∫sinx/(1-sinx)dx
1个回答

参考以下

∫sinx/(1+sinx)dx

=∫(1+sinx-1)/(1+sinx)dx

=∫[1-1/(1+sinx)]dx

=∫dx-∫dx/(1+sinx)

=x-∫dx/[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]

=x-∫dx/[sin(x/2)+cos(x/2)]²

=x-∫sec²(x/2)/[tan(x/2)+1]²dx

=x-∫d[tan(x/2)]/[tan(x/2)+1]²dx

=x-∫d[tan(x/2)+1]/[tan(x/2)+1]²dx

=x+1/[tan(x/2)+1]+C (C是积分常数)