Indefinite integral ∫ 1 2 − cos ⁡ ( x ) d x has discontinuities. How to fix?Using the standard tangent half-angle substitution, we get ∫ 1 2 − cos ⁡ ( x ) d x = 2 3 tan − 1 ⁡ ( 3 tan ⁡ x 2 ) + CThe resulting antiderivatives are piecewise continuous functions with discontinuities at x = ( 2 k + 1 ) π. However, the integrand is continuous everywhere, so any antiderivative must be continuous everywhere (??) according to FTC. So, how do we deal with this problem? What is the correct form of the antiderivatives? Is there a form that avoids the discontinuities?

Question

Indefinite integral   ∫      1          2      −      cos      ⁡      (      x      )          d  x has discontinuities. How to fix?Using the standard tangent half-angle substitution, we get  ∫      1          2      −      cos      ⁡      (      x      )          d  x  =      2          3            tan          −      1        ⁡  (      3    tan  ⁡      x    2    )  +  CThe resulting antiderivatives are piecewise continuous functions with discontinuities at   x  =  (  2  k  +  1  )  π. However, the integrand is continuous everywhere, so any antiderivative must be continuous everywhere (??) according to FTC. So, how do we deal with this problem? What is the correct form of the antiderivatives? Is there a form that avoids the discontinuities?

Superina0xb4i · Accepted Answer

Step 1Although I&#039;m really late with the answer, here&#039;s something other people might find interesting.There is a continous elementary antiderivative,  F  (  x  )  =      2          3            (          x      2        +          tan              −        1              ⁡          (                                    (                          3                        −            1            )                    tan          ⁡                      x            2                                    1          +                      3                                tan            2                    ⁡                      x            2                              )        )  but finding it may be somewhat inconvenient. Here&#039;s the how.First of all, we carefully look at the function       tan          −      1        ⁡      (    tan    ⁡    x    )  . Quite interesting plot, isn&#039;t it?So, we see (check it!) that the floor function can be written as      f    l    o    o    r    (  x  )  =            2      x      −      1        2    −      1    π        tan          −      1        ⁡      (    tan    ⁡          (                        2          x          −          1                2            π      )        )  Next, we find the precise floor expression that eliminates the discontinuities in your antiderivative. That part of the work is already done in one of the answers,      ⌊          1      2              (              x        π            +      1      )        ⌋    =      1    π        (          x      2        −          tan              −        1              ⁡    (    tan    ⁡    (          x      2        )    )    )  Step 2Then you add this staircase function to your antiderivative, and end up with  F  (  x  )  =      2          3            [          tan              −        1              ⁡          (              3            tan      ⁡              x        2            )        +          x      2        −          tan              −        1              ⁡          (      tan      ⁡              x        2            )        ]  which is already cool if you don&#039;t mind stuff like       tan          −      1        ⁡  (  tan  ⁡  (  x  )  ). But since we can combine inverse tangent sums by using the formula       tan          −      1        ⁡  (  a  )  +      tan          −      1        ⁡  (  b  )  =      tan          −      1        ⁡      (                  a        +        b                    1        −        a        b              )  , we can go even further!Combining inverse tangents and simplifying the resulting expression, we finally get a fully continuous antiderivative F that we can use to calculate definite integrals without paying attention to the limits.

lurtzslikgtgjd · Accepted Answer

Step 1Suppose you want to it from   x  =  0 to   2  π using   tan  ⁡  (  x      /    2  )  =  t. A transformation t(x) needs to be continuous and monotonic (without max/min). But   (  x  )  =  tan  ⁡  (  x      /    2  ) is discontinuous at   x  =  π, so it is a bad substitution. A bad substitution can be made to work but breaking the domain of integration at the discontinuities.One way is   I  =      ∫          0              2      π                  d      x              2      −      cos      ⁡      x        =      ∫                  0        +                            π        −                        d      x              2      −      cos      ⁡      x        +      ∫                  π        +                    2              π        −                        d      x              2      −      cos      ⁡      x        .Step 2This way   tan  ⁡  (  x      /    2  )            |              x      =              π        −              =  +  ∞ and   tan  ⁡  (  x      /    2  )            |                      π        +              =  −  ∞  .. So you get   I  =      2          3        [  (      π    2    −  0  )  +  (  0  −            −      π        2    )  ]  =            2      π              3        ..Else, the other way isYou may directly use the domain reduction by       ∫          0              2      a        f  (  x  )  d  x  =  2      ∫          0              a        f  (  x  )  d  x  ,     i  f     f  (  2  a  −  x  )  =  f  (  x  )  .You get   I  =  2      ∫          0                      π        −                        d      x              2      −      cos      ⁡      x        .

Indefinite integral ∫ 1 2 −

Answered question

Answer & Explanation

New Questions in Calculus 1