cosx)(1-sinx)]/[(1-(sinx)^2]}dx
=∫[(x-xsinx-cosx+sinxcosx)/(cosx)^2]dx
=∫x(secx)^2 dx - ∫x tanx secx dx - ∫secx dx + ∫tanx dx
=∫x d(tanx) - ∫x d(secx) - ∫secx dx + ∫tanx dx
=(xtanx -∫tanx dx)-(xsecx -∫secx)dx -∫secx dx +∫tanx dx
=xtanx - xsecx + C