How do you calculate ∫ 0 5 π / 2 d x 2 + cos ⁡ x I tried all available substitutions including tangent half angle, but all these substitutions do not distinguish between π / 2 and 5 π / 2. I tried splitting up the integral into two parts, with one part from 0 to 2 π, but again this had problems as 0 and 2 π were basically "the same"

Question

How do you calculate      ∫          0              5      π              /            2                  d      x              2      +      cos      ⁡      x      I tried all available substitutions including tangent half angle, but all these substitutions do not distinguish between   π      /    2 and   5  π      /    2. I tried splitting up the integral into two parts, with one part from 0 to   2  π, but again this had problems as 0 and   2  π were basically &quot;the same&quot;

1s1zubv · Accepted Answer

Substitute (Weierstrass):  t  =  tan  ⁡      x    2      ⟹    cos  ⁡  x  =            1      −              t        2                    1      +              t        2                ,      d  x  =      2          1      +              t        2              d  t    ⟹  by the period of cosx (and also of tanx , in fact), we get      ∫    0          5      π              /            2        =      ∫    0          π                  d      x              2      +      cos      ⁡      x        +      ∫          π              2      π                  d      x              2      +      cos      ⁡      x        +      ∫          2      π              5      π              /            2                  d      x              2      +      cos      ⁡      x        =  =      ∫    0    ∞              2      d      t              (      1      +              t        2            )              (        2        +                              1            −                          t              2                                            1            +                          t              2                                      )              +      ∫          −      ∞        0              2      d      t              (      1      +              t        2            )              (        2        +                              1            −                          t              2                                            1            +                          t              2                                      )              +      ∫    0          1                  2      d      t              (      1      +              t        2            )              (        2        +                              1            −                          t              2                                            1            +                          t              2                                      )              =  =  2      ∫          −      ∞        ∞              d      t              3      +              t        2              +  2      ∫    0    1              d      t              3      +              t        2              =      2          3            (          ∫              −        ∞            ∞                                1                      3                          d        t                    1        +                              (                          t                              3                                      )                    2                      )    +      2          3            (          ∫      0      1                                1                      3                          d        t                    1        +                              (                          t                              3                                      )                    2                      )    =  =                          2                  3                    arctan      ⁡              t                  3                    |              −      ∞        ∞    +                          2                  3                    arctan      ⁡              t                  3                    |        0    1    =      2          3            (    π    +          π      6        )    =            7      π              3              3

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