What is the value of integral ∫ 0 1 log ⁡ [ ( 1 + x ) 1 / 2 + ( 1 − x ) 1 / 2 ] d x? I'm applying integration by parts but couldn't find any substitution after couple of steps

Question

What is the value of integral      ∫    0    1    log  ⁡  [  (  1  +  x      )          1              /            2        +  (  1  −  x      )          1              /            2        ]  d  x? I&#039;m applying integration by parts but couldn&#039;t find any substitution after couple of steps

amilazamiyn · Accepted Answer

Let  I  =      ∫          0              1        ln  ⁡            (            1    +    x    +      1    −    x              )        ⋅  1  d  xIntegration by parts  I  =  log  ⁡            (            1    +    x    +      1    −    x              )        ⋅  x                    |                    0              1        +      1    2        ∫          0              1                          1        +        x            −              1        −        x                            1        +        x            +              1        −        x              ⋅      1          (              1        +        x            ⋅              1        −        x            )        x  d  x  I  =            ln      ⁡      2        2    +      1    2        ∫          0              1                  x      2              (      1      +              1        −                  x          2                    )      ⋅              1        −                  x          2                      d  xPut   x  =  sin  ⁡  θ Then   d  x  =  cos  ⁡  θ  d  θ  I  =            ln      ⁡      2        2    +      1    2        ∫          0                      π        2                                sin        2            ⁡      θ      ⋅      cos      ⁡      θ              (      1      +      cos      ⁡      θ      )      ⋅      cos      ⁡      θ        d  θ  I  =            ln      ⁡      2        2    +      1    2        ∫          0                      π        2              (  1  −  cos  ⁡  θ  )  d  θ  =            ln      ⁡      2        2    +      π    4    −      1    2    .

Aarav Atkins · Accepted Answer

If   x  ∈  (  0  ,  1  ) we have       1    −    x    +      1    +    x    =      2    +    2          1      −              x        2            , hence      ∫          0              1        log  ⁡      (          1      +      x        +          1      −      x        )      d  x  =      1    2        (    log    ⁡    2    +          ∫              0                    1              log    ⁡          (      1      +              1        −                  x          2                    )            d    x    )    .On the other hand,                              ∫                      0                                1                          log        ⁡                  (          1          +                      1            −                          x              2                                )                        d        x                            =                              ∫                      0                                π                          /                        2                          cos        ⁡        (        θ        )        log        ⁡        (        1        +        cos        ⁡        θ        )                d        θ                            (        I        B        P        )                                    =                              ∫                      0                                π                          /                        2                                                              sin              2                        ⁡            θ                                1            +            cos            ⁡            θ                                  d        θ                                          =                              ∫                      0                                π                          /                        2                                    (          1          −          cos          ⁡          θ          )                        d        θ        =                              π            −            2                    2                    hence:      ∫          0              1        log  ⁡      (          1      +      x        +          1      −      x        )      d  x  =                    π        −        2        +        2        log        ⁡        2            4        .

What is the value of integral int 10log[(1+x)1/2+(1−x)1/2]dx

Answered question

Answer & Explanation

New Questions in Calculus 1