How to evaluate ∫ 0 π 2 d x a + cos ⁡ x using contour integration?I`m used Cauchy Residue theorem: ∫ 0 π 2 d x a + cos ⁡ x = ∫ 0 π 2 d x a + e i x + e − i x 2 = 2 ∫ 0 π 2 e i x d x 2 a e i x + e 2 i x + 1 Let z = e i x , so d z = i e i x d x . = 2 i ∫ | z | = 1 d z z 2 + 2 a z + 1 = 2 i ∫ | z | = 1 d z ( z − z 1 ) ( z − z 2 ) where the circle |z|=1My problem is how to proceed further from here?

Question

How to evaluate       ∫          0                      π        2                                d            x              a      +      cos      ⁡      x       using contour integration?I`m used Cauchy Residue theorem:                              ∫          0                                    π              2                                                                          d                        x                                a            +            cos            ⁡            x                                              =                  ∫          0                                    π              2                                                                          d                        x                                a            +                                                            e                                      i                    x                                                  +                                  e                                      −                    i                    x                                                              2                                                                        =        2                  ∫          0                                    π              2                                                                          e                              i                x                                                                 d                        x                                2            a                          e                              i                x                                      +                          e                              2                i                x                                      +            1                                            Let         z        =                  e                      i            x                          ,         so                   d                z        =        i                  e                      i            x                                             d                x        .                                          =                  2          i                          ∫                                    |                        z                          |                        =            1                                                              d                        z                                              z              2                        +            2            a            z            +            1                                                            =                  2          i                          ∫                                    |                        z                          |                        =            1                                                              d                        z                                (            z            −                          z              1                        )            (            z            −                          z              2                        )                              where the circle |z|=1My problem is how to proceed further from here?

meexeniexia17h · Accepted Answer

Your substitution gives                              ∫          1          i                                      −            2            i            d            z                                (            z            +            a                          )              2                        +            1            −                          a              2                                                          =                              [                                          −                2                i                                            1                −                                  a                  2                                                      arctan            ⁡                                          z                +                a                                            1                −                                  a                  2                                                      ]                    1          i                                                  =                  2                      1            −                          a              2                                      artanh        ⁡                                            1              −              a                                      1              +              a                                      .            It&#039;s technically a contour integral; it&#039;s just the contour isn&#039;t closed, so it&#039;s a &quot;normal&quot; integration evaluation technique, rather than one using the residue theorem.

kaltEvallwsr · Accepted Answer

I doubt you can use contour integration to handle this since you can&#039;t obtain a closed curve. Noting  cos  ⁡  x  =            1      −              tan        2            ⁡      (              x        2            )              1      +              tan        2            ⁡      (              x        2            )      one has, under   tan  ⁡  (      x    2    )  →  t                                          ∫                      0                                              π              2                                                                                          d                            x                                      a              +              cos              ⁡              x                                                                        =                              ∫                      0                                              π              2                                                                                          d                            x                                      a              +                                                1                  −                                      tan                    2                                    ⁡                  (                                      x                    2                                    )                                                  1                  +                                      tan                    2                                    ⁡                  (                                      x                    2                                    )                                                                                                      =                              ∫                      0                                              π              2                                                                          (              1              +                              tan                2                            ⁡              (                              x                2                            )              )                              d                            x                                      a              (              1              +                              tan                2                            ⁡              (                              x                2                            )              )              +              1              −                              tan                2                            ⁡              (                              x                2                            )                                                                        =                              ∫                      0                                              π              2                                                                          (              1              +                              tan                2                            ⁡              (                              x                2                            )              )                              d                            x                                      (              a              +              1              )              +              (              a              −              1              )                              tan                2                            ⁡              (                              x                2                            )              )                                                                        =                              ∫          0          1                          2                      (            a            +            1            )            +            (            a            −            1            )                          t              2                                                d                t                                          =                              2                                    a              2                        −            1                          arctan        ⁡                              (                                                              a              −              1                                      a              +              1                                                            )                          .

How to evaluate int_0^(pi) (dx)/(a+cos x) using contour integration?

Answered question

Answer & Explanation

New Questions in Trigonometry