Question

请教高数题：secX的不定积分该怎么求

Answer 1

不定积分

Answer 2

• ∫dx/cosx

  =∫dx·{1+[tan(x/2)]²}/{1-[tan(x/2)]²}  （万能公式）

  =∫dx/(1+t²)/(1-t²)                           (令tan(x/2)=t

  =∫2dt/(1+t²)·(1+t²)/(1-t²)               (dx=½dt/(1+t²)

  =∫2dt/(1-t²)

  =∫2dt/[(1+t)(1-t)

  =∫[1/(1-t)-1/(t+1)]dt

  =∫dt/(1-t)-∫dt/(1+t)

  =-ln|1-t|-ln|1+t|+C

  =-ln|1-t²|+C
  =-ln|1-[tan(x/2)]²|+C
  

Answer 3