Evaluate ∫ 1 1 + 3 sin 2 ⁡ x d x (Making antiderivative continuous.)Evaluate ∫ 1 1 + 3 sin 2 ⁡ x d xI know that this has an antiderivative on R .I can use the trig. substitution t = tan ⁡ x on ( − π 2 + k π , π 2 + k π ) x = arctan ⁡ t d x = 1 1 + t 2 d tTo get ∫ 1 1 + 3 sin 2 ⁡ x d x = ∫ 1 1 + 3 t 2 t 2 + 1 ⋅ 1 1 + t 2 d t= ∫ 1 1 + ( 2 t ) 2 d t = 1 2 ∫ 1 1 + u 2 d u = 1 2 arctan ⁡ ( 2 tan ⁡ x ) + CBut this only appplies on the interval ( − π 2 + k π , π 2 + k π ), whereas the original function should have an antiderivative on R . So I have to make it continuous, but I don't know how.(I do know that we did a similar thing for ∫ | x | d x, but I lost my notes and don't know how doing this is called in english so I can't google it. We called it "gluing" antiderivatives. And I remember that we utilized limits in some fashion related to the integration constants)

Question

Evaluate   ∫      1          1      +      3              sin        2            ⁡      x        d  x (Making antiderivative continuous.)Evaluate   ∫      1          1      +      3              sin        2            ⁡      x        d  xI know that this has an antiderivative on       R  .I can use the trig. substitution   t  =  tan  ⁡  x on   (  −      π    2    +  k  π  ,      π    2    +  k  π  )  x  =  arctan  ⁡  t  d  x  =      1          1      +              t        2              d  tTo get   ∫      1          1      +      3              sin        2            ⁡      x        d  x =   ∫      1          1      +      3                        t          2                                      t            2                    +          1                      ⋅      1          1      +              t        2              d  t=   ∫      1          1      +      (      2      t              )        2              d  t  =      1    2    ∫            1              1        +                  u          2                      d    u    =      1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +  CBut this only appplies on the interval   (  −      π    2    +  k  π  ,      π    2    +  k  π  ), whereas the original function should have an antiderivative on       R  . So I have to make it continuous, but I don&#039;t know how.(I do know that we did a similar thing for   ∫      |    x      |       d  x, but I lost my notes and don&#039;t know how doing this is called in english so I can&#039;t google it. We called it &quot;gluing&quot; antiderivatives. And I remember that we utilized limits in some fashion related to the integration constants)

Dreforganzv · Accepted Answer

Step 1In each interval   (  −      π    2    +  k  π  ,      π    2    +  k  π  ) the solution will be   F  (  x  )  =      1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +      C    k  . To make it continuous for   x  ∈      R   you need that for every   k  ∈      Z  .      lim          x      →      (      k      π      +              π        2                    )        −              F  (  x  )  =      lim          x      →      (      k      π      +              π        2                    )        +              F  (  x  )that is       lim          x      →      (      k      π      +              π        2                    )        −                  1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +      C    k    =      lim          x      →      (      k      π      +              π        2                    )        +                  1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +      C          k      +      1            lim          x      →      (              π        2                    )        −                  1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +      C    k    =      lim          x      →      (      −              π        2                    )        +                  1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +      C          k      +      1            lim          t      →      +      ∞            1    2    arctan  ⁡  (  2  t  )  +      C    k    =      lim          t      →      −      ∞            1    2    arctan  ⁡  (  2  t  )  +      C          k      +      1            π    4    +      C    k    =  −      π    4    +      C          k      +      1            C    k    =      C    0    +            k      π        2  Step 2You have then   F  (  x  )  =      1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +            k      π        2    +      C    0      for   x  ∈  (  −      π    2    +  k  π  ,      π    2    +  k  π  )or equivalently   F  (  x  )  =      1    2    arctan  ⁡  (  2  tan  ⁡  x  )  +  ⌊      x    π    +      1    2    ⌋      π    2    +      C    0

Davin Fields · Accepted Answer

Step 1Actually, the answer should have been       1    2    arctan  ⁡  (  2  tan  ⁡  x  ).Now, define   F  :  [  0  ,  ∞  )  ⟶      R   as   x  ↦      {                                        1            2                    arctan          ⁡          (          2          tan          ⁡          x          )                           if           x          ∈                      [            0            ,                          π              2                        )                                                            π            4                                     if           x          =                      π            2                                                            1            2                    arctan          ⁡          (          2          tan          ⁡          x          )          +                      π            2                                     if           x          ∈                      (                          π              2                        ,                                          3                π                            2                        )                                                                          3              π                        4                                     if           x          =                                    3              π                        2                                                            1            2                    arctan          ⁡          (          2          tan          ⁡          x          )          +          π                           if           x          ∈                      (                                          3                π                            2                        ,                                          5                π                            2                        )                                                ⋮                        Step 2Finally, extend F to       R   by doing   F  (  x  )  =  −  F  (  −  x  ) is   x  &amp;lt;  0. And now F is the primitive of   x  ↦      1          1      +      3              sin        2            ⁡      x       such that   F  (  0  )  =  0.

Evaluate ∫ 1 1 + 3 sin

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