Finding an antiderivativeI need to find the following: ∫ 1 − x 1 + x ⋅ 1 x d xFirstly, it has to be that x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ]. From this, it is implied that 1 + x &gt; 0. Also, it holds that 1 − x ≥ 0.What I did is that I multiplied both numerator and denominator by 1 + x . Thus, we have: ∫ 1 − x 2 1 + x ⋅ 1 x d x = x = sin ⁡ t ∫ cos ⁡ t 1 + sin ⁡ t ⋅ 1 sin ⁡ t ⋅ cos ⁡ t d t = ∫ cos 2 ⁡ t ( 1 + sin ⁡ t ) sin ⁡ t d t = ∫ 1 − sin 2 ⁡ t ( 1 + sin ⁡ t ) sin ⁡ t d t = ∫ ( 1 − sin ⁡ t ) ( 1 + sin ⁡ t ) ( 1 + sin ⁡ t ) sin ⁡ t d t = ∫ 1 sin ⁡ t − 1 d t . Thus, the final answer is: log ⁡ | tan ⁡ ( arcsin ⁡ x 2 ) | − arcsin ⁡ x + C ,,since ∫ 1 sin ⁡ t d t = log ⁡ | tan ⁡ ( t 2 ) | + c ..The answer given in the textbook is: log ⁡ | 1 − x − 1 + x 1 − x + 1 + x | + 2 arctan ⁡ ( 1 − x 1 + x ) + c o n s t a n tI have derived that the arguments in the logarithms are the same, but it seems difficult to connect arc sin x with 2 arctan ⁡ ( 1 − x 1 + x ) ..

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Finding an antiderivativeI need to find the following:  ∫                    1        −        x                    1        +        x              ⋅      1    x          d    xFirstly, it has to be that   x  ∈  (  −  1  ,  0  )  ∪  (  0  ,  1  ]. From this, it is implied that   1  +  x  &amp;gt;  0. Also, it holds that   1  −  x  ≥  0.What I did is that I multiplied both numerator and denominator by       1    +    x  . Thus, we have:  ∫            1      −              x        2                    1      +      x        ⋅      1    x          d    x                                                              =                          x              =              sin              ⁡              t                                ∫                                    cos              ⁡              t                                      1              +              sin              ⁡              t                                ⋅                      1                          sin              ⁡              t                                ⋅          cos          ⁡          t                                d                    t                                            =                  ∫                                                    cos                2                            ⁡              t                                      (              1              +              sin              ⁡              t              )              sin              ⁡              t                                                      d                    t                                            =                  ∫                                    1              −                              sin                2                            ⁡              t                                      (              1              +              sin              ⁡              t              )              sin              ⁡              t                                                      d                    t                                            =                  ∫                                    (              1              −              sin              ⁡              t              )              (              1              +              sin              ⁡              t              )                                      (              1              +              sin              ⁡              t              )              sin              ⁡              t                                                      d                    t          =                      ∫                          1                              sin                ⁡                t                                      −            1                                      d                        t            .                              Thus, the final answer is:  log  ⁡      |    tan    ⁡          (                        arcsin          ⁡          x                2            )        |    −  arcsin  ⁡  x  +  C  ,,since   ∫      1          sin      ⁡      t              d    t  =  log  ⁡      |    tan    ⁡          (              t        2            )        |    +  c  ..The answer given in the textbook is:  log  ⁡      |                            1          −          x                −                  1          +          x                                      1          −          x                +                  1          +          x                      |    +  2  arctan  ⁡      (                            1          −          x                          1          +          x                      )    +  c  o  n  s  t  a  n  tI have derived that the arguments in the logarithms are the same, but it seems difficult to connect arc sin x with   2  arctan  ⁡      (                            1          −          x                          1          +          x                      )    ..

Danika Rojas · Accepted Answer

Step 1Let   x  =  cos  ⁡  2  y    ⟹    0  ≤  2  y  ≤  π                              1          −          x                          1          +          x                      =  tan  ⁡  y    ⟹    2  arctan  ⁡                              1          −          x                          1          +          x                      =  2  y  =  arccos  ⁡  x  =            π      2        −  arcsin  ⁡  xStep 2                              1          −          x                          1          +          x                      =  u    ⟹    x  =                    1        −                  u          2                            1        +                  u          2

Finding an antiderivative I need to find the following: ∫ <msqrt>

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New Questions in Calculus 1