Question

求不定积分∫secxdx

Answer 1

三组答案：

Answer 2

解：∫secxdx
=∫dx/cosx
=∫cosxdx/cos²x
=∫dsinx/cos²x
=∫dsinx/(1-sin²x)
=(1/2)[∫dsinx/(sinx+1)-∫dsinx/(sinx-1)]
=(1/2)(ln|sinx+1|-ln|sinx-1|)+C
=(1/2)ln|(sinx+1)/(sinx-1)|+C
(对数里分子分母都乘以sinx+1)
=(1/2)ln|(sinx+1)²/cos²x|+C
=ln|(sinx+1)/cosx|+C
=ln|tanx+secx|+C，望采纳，谢谢。

Answer 3

∫secxdx=∫1/cosxdx=∫1/cos²xdsinx
=∫1/(1-sin²x)dsinx=1/2∫1/(1+sinx)+1/(1-sinx)dsinx
=1/2ln|(1-sinx)/(1-sinx)|+C