How to reconcile the fact that the antiderivative of sin(x) cos(x) has two possible answers?So here is the problem. ∫ sin ⁡ ( x ) cos ⁡ ( x ) d x = 1 2 sin 2 ⁡ ( x ) + c 1 = − 1 2 cos 2 ⁡ ( x ) + c 2 This fact doesn't make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren't only different by a constant)?What are some other functions that have more than one antiderivatives (not only different by a constant)?

Question

How to reconcile the fact that the antiderivative of sin(x) cos(x) has two possible answers?So here is the problem.  ∫  sin  ⁡  (  x  )  cos  ⁡  (  x  )  d  x  =      1    2        sin    2    ⁡  (  x  )  +      c    1      =  −      1    2        cos    2    ⁡  (  x  )  +      c    2  This fact doesn&#039;t make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren&#039;t only different by a constant)?What are some other functions that have more than one antiderivatives (not only different by a constant)?

Donavan Scott · Accepted Answer

Explanation:You wrote:(notice that these answers aren&#039;t only different by a constant)But they do differ by a constant. If you subtract one solution from the other, you have  (      1    2        sin    2    ⁡  (  x  )  +      c    1    )  −  (  −      1    2        cos    2    ⁡  (  x  )  +      c    2    )which is equal to       1    2    (      sin    2    ⁡  (  x  )  +      cos    2    ⁡  (  x  )  )  +      c    1    −      c    2   and using the trig identity       sin    2    ⁡  (  x  )  +      cos    2    ⁡  (  x  )  =  1, all of this is equal to just       1    2    +      c    1    −      c    2   which is a constant.

Oakey1w · Accepted Answer

Explanation:Since          sin    2    ⁡  (  x  )  +      cos    2    ⁡  (  x  )  =  1, if we let       c    1    =      c    2    −      1    2   so we&#039;d have                              1          2                          sin          2                ⁡        (        x        )        +                  c          1                                    =                  1          2                          sin          2                ⁡        (        x        )        +        (                  c          2                −                  1          2                )                                          =                  1          2                          sin          2                ⁡        (        x        )        +        (                  c          2                −                  1          2                )                                          =                  1          2                (        1        −                  cos          2                ⁡        (        x        )        )        +        (                  c          2                −                  1          2                )                                          =                  c          2                −                  1          2                          cos          2                ⁡        (        x        )