∫(tanx-1)^2dx=∫(tan^2(x)+1-2tanx)dx
=∫(sec^2-2tanx)dx
=∫(1/cos^2(x) - 2tanx)dx
=∫(2/(1+cos2x) - 2tanx)dx
=∫(1/(1+cos2x))d(2x) -2∫tanxdx
现在求∫(1/1+cos2x)d(2x)
分部积分后可得
tan(x) + constant
∫tanxdx
=-log(cos(x)) + constant
原式 = tan(x) + 2log(cos(x)) + constant
=