Integral ∫ d x sin 2 ⁡ ( x ) + sin ⁡ ( 2 x )

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Integral   ∫            d      x                      sin        2            ⁡      (      x      )      +      sin      ⁡      (      2      x      )

kejohananws · Accepted Answer

∫                              d            x                                              sin              2                        ⁡            (            x            )            +            sin            ⁡            (            2            x            )                                              =        ∫                                            sec              2                        ⁡            (            x            )                                              sin              2                        ⁡            (            x            )                          sec              2                        ⁡            (            x            )            +            2            sin            ⁡            (            x            )            cos            ⁡            (            x            )                          sec              2                        ⁡            (            x            )                          d        x                                          =        ∫                                            sec              2                        ⁡            (            x            )                                              tan              2                        ⁡            (            x            )            +            2            tan            ⁡            (            x            )                          d        x            Letting   u  :=  tan  ⁡  (  x  ), then   d  u  =      sec    2    ⁡  (  x  )  d  x gives                    ∫                              d            x                                              sin              2                        ⁡            (            x            )            +            sin            ⁡            (            2            x            )                                              =        ∫                              d            u                                              u              2                        +            2            u                                                            =        ∫                              d            u                                (            u            +            1                          )              2                        −            1                              Letting z:=u+1, then dz=du gives                    ∫                              d            x                                              sin              2                        ⁡            (            x            )            +            sin            ⁡            (            2            x            )                                              =        ∫                              d            z                                              z              2                        −            1                                                            =        −        artanh         (        z        )        +        C                                          =        −        artanh         (        u        +        1        )        +        C            Reersing the final substitution gives  ∫            d      x                      sin        2            ⁡      (      x      )      +      sin      ⁡      (      2      x      )        =  −  artanh   (  tan  ⁡  (  x  )  +  1  )  +  C

Integral ∫ d x </mrow> sin

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