How to integrate ∫ 0 a d x x + a 2 − x 2 ?

Question

How to integrate       ∫    0    a                                                                      d            x                                                                                      x                         +                                                       a                2                                           −                                             x                2                                                        ?

drumette824ed · Accepted Answer

I  =      ∫    0    a                                                                      d            x                                                                                      x                         +                                                       a                2                                           −                                             x                2                                                        Let       x    2    +      y    2    =      a    2      ⟹    d  x  =  −      y                  a        2            −              y        2              d  yThus:  I  =      ∫    0    a                                                                      d            y                                                                                      y                         +                                                       a                2                                           −                                             y                2                                                              y                  a        2            −              y        2            Adding these together,  2  I  =      ∫    0    a                                                                      d            x                                                                                      x                         +                                                       a                2                                           −                                             x                2                                                              (    1    +          x                        a          2                −                  x          2                      )    =      ∫    0    a              d      x                      a        2            −              x        2              =      sin          −      1        ⁡      (          x      a        )        |    0    a    =      π    2  Thus,   I  =      π    4

Kendrick Hampton · Accepted Answer

Assuming a&amp;gt;0 and applying the substitutions x=az,   z  =  sin  ⁡  θ:  I  (  a  )  =      ∫          0              a                  d      x              x      +                        a          2                −                  x          2                      =      ∫          0              1                  d      z              z      +              1        −                  z          2                      =      ∫          0              π              /            2                  cos      ⁡      θ              sin      ⁡      θ      +      cos      ⁡      θ          d  θbut due to the substitution   θ  =      π    2    −  φ we also have   θ  =      π    2    −  φ, hence:  2  ⋅  I  (  a  )  =      ∫          0              π              /            2                  sin      ⁡      θ      +      cos      ⁡      θ              sin      ⁡      θ      +      cos      ⁡      θ          d  θ  =      π    2  and                               I          (          a          )          =                                                    π                4                                                         holds by symmetry.

How to integrate <msubsup> ∫ 0 a </msubsup> <mrow class="M

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