R ( sin ⁡ x , cos ⁡ x ) d x this type of integral where R ( u , v ) is a rational function can be written as an integral of a rational function with t = tan ⁡ x 2 . If it is known that R ( − u , v ) = − R ( u , v ) or R ( u , − v ) = − R ( u , v ) or R ( u , − v ) = − R ( u , v ) then it is suggested to do t = cos ⁡ x t = sin ⁡ x t = tan ⁡ x change. My question is what are u and v here? Can you please bring specific example.

Question

R    (    sin    ⁡    x    ,    cos    ⁡    x    )        d    x    this type of integral where     R    (    u    ,    v    )   is a rational function can be written as an integral of a rational function with     t    =    tan    ⁡          x      2      .    If it is known that     R    (    −    u    ,    v    )    =    −    R    (    u    ,    v    )   or     R    (    u    ,    −    v    )    =    −    R    (    u    ,    v    )   or     R    (    u    ,    −    v    )    =    −    R    (    u    ,    v    )   then it is suggested to do     t    =    cos    ⁡    x          t    =    sin    ⁡    x          t    =    tan    ⁡    x   change.    My question is what are     u   and     v   here? Can you please bring specific example.

Hezekiah House · Accepted Answer

If you make a substitution   t  =  tan  ⁡  (      x    2    ), then you can rewrite   sin  ⁡  (  x  )  ,  cos  ⁡  (  x  ) in terms of   t. Indeed,  sin  ⁡  (  x  )  =  2  sin  ⁡  (      x    2    )  cos  ⁡  (      x    2    )  =            2      sin      ⁡      (              x        2            )      cos      ⁡      (              x        2            )                      sin        2            ⁡      (              x        2            )      +              cos        2            ⁡      (              x        2            )        =  2            tan      ⁡      (              x        2            )                      tan        2            ⁡      (              x        2            )      +      1        =            2      t                      t        2            +      1      where both time we used the fact that       sin    2    ⁡  (  α  )  +      cos    2    ⁡  (  α  )  =  1 for any   α  ∈      R  .Moreover, after substitution   t  =  tan  ⁡  (      x    2    ) you get   d  t  =      1          2              cos        2            ⁡      (              x        2            )        d  x  =      1    2              (        1  +      tan    2    ⁡  (      x    2    )            )        d  x In other words       2          1      +              t        2              d  t  =  d  xSo integral  ∫  R  (  cos  ⁡  (  x  )  ,  sin  ⁡  (  x  )  )  d  xafter substitution   t  =  tan  ⁡  (      x    2    ) becomes  ∫  R            (                  1      −              t        2                    1      +              t        2              ,            2      t                      t        2            +      1                  )            2          1      +              t        2              d  tSpecific example: Let&#039;s say we want to calculate   ∫            sin      ⁡      (      x      )              1      +      cos      ⁡      (      x      )        d  x. Although it&#039;s easier after substitution   t  =  cos  ⁡  (  x  ), we can try to use our technique of substituting   t  =  tan  ⁡  (      x    2    ). In our example   R  (  a  ,  b  )  =      b          1      +      a      . After substitution we should get  ∫  R            (                  1      −              t        2                    1      +              t        2              ,            2      t                      t        2            +      1                  )            2          1      +              t        2              d  t  =  ∫            2      t              1      +              t        2              ⋅      1          1      +                        1          −                      t            2                                    1          +                      t            2                                    2          1      +              t        2              d  tMost of it cancels out and we arrive at  ∫            2      t              1      +              t        2              d  t  =  ln  ⁡  (  1  +      t    2    )  +  C  =  ln  ⁡  (  1  +      tan    2    ⁡  (      x    2    )  )  +  C

R ( sin ⁡ x , cos ⁡ x ) <mspace

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