Re: [積分] 證明題!

[LuisSantos (^______^)]

※ 引述《tarradiddle (他..他...他牛的!)》之銘言：
: 請問有沒有人會證
: ∫sec(x)dx = ln|sec(x)+tan(x)| + c
: 謝謝



   ∫sec(x) dx

        1
 = ∫------- dx
      cos(x)

        cos(x)
 = ∫------------ dx
      (cos(x))^2

        cos(x)
 = ∫---------------- dx
      1 - (sin(x))^2

              cos(x)
 = ∫-------------------------- dx
      (1 + sin(x))(1 - sin(x))

               1         1              1
 = ∫(cos(x))(---)(------------ + ------------) dx
               2    1 + sin(x)     1 - sin(x)

     1         cos(x)         cos(x)
 = (---)(∫------------ + ------------ dx)
     2      1 + sin(x)     1 - sin(x)

     1
 = (---)(ln|1 + sin(x)| - ln|1 - sin(x)|) + c
     2

     1     | 1 + sin(x) |
 = (---)(ln|------------|) + c
     2     | 1 - sin(x) |

     1     | (1 + sin(x))(1 + sin(x)) |
 = (---)(ln|--------------------------|) + c
     2     | (1 - sin(x))(1 + sin(x)) |

     1     | (1 + sin(x))^2 |
 = (---)(ln|----------------|) + c
     2     | 1 - (sin(x))^2 |

     1     | (1+sin(x))^2  |
 = (---)(ln|---------------|) + c
     2     | (cos(x))^2    |

     1     |  1 + sin(x)    |
 = (---)(ln|(------------)^2|) + c
     2     |   cos(x)       |

     1        | 1 + sin(x) |
 = (---)(2)(ln|------------|) + c
     2        |   cos(x)   |


 = ln|sec(x) + tan(x)| + c

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