求不定积分∫secx dx似乎不定积分很难·有没有什么诀窍

1个回答

  • ∫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