求不定积分∫secxdx

2024-11-23 12:50:14
推荐回答(3个)
回答1:

三组答案:

回答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。

回答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