Re: [積分]

[Frobenius (i^(-i)= e^(π/2))]

          1
   ∫---------- dx
      1 + cosx

         (1 - cosx)               (1 - cosx)
 = ∫-------------------- dx = ∫----------- dx
     (1 + cosx)(1 - cosx)          (sinx)^2


 = ∫(cscx)^2 - (cscx)(cotx) dx = - cotx + cscx + C

     cosx     1         1 - cosx         1 - cosx
 = - ---- + ---- + C = ---------- = ------------------- + C
     sinx   sinx          sinx       2sin(x/2)cos(x/2)

    sin(x/2)sin(x/2)
 = ------------------ + C = tan(x/2) + C
    sin(x/2)cos(x/2)

          1                               2du            1-u^2
   ∫---------- dx ( u = tan(x/2)，dx = -------，cosx = ------- )
      1 + cosx                           1+u^2           1+u^2

         1         2du              2du              2
 = ∫----------- ------- = ∫------------------ = ∫--- du = ∫du
          1-u^2              (1+u^2) + (1-u^2)       2
     (1 + -----) (1+u^2)
          1+u^2

 = tan(x/2) + C

   ∫ 〔(√3 ) sinx + cosx 〕^2 dx

 = ∫ 4〔sinx (√3/2 ) + cosx (1/2) 〕^2 dx

 = ∫ 4〔sin(x + π/6)〕^2 dx

 = ∫ 2 2〔sin(x + π/6)〕^2 dx = 2∫ 2〔sin(x + π/6)〕^2 dx

 = 2 ∫〔1 - cos(2x + π/3)〕dx = ∫ 2 dx - ∫ cos(2x + π/3) d(2x + π/3)

 = 2x - sin(2x + π/3) + C

 = 2x - sin(2x) (1/2) - cos(2x) (√3/2 ) + C

 = 2x - (√3/2) cos(2x) - (1/2) sin(2x) + C

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