Finding ∫ 0 2 π ln ⁡ | 2 sin ⁡ ( x ) + 1 | d x This is what I have tried:Let I = ∫ 0 2 π ln ⁡ | 2 sin ⁡ ( x ) + 1 | d x = ∫ 0 2 π ln ⁡ | − 2 sin ⁡ x + 1 | d x 2 I = ∫ 0 2 π ln ⁡ | 1 − 2 sin 2 ⁡ ( x ) | d x = 4 ∫ 0 π 2 ln ⁡ | 1 − 2 sin 2 ⁡ ( x ) | d x Any ideas how to solve this?

Question

Finding             ∫              0                    2        π              ln    ⁡          |        2    sin    ⁡    (    x    )    +    1          |        d    x  This is what I have tried:Let       I    =          ∫              0                    2        π              ln    ⁡          |        2    sin    ⁡    (    x    )    +    1          |        d    x    =          ∫              0                    2        π              ln    ⁡          |        −    2    sin    ⁡    x    +    1          |        d    x        2    I    =          ∫              0                    2        π              ln    ⁡          |        1    −    2          sin      2        ⁡    (    x    )          |        d    x    =    4          ∫              0                              π          2                      ln    ⁡          |        1    −    2          sin      2        ⁡    (    x    )          |        d    x  Any ideas how to solve this?

Aidyn Mccarthy · Accepted Answer

Let I denote the integral given by                    I                            =                  ∫          0                      2            π                          log        ⁡        (                  |                2        sin        ⁡        (        x        )        −        1                  |                )                d        x                                          (1)                                  =                  ∫          0                      2            π                          log        ⁡        (                  |                2        cos        ⁡        (        x        )        −        1                  |                )                d        x            Enforcing the substitution   z  =      e          i      x       in (1), we find                    I                            =                  ∮                                    |                        z                          |                        =            1                                                log            ⁡            (                          |                        z            +                          z                              −                1                                      −            1                          |                        )                                i            z                                  d        z                                          (2)                                  =                  ∮                                    |                        z                          |                        =            1                                                log            ⁡            (                          |                                      z              2                        −            z            +            1                          |                        )            −                                          log                ⁡                (                                  |                                z                                  |                                )                                            0                                                          i            z                                  d        z            Inasmuch as I is purely real and   i  arg  ⁡  (      z    2    −  z  +  1  ) is purely imaginary, then we must have from (2)                    (3)                    I        =        Re                  (                      ∮                                          |                            z                              |                            =              1                                                          log              ⁡              (                              z                2                            −              z              +              1              )                                      i              z                                          d          z          )                    Note that             log      ⁡      (              z        2            −      z      +      1      )        z   is analytic inside and on the C path, which deforms the unit circle to eliminate branch points with small semicircular dimples. The contribution to the value of the integral from integration around dimples disappears when the radii of the semicircular dimples tend to 0. Therefore, the integral in (3) is equal to 0, and the job is done.

Find int_0^(2pi) ln |2 sin(x)+1|dx

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