Integral of a trig function divided by the square root of a polynomial: ∫ a b sin ⁡ x ( x − a ) ( b − x ) d x

Question

Integral of a trig function divided by the square root of a polynomial:       ∫    a    b              sin      ⁡      x              (      x      −      a      )      (      b      −      x      )        d  x

xxsailojaixxv5 · Accepted Answer

Rewrite as                              ∫          0                      b            −            a                          d        x                              sin            ⁡                          (              x              +              a              )                                            x            (            b            −            a            −            x            )                                              =                  ∫          0          1                d        u                              sin            ⁡                          [              (              b              −              a              )              u              +              a              ]                                            u            (            1            −            u            )                                                            =        cos        ⁡                  a                          ∫          0          1                d        u                              sin            ⁡                          [              (              b              −              a              )              u              ]                                            u            (            1            −            u            )                          +        sin        ⁡                  a                          ∫          0          1                d        u                              cos            ⁡                          [              (              b              −              a              )              u              ]                                            u            (            1            −            u            )                                                            =        2        cos        ⁡                  a                          ∫          0          1                d        u                              sin            ⁡                          [              (              b              −              a              )                              u                2                            ]                                            1            −                          u              2                                      +        2        sin        ⁡                  a                          ∫          0          1                d        u                              cos            ⁡                          [              (              b              −              a              )                              u                2                            ]                                            1            −                          u              2                                          Now, consider                              ∫          0          1                d        u                (        1        −                  u          2                          )                      −            1                          /                        2                                    e                      i            (            b            −            a            )                          u              2                                                          =                  ∫          0                      π                          /                        2                          d        t                          e                      i            (            b            −            a            )                          sin              2                        ⁡                          t                                                                        =                  e                      i            (            b            −            a            )                          /                        2                                    ∫          0                      π                          /                        2                          d        t                          e                      −            i            [            (            b            −            a            )                          /                        2            ]            cos            ⁡                          2              t                                                                        =                  1          2                          e                      i            (            b            −            a            )                          /                        2                                    ∫          0                      π                          d        t                          e                      −            i            [            (            b            −            a            )                          /                        2            ]            cos            ⁡                          t                                                                        =                  π          2                          e                      i            (            b            −            a            )                          /                        2                                    J          0                          (                                    b              −              a                        2                    )                    Putting this altogether, the integral is  π      J    0        (                  b        −        a            2        )        [    cos    ⁡          a        sin    ⁡                  (                              b            −            a                    2                )              +    sin    ⁡          a        cos    ⁡                  (                              b            −            a                    2                )              ]  or      ∫    a    b    d  x            sin      ⁡              x                    (      x      −      a      )      (      b      −      x      )        =  π      J    0        (                  b        −        a            2        )    sin  ⁡            (                        a          +          b                2            )

Integral of a trig function divided by the square root of a polynomial: <msubsup> ∫<

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