Jump discontinuity in antiderivativeI am dealing with integral I = ∫ 0 π d x sin 2 ⁡ x a 2 + sin 2 ⁡ x .I know the answer is I = π ( 1 − a 1 + a 2 ) ..However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like I = [ x − a tan − 1 ⁡ ( 1 + a 2 a tan ⁡ x ) 1 + a 2 ] x = 0 x = π .but at first sight this would yield merely π. I can see that the right step to recover the correct solution is to take tan − 1 ⁡ ( 1 + a 2 a tan ⁡ x ) | x = π = π however how do I justify this (besides that it yields the correct answer)?

Question

Jump discontinuity in antiderivativeI am dealing with integral   I  =      ∫    0          π        d  x                    sin        2            ⁡      x                      a        2            +              sin        2            ⁡      x        .I know the answer is   I  =  π  (  1  −      a          1      +              a        2              )  ..However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like   I  =            [      x      −                        a                      tan                          −              1                                ⁡                      (                                          1                +                                  a                  2                                            a                        tan            ⁡            x            )                                    1          +                      a            2                              ]              x      =      0              x      =      π      .but at first sight this would yield merely   π. I can see that the right step to recover the correct solution is to take       tan          −      1        ⁡      (                  1        +                  a          2                    a        tan    ⁡    x    )                      |                    x      =      π        =  π however how do I justify this (besides that it yields the correct answer)?

Blaine Foster · Accepted Answer

Step 1First notice that the integral exists for all   a  ∈      R  , and from symmetry  I  (  a  )  =  2      ∫    0          π              /            2                          sin        2            ⁡      x                      a        2            +              sin        2            ⁡      x        d  x    =  π  −  2      a    2        ∫    0          π              /            2                  d      x                      a        2            +              sin        2            ⁡      x      Then we have that       sin    2    ⁡  x  =      1    2    (  1  −  cos  ⁡  2  x  ) so  I  (  a  )  =  π  −  4      a    2        ∫    0          π              /            2                  d      x              2              a        2            +      1      −      cos      ⁡      2      x          =  π  +  2      a    2        ∫    0    π              d      x              cos      ⁡      x      −      2              a        2            −      1      Step 2Then from the comments here, we have that      ∫    0    π              d      x              b      +      a      cos      ⁡      x        =      π                  b        2            −              a        2            (I can show you a proof if you&#039;d like). So of course  I  (  a  )  =  π  +            2              a        2            π              (      2              a        2            +      1              )        2            −      1        =  π  +            a      π                      a        2            +      1      And from the definition of I(a), we have that   I  (  a  )  =  I  (  −  a  ) so your desired result:      1    π    I  (  a  )  =  1  −      a                  a        2            +      1

Jump discontinuity in antiderivative I am dealing with integral I = <msubsup> &#x222

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