Evaluating ∫ d x cos ⁡ x − 1 I was wondering if my solution to the integral: ∫ d x 1 − cos ⁡ x is legit? ∫ 1 cos ⁡ x − 1 d x = ∫ − 1 2 ⋅ 1 sin 2 ⁡ ( x 2 ) = 1 2 ∫ − csc ⁡ ( 1 2 x ) d x = cot ⁡ ( 1 2 x ) + C ( cot ⁡ ( 1 2 x ) ) ′ = − csc 2 ⁡ ( 1 2 x ) ⋅ 1 2 = − 1 2 csc 2 ⁡ ( 1 2 x ) cos ⁡ x − 1 = cos ⁡ x − cos ⁡ ( 0 ) = − 2 sin ⁡ ( x + 0 2 ) sin ⁡ ( x − 0 2 ) = − 2 sin 2 ⁡ ( x 2 ) My solution is based around the fact that the derivative of cot ⁡ x is − csc 2 ⁡ x. I basically converted cos ⁡ x − 1 to cos ⁡ x − cos ⁡ 0 and from there used the cos ⁡ a − cos ⁡ b trig identity.

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Evaluating   ∫            d      x              cos      ⁡      x      −      1      I was wondering if my solution to the integral:  ∫            d      x              1      −      cos      ⁡      x      is legit?  ∫      1          cos      ⁡      x      −      1              d    x  =  ∫  −      1    2    ⋅      1                  sin        2            ⁡              (                  x          2                )              =      1    2    ∫  −  csc  ⁡      (          1      2        x    )          d    x  =                              cot          ⁡                      (                          1              2                        x            )                    +          C                                  (      cot      ⁡              (                  1          2                x        )            )        ′    =  −      csc    2    ⁡      (          1      2        x    )    ⋅      1    2    =                              −                      1            2                                csc            2                    ⁡                      (                          1              2                        x            )                                  cos  ⁡  x  −  1  =  cos  ⁡  x  −  cos  ⁡  (  0  )  =  −  2  sin  ⁡      (                  x        +        0            2        )    sin  ⁡      (                  x        −        0            2        )    =  −  2      sin    2    ⁡      (          x      2        )  My solution is based around the fact that the derivative of   cot  ⁡  x is   −      csc    2    ⁡  x. I basically converted   cos  ⁡  x  −  1 to   cos  ⁡  x  −  cos  ⁡  0 and from there used the   cos  ⁡  a  −  cos  ⁡  b trig identity.

lelestalis80d · Accepted Answer

Your logic is good, and your answer is correct. In the video, they find that    ∫            d      x              1      −      cos      ⁡      x        =  csc  ⁡  x  +  cot  ⁡  x  +  C, and you found that   ∫            d      x              1      −      cos      ⁡      x        =  cot  ⁡      x    2    +  C, and you can show that these are equivalent:                    csc        ⁡        x        +        cot        ⁡        x                            =                              1                      sin            ⁡            x                          +                              cos            ⁡            x                                sin            ⁡            x                                                            =                                          1            +            cos            ⁡            x                                sin            ⁡            x                                                            =                                          1            +            (            2                          cos              2                        ⁡                          x              2                        −            1            )                                2            sin            ⁡                          x              2                        cos            ⁡                          x              2                                                                        =                                          2                          cos              2                        ⁡                          x              2                                            2            sin            ⁡                          x              2                        cos            ⁡                          x              2                                                                        =                                          cos            ⁡                          x              2                                            sin            ⁡                          x              2                                                                        =                    cot        ⁡                  x          2                    (Note that there was no guarantee that the   +  C in both integrals would be the same, but in this case it is.)

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