How do you integrate 1 ( 1 + cos ⁡ x ) ( 1 + sin ⁡ x )

Question

How do you integrate             1              (        1        +        cos        ⁡        x        )        (        1        +        sin        ⁡        x        )

vard6vv · Accepted Answer

Let      I    =    ∫          1              (        1        +        sin        ⁡        x        )        ⋅        (        1        +        cos        ⁡        x        )              d    x    =          1      2        ∫          1                                    (            sin            ⁡                          x              2                        +            cos            ⁡                          x              2                        )                    2                ⋅                  cos          2                ⁡                  x          2                      d    x  Above we used      ∙        1    +    sin    ⁡    x    =          sin      2        ⁡          x      2        +          cos      2        ⁡          x      2        +    2    sin    ⁡          x      2        ⋅    cos    ⁡          x      2      And      ∙        1    +    cos    ⁡    x    =    2          cos      2        ⁡          x      2        .  So we get      I    =          1      2        ∫                            sec          2                ⁡                  x          2                                      (          sin          ⁡                      x            2                    +          cos          ⁡                      x            2                    )                2              d    x    =          1      2        ∫                            sec          2                ⁡                  x          2                ⋅                  (          1          +                      tan            2                    ⁡                      x            2                    )                                      (          tan          ⁡                      x            2                    +          1          )                2            Now Put       tan    ⁡          x      2        =    t        ,  , Then             1      2              sec      2        ⁡          x      2        d    x    =    d    t  So we get      I    =    ∫                  1        +                  t          2                            (        1        +        t                  )          2                      d    x    =    ∫                  1        +                  t          2                +        2        t        −        2        t                    (        1        +        t                  )          2                      d    t    =    ∫    1    d    t    −    2    ∫          t              (        1        +        t                  )          2                      d    t  Now   (  1  +  t  )  =  u    ,, Then   d  t  =  d  uSo we get      I    =    t    −    2    ∫                  u        −        1                    u        2              d    u    =    t    −    2    ln    ⁡          |        u          |        −          2      u        +          C      So we get      I    =    t    −    2    ln    ⁡          |        1    +    t          |        −          2              1        +        t              +          C      So we get      I    =    tan    ⁡          x      2        −    2    ln    ⁡          |      1      +      tan      ⁡              x        2            |        −          2              1        +        tan        ⁡                  x          2                      +          C

Bailee Landry · Accepted Answer

HINT:            1              (        1        +        cos        ⁡        x        )        (        1        +        sin        ⁡        x        )              =                    (        1        −        sin        ⁡        x        )        (        1        −        cos        ⁡        x        )                              cos          2                ⁡        x                  sin          2                ⁡        x              =      sec    2    ⁡  x      csc    2    ⁡  x  +  2  csc  ⁡  2  x  −                    sin        ⁡        x        +        cos        ⁡        x                              cos          2                ⁡        x                  sin          2                ⁡        x            Now,       sec    2    ⁡  x      csc    2    ⁡  x  =      sec    2    ⁡  x  +      csc    2    ⁡  xFor       ∫                            sin          ⁡          x                                      cos            2                    ⁡          x                      sin            2                    ⁡          x                      d    x    =    ∫                            sin          ⁡          x                                      cos            2                    ⁡          x          (          1          −                      cos            2                    ⁡          x          )                      d    x  set   cos  ⁡  x  =  uSimilarly for       ∫                            cos          ⁡          x                                      cos            2                    ⁡          x                      sin            2                    ⁡          x                      d    x

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