Where is the absolute value when computing antiderivatives?Here is a typical second-semester single-variable calculus question: ∫ 1 1 − x 2 d xStudents are probably taught to just memorize the result of this since the derivative of arcsin(x) is taught as a rule to memorize. However, if we were to actually try and find an antiderivative, we might let x = sin ⁡ θ ⟹ d x = cos ⁡ θ d θso the integral may be rewritten as ∫ cos ⁡ θ 1 − sin 2 ⁡ θ d θ = ∫ cos ⁡ θ cos 2 ⁡ θ d θAt this point, students then simplify the denominator to just cos ⁡ θ, which boils the integral down to ∫ 1 d θ = θ + C = arcsin ⁡ x + Cwhich is the correct antiderivative. However, by definition, x 2 = | x | , implying that the integral above should really be simplified to ∫ cos ⁡ θ | cos ⁡ θ | d θ = ∫ ± 1 d θ depending on the interval for θ. At this point, it looks like the answer that we will eventually arrive at is different from what we know the correct answer to be.

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Where is the absolute value when computing antiderivatives?Here is a typical second-semester single-variable calculus question:  ∫      1          1      −              x        2                d  xStudents are probably taught to just memorize the result of this since the derivative of arcsin(x) is taught as a rule to memorize. However, if we were to actually try and find an antiderivative, we might let   x  =  sin  ⁡  θ      ⟹      d  x  =  cos  ⁡  θ    d  θso the integral may be rewritten as   ∫            cos      ⁡      θ              1      −              sin        2            ⁡      θ          d  θ  =  ∫            cos      ⁡      θ                      cos        2            ⁡      θ          d  θAt this point, students then simplify the denominator to just   cos  ⁡  θ, which boils the integral down to  ∫  1    d  θ  =  θ  +  C  =  arcsin  ⁡  x  +  Cwhich is the correct antiderivative. However, by definition,             x      2        =      |    x      |  , implying that the integral above should really be simplified to   ∫            cos      ⁡      θ                      |            cos      ⁡      θ              |                d  θ  =  ∫  ±  1    d  θ depending on the interval for   θ. At this point, it looks like the answer that we will eventually arrive at is different from what we know the correct answer to be.

Aiden Norman · Accepted Answer

Step 1This is a good question. Note that the domain for the arcsine function,   arcsin  ⁡  (  x  )  =      ∫    0    x        1          1      −              t        2                d  t, is       |    x      |    ≤  1.Hence, upon enforcing the substitution   t  =  sin  ⁡  (  θ  ), then for       |    θ      |    ≤  π      /    2,   cos  ⁡  (  θ  )  ≥  0.Step 2We could have chosen alternatively   θ  ∈  (  −  π      /    2  +  n  π  ,  π      /    2  +  n  π  ) from which we see that   sgn      (    cos    ⁡    (    θ    )    )    =      {                            1                          ,          n                              even                                                          −          1                          ,          n                              odd                        For odd values of n, the substitution   x  =  sin  ⁡  (  θ  ) yields                    ∫                  1                      1            −                          x              2                                              d        x                            =        −        θ        +        C                                                        =        −        arcsin        ⁡        (        x        )        +        C                                                        =        n        π        −        arcsin        ⁡        (        x        )        +                  C          ′                                                                =        arcsin        ⁡        (        x        )        +                  C          ′                    And hence, the antiderivative   ∫      1          1      −              x        2                d  x  =  arcsin  ⁡  (  x  )  +  C is valid regardless of whether n is even or odd.

Cory Patrick · Accepted Answer

Explanation:By the FTC, the answer is   F  (  x  )  =      ∫    0    x        1          1      −              t        2                d  t  +  C for       |    x      |    &amp;lt;  1.. But can we write F(x) in terms of functions that we understand better? Yes, in this case, by using the basic substitution theorem, which is usually lost in the haze by teaching assistants, textbook writers, and too many professors as well. The real answer to your question lies in understanding this theorem thoroughly.

Where is the absolute value when computing antiderivatives? Here is a typical second-semester singl

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