∫ e − 3 t c o s ( 2 − 3 t ) d tI have been asked to evaluate that using complex exponential/euler's method. I have done many similar questions but all of them had something like (cos3x), sin(5t) etc. This is the first time I have come across a question where its of the format (cos a+bx) and cannot understand how to deal with the extra term. I have been stuck for a while now and have thus decided to ask the community for help, please help me out.

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∫      e          −      3      t        c  o  s  (  2  −      3    t  )  d  tI have been asked to evaluate that using complex exponential/euler&#039;s method. I have done many similar questions but all of them had something like (cos3x), sin(5t) etc. This is the first time I have come across a question where its of the format (cos a+bx) and cannot understand how to deal with the extra term. I have been stuck for a while now and have thus decided to ask the community for help, please help me out.

luluna81mxmbk · Accepted Answer

It&#039;s just like a complex constant thats all  ∫      e          −      3      t        c  o  s  (  2  −      3    t  )  d  t  ∫            e              −        3        t              2    (      e          +      i      (      2      −              3            t      )        +      e          −      i      (      2      −              3            t      )      )        )  d  t  ∫            e              −        3        t              2        e          +      i      (      2      −              3            t      )        d  t  +  ∫            e              −        3        t              2        e          −      i      (      2      −              3            t      )        d  t      e          2      i        ∫            e              −        3        t              2        e          −      i              3            t        d  t  +      e          −      2      i        ∫            e              −        3        t              2        e          i              3            t        d  t            e              2        i              2    ∫            e              −        t        (        3        +        i                  3                )              d  t  +            e              −        2        i              2    ∫            e              −        t        (        3        −        i                  3                )              d  t  )Then integrate ...

Alice Chen · Accepted Answer

Consider the more general case  I  =  ∫      e          a      t        cos  ⁡  (  b  +  c  t  )  d  tand start with  b  +  c  t  =  x    ⟹    t  =            x      −      b        c      ⟹    d  t  =            d      x        c  which makes  I  =            e              −                              a            b                    c                      c    ∫      e                  a        c            x        cos  ⁡  (  x  )    d  xAt this point, you could consider  J  =            e              −                              a            b                    c                      c    ∫      e                  a        c            x        sin  ⁡  (  x  )    d  xwhich makes  I  +  i  J  =            e              −                              a            b                    c                      c    ∫      e          (              a        c            +      i      )      x          d  x  =            e              −                              a            b                    c                      c        c          a      +      i      c            e                            x          (          a          +          i          c          )                c              I  −  i  J  =            e              −                              a            b                    c                      c    ∫      e          (              a        c            −      i      )      x          d  x  =            e              −                              a            b                    c                      c        c          a      −      i      c            e                            x          (          a          −          i          c          )                c            and combine them to get after simplifications  I  =            a      cos      ⁡      (      x      )      +      c      sin      ⁡      (      x      )                      a        2            +              c        2                    e                            a          (          x          −          b          )                c              J  =            a      sin      ⁡      (      x      )      −      c      cos      ⁡      (      x      )                      a        2            +              c        2                    e                            a          (          x          −          b          )                c            Back to   t, this would give  I  =  ∫      e          a      t        cos  ⁡  (  b  +  c  t  )    d  t  =            a      cos      ⁡      (      b      +      c      t      )      +      c      sin      ⁡      (      b      +      c      t      )                      a        2            +              c        2                    e          a      t        J  =  ∫      e          a      t        sin  ⁡  (  b  +  c  t  )    d  t  =            a      sin      ⁡      (      b      +      c      t      )      −      c      cos      ⁡      (      b      +      c      t      )                      a        2            +              c        2                    e          a      t      EditWe could even make it more complex and using exactly the same procedure get  I  =  ∫      e          a      t      +      d        cos  ⁡  (  b  +  c  t  )    d  t  =            a      cos      ⁡      (      b      +      c      t      )      +      c      sin      ⁡      (      b      +      c      t      )                      a        2            +              c        2                    e          a      t      +      d        J  =  ∫      e          a      t      +      d        sin  ⁡  (  b  +  c  t  )    d  t  =            a      sin      ⁡      (      b      +      c      t      )      −      c      cos      ⁡      (      b      +      c      t      )                      a        2            +              c        2                    e          a      t      +      d

int e^(−3t)cos(2−sqrt(3t))dt using Complex exponential/Euler's Method to integrate, bizarre case

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