Antiderivative of | sin ⁡ ( x ) | Because ∫ 0 π sin ⁡ ( x ) d x = 2 ,, then ∫ 0 16 π | sin ⁡ ( x ) | d x = 32..And Wolfram Alpha agrees to this, but when I ask for the indefinite integral ∫ | sin ⁡ ( x ) | d x ,,Wolfram gives me − cos ⁡ ( x ) s g n ( sin ⁡ ( x ) ) + c ..However, [ − cos ⁡ ( x ) s g n ( sin ⁡ ( x ) ) ] 0 16 π = 0 ,,So what's going on here? What is the antiderivative of | sin ⁡ ( x ) | ?

Question

Antiderivative of       |    sin  ⁡  (  x  )      |  Because       ∫    0          π        sin  ⁡  (  x  )        d    x  =  2  ,, then       ∫    0          16      π            |    sin  ⁡  (  x  )      |          d    x  =  32..And Wolfram Alpha agrees to this, but when I ask for the indefinite integral   ∫      |    sin  ⁡  (  x  )      |          d    x  ,,Wolfram gives me   −  cos  ⁡  (  x  )        s    g    n    (  sin  ⁡  (  x  )  )  +  c  ..However,   [  −  cos  ⁡  (  x  )        s    g    n    (  sin  ⁡  (  x  )  )      ]    0          16      π        =  0  ,,So what&#039;s going on here? What is the antiderivative of       |    sin  ⁡  (  x  )      |  ?

Mateo Carson · Accepted Answer

Step 1Let   f  (  x  )  =      ∫    0    x        |    sin  ⁡  x      |      d  x  =  (  1  −  cos  ⁡  x  )  ,  0  ≤  x  ≤  π  ..since the integrand is   π- periodic, we can extend the formula for   f  (  x  )  =  f  (  π  )  +  f  (  x  −  π  )  ,  π  ≤  x  ≤  2  πStep 2And so on. you can verify that   f  (  n  π  )  =  2  n   for all integer   n  ..in particular   f  (  16  π  )  =  32..

Savanah Boone · Accepted Answer

Step 1For   x  ∈  [  n  π  ,  (  n  +  1  )  π  ], we get                    (1)                              ∫          0                      n            π                                    |                sin        ⁡        (        t        )                  |                                  d                t        =        2        n            and                               ∫                      n            π                    x                          |                sin        ⁡        (        t        )                  |                                  d                t                            =                  ∫          0                      x            −            n            π                          sin        ⁡        (        t        )                          d                t                            (2)                                  =        1        −        cos        ⁡        (        x        −        n        π        )            Step 2Piecing (1) and (2) together yields                     (3)                              ∫          0          x                          |                sin        ⁡        (        t        )                  |                                  d                t        =        1        −        cos        ⁡        (        x        −        π                                                            ⌊                x                                  /                                π                ⌋                            ⏞                                n                )        +        2                                                            ⌊                x                                  /                                π                ⌋                            ⏞                                n