sec x的反导数,高手请回答.

简单一句,即是求不定积分咯~
∫secx dx
=∫1/cosx dx
=∫cosx/cos²x dx
=∫d(sinx)/(1-sin²x),u=sinx
=∫du/(1-u²)
=∫du/[(1+u)(1-u)]
令1/[(1+u)(1-u)]=A/(1+u)+B/(1-u)
1=A(1-u)+B(1+u)
当u=-1,1=2A→A=1/2
当u=1,1=2B→B=1/2
∴1/[(1+u)(1-u)]=1/[2(1+u)]+1/[2(1-u)]
∴∫du/[(1+u)(1-u)]
=(1/2)∫d(1+u)/(1+u)-(1/2)∫d(1-u)/(1-u)
=(1/2)ln|1+u|-(1/2)ln|1-u|+C
=(1/2)ln|(1+sinx)/(1-sinx)|+C
=(1/2)ln|(1+sinx)²/cos²x|+C
=ln|(1+sinx)/cosx|+C
=ln|secx+tanx|+C