求不定积分∫secx dx似乎不定积分很难·有没有什么诀窍

∫secx dx=∫(dx)/cosx=∫(cosx/cos²x)dx
=∫(d sinx)/(1-sin²x)
=(1/2)ln│(1+sinx)/(1-sinx)│+C
=(1/2)ln(1+sinx)²/(1-sin²x)+C
=(1/2)ln[(1+sinx)/cosx]²+C
=ln│secx+tanx│+C
详细的:
∫secxdx
=∫sec²x/secxdx
=∫cosx/cos²xdx
=∫1/cos²xdsinx
=∫1/(1-sin²x)dsinx
=-∫1/(sinx+1)(sinx-1)dsinx
=-∫[1/(sinx-1)-1/(sinx+1)]/2dsinx
=-[∫1/(sinx-1)dsinx-∫1/(sinx+1)dsinx]/2
=[∫1/(sinx+1)d(sinx+1)-∫1/(sinx-1)d(sinx-1)]/2
=(ln|sinx+1|-ln|sinx-1|)/2+C
=ln√|(sinx+1)/(sinx-1)|+C
=ln√|(sinx+1)²/(sinx+1)(sinx-1)|+C
=ln√|(sinx+1)²/(sin²x-1)|+C
=ln√|-(sinx+1)²/cos²x|+C
=ln|(sinx+1)/cosx|+C
=ln|tanx+1/cosx|+C
=ln|secx+tanx|+C