Is there any short method to solve the integral I = ∫ sin ⁡ 3 x ⋅ sin ⁡ 5 x 2 sin ⁡ x 2 d x

Question

Is there any short method to solve the integral  I  =  ∫                              sin          ⁡          3          x                ⋅        sin        ⁡                                            5              x                        2                                      sin        ⁡                              x            2                                d  x

Kaydence Washington · Accepted Answer

sin      ⁡      3      x      sin      ⁡              5        2            x              sin      ⁡                        x          2                      =            2      sin      ⁡              x        2            cos      ⁡              x        2            (      3      −      4              sin        2            ⁡      x      )      sin      ⁡              5        2            x              sin      ⁡                        x          2                      =  =  (  sin  ⁡  2  x  +  sin  ⁡  3  x  )  (  1  +  2  cos  ⁡  2  x  )  =  sin  ⁡  2  x  +  sin  ⁡  3  x  +  sin  ⁡  4  x  +  sin  ⁡  5  x  +  sin  ⁡      x

Zion Wheeler · Accepted Answer

Using   sin  ⁡  θ  =                    e                  i          θ                    −              e                  i          θ                            2      i      , one has                                                                                    sin                ⁡                3                x                            ⋅              sin              ⁡                                                                    5                    x                                    2                                                                    sin              ⁡                                                x                  2                                                                                                      =                              1                      2            i                                                                              (                                  e                                      3                    i                    x                                                  −                                  e                                      −                    3                    i                    x                                                  )                            ⋅              (                              e                                  5                  i                  x                                      /                                    2                                            −                              e                                  −                  5                  i                  x                                      /                                    2                                            )                                                      e                                  i                  x                                      /                                    2                                            −                              e                                  −                  i                  x                                      /                                    2                                                                                                      =                              1                      2            i                                                                              (                                  e                                      3                    i                    x                                                  −                                  e                                      −                    3                    i                    x                                                  )                            ⋅              (                              e                                  3                  i                  x                                            −                              e                                  −                  2                  i                  x                                            )                                                      e                                  i                  x                                            −              1                                                                        =                              1                      2            i                          e                              5                i                x                                                                                                        (                                  e                                      6                    i                    x                                                  −                1                )                            (                              e                                  5                  i                  x                                            −              1              )                                                      e                                  i                  x                                            −              1                                                                        =                              1                      2            i                          e                              5                i                x                                                    (                  e                      6            i            x                          −        1        )        (                  e                      4            i            x                          +                  e                      3            i            x                          +                  e                      2            i            x                          +                  e                      i            x                          +        1        )                                          =                              1                      2            i                          e                              5                i                x                                                    (                  e                      10            i            x                          +                  e                      9            i            x                          +                  e                      8            i            x                          +                  e                      7            i            x                          +                  e                      6            i            x                          −                  e                      4            i            x                          −                  e                      3            i            x                          −                  e                      2            i            x                          −                  e                      i            x                          −        1        )                                          =                              1                      2            i                          (                  e                      5            i            x                          +                  e                      4            i            x                          +                  e                      3            i            x                          +                  e                      2            i            x                          +                  e                      i            x                          −                  e                      −            i            x                          −                  e                      −            2            i            x                          −                  e                      −            3            i            x                          −                  e                      −            4            i            x                          −                  e                      −            5            i            x                          )                                          =                    sin        ⁡        (        5        x        )        +        sin        ⁡        (        4        x        )        +        sin        ⁡        (        3        x        )        +        sin        ⁡        (        2        x        )        +        sin        ⁡        x        .

Is there any short method to solve the integral I = ∫ <mrow class="MJ

Answered question

Answer & Explanation

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