Investigate to relationship ∫ − 2 cos ⁡ x d x ( 1 − sin ⁡ x ) 2 shares with ( 1 + sin ⁡ x 1 − sin ⁡ x )I have used integration by recognition to go about it showing that, d d x ( 1 + sin ⁡ x 1 − sin ⁡ x ) = − 2 cos ⁡ x ( 1 − sin ⁡ x ) 2 However, when I tried integrating − 2 c o s x ( 1 − s i n x ) 2 I don't arrive back at 1 + s i n x 1 − s i n x So, how do you integrate − 2 cos ⁡ x ( 1 − sin ⁡ x ) 2 ?

Question

Investigate to relationship       ∫                            −          2          cos          ⁡          x          d          x                          (          1          −          sin          ⁡          x                      )            2                               shares with   (                    1        +        sin        ⁡        x                    1        −        sin        ⁡        x              )I have used integration by recognition to go about it showing that,            d              d        x              (                    1        +        sin        ⁡        x                    1        −        sin        ⁡        x              )  =                    −        2        cos        ⁡        x                    (        1        −        sin        ⁡        x                  )          2                    However, when I tried integrating                     −        2        c        o        s        x                    (        1        −        s        i        n        x                  )          2                     I don&#039;t arrive back at                     1        +        s        i        n        x                    1        −        s        i        n        x            So, how do you integrate                     −        2        cos        ⁡        x                    (        1        −        sin        ⁡        x                  )          2                    ?

Janiyah Patton · Accepted Answer

hint:Still go thru the traditional steps:   u  =  1  −  sin  ⁡  x is the first step. Can you take it from here?

malalawak44 · Accepted Answer

Let      I    =    −    ∫                  2        cos        ⁡        x                    (        1        −        sin        ⁡        x                  )          2                      d    x        ,  Now Let   (  1  −  sin  ⁡  x  )  =  t    ,, Then   cos  ⁡  d  x  =  d  tSo Integral      I    =    −    2    ∫          1              t        2              d    t    =          2      t        +          C        =          2              (        1        −        sin        ⁡        x        )              +          C        =                  (        1        +        sin        ⁡        x        )        +        (        1        −        sin        ⁡        x        )                    (        1        −        sin        ⁡        x        )              +          C      So we get      I    =                  1        +        sin        ⁡        x                    1        −        sin        ⁡        x              +    1    +          C        =                  1        +        sin        ⁡        x                    1        −        sin        ⁡        x              +                  C        ′                  ,  ,Where             C      ′        =  1  +      C

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