Solve ∫ cos 2 ⁡ x sin ⁡ 2 x d xI solved it using the identity cos 2 ⁡ x = 1 2 ( 1 + cos ⁡ ( 2 x ) ) and the substitution u = 2 x ∫ cos 2 ⁡ x sin ⁡ 2 x d x = 1 2 ∫ ( 1 + cos ⁡ 2 x ) sin ⁡ 2 x d x ( u = 2 x ) 1 4 ∫ ( sin ⁡ u + sin ⁡ u cos ⁡ u ) d u ( t = sin ⁡ u ) = 1 4 ( cos ⁡ u + ∫ t d t ) = cos ⁡ 2 x 4 + sin 2 ⁡ 2 x 8 + CI can't find an error in my procedure.

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Solve  ∫      cos    2    ⁡  x  sin  ⁡  2  x     d  xI solved it using the identity       cos    2    ⁡  x  =      1    2    (  1  +  cos  ⁡  (  2  x  )  ) and the substitution   u  =  2  x  ∫      cos    2    ⁡  x  sin  ⁡  2  x     d  x  =      1    2    ∫  (  1  +  cos  ⁡  2  x  )  sin  ⁡  2  x     d  x                              (                      u            =            2            x                    )                                      1          4                ∫        (        sin        ⁡        u        +        sin        ⁡        u        cos        ⁡        u        )                 d        u                                          (                      t            =            sin            ⁡            u                    )                            =                  1          4                (        cos        ⁡        u        +        ∫        t                 d        t        )              =            cos      ⁡      2      x        4    +                    sin        2            ⁡      2      x        8    +  CI can&#039;t find an error in my procedure.

Waldruhylm · Accepted Answer

You should get:  ∫  sin  ⁡  (  u  )    d  u  =  −  cos  ⁡  u  +  C  =  −  cos  ⁡  (  2  x  )  +  C  .You lost the minus sign.So your closed form, without your constant, is            −      cos      ⁡      2      x        4    +                    sin        2            ⁡      2      x        8    =            −      2      cos      ⁡      (      2      x      )      +      1      −              cos        2            ⁡      (      2      x      )        8    =            2      −      (      cos      ⁡      (      2      x      )      +      1              )        2              8  Then substitute   cos  ⁡  (  2  x  )  =  2      cos    2    ⁡  x  −  1  ,, and you get            2      −      4              cos        4            ⁡      x        8    =  −                    cos        2            ⁡      x        2    +      1    4    ,which differs from the given answer by a constant.So your approach works to give an equivalent correct answer, but you needed to fix the two integral you used.

Solve int cos^2 x sin 2x dx

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