integral of $cosxcos(sinx)dx$

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I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$

$ucos(sinx)dx$



$u [-sin(sinx)cosx(1)]+C$



$frac-(sin(sinx)).(cosx)-sinx + C$



tentative answer: $sin(cosx) + C$



Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.







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  • 1




    Why not u=sinx?
    – imranfat
    May 5 '14 at 21:24










  • try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
    – user137481
    May 5 '14 at 21:25










  • Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
    – Michael Hardy
    May 5 '14 at 22:00














up vote
2
down vote

favorite












I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$

$ucos(sinx)dx$



$u [-sin(sinx)cosx(1)]+C$



$frac-(sin(sinx)).(cosx)-sinx + C$



tentative answer: $sin(cosx) + C$



Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.







share|cite|improve this question


















  • 1




    Why not u=sinx?
    – imranfat
    May 5 '14 at 21:24










  • try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
    – user137481
    May 5 '14 at 21:25










  • Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
    – Michael Hardy
    May 5 '14 at 22:00












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$

$ucos(sinx)dx$



$u [-sin(sinx)cosx(1)]+C$



$frac-(sin(sinx)).(cosx)-sinx + C$



tentative answer: $sin(cosx) + C$



Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.







share|cite|improve this question














I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$

$ucos(sinx)dx$



$u [-sin(sinx)cosx(1)]+C$



$frac-(sin(sinx)).(cosx)-sinx + C$



tentative answer: $sin(cosx) + C$



Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.









share|cite|improve this question













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edited Jun 3 '16 at 5:44









user 170039

10.3k42361




10.3k42361










asked May 5 '14 at 21:20









Gabe Carr

162




162







  • 1




    Why not u=sinx?
    – imranfat
    May 5 '14 at 21:24










  • try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
    – user137481
    May 5 '14 at 21:25










  • Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
    – Michael Hardy
    May 5 '14 at 22:00












  • 1




    Why not u=sinx?
    – imranfat
    May 5 '14 at 21:24










  • try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
    – user137481
    May 5 '14 at 21:25










  • Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
    – Michael Hardy
    May 5 '14 at 22:00







1




1




Why not u=sinx?
– imranfat
May 5 '14 at 21:24




Why not u=sinx?
– imranfat
May 5 '14 at 21:24












try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
– user137481
May 5 '14 at 21:25




try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
– user137481
May 5 '14 at 21:25












Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
– Michael Hardy
May 5 '14 at 22:00




Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
– Michael Hardy
May 5 '14 at 22:00










3 Answers
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up vote
1
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Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.



Then we have the integral $intcos(u)du$.






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    up vote
    0
    down vote













    Hint:



    $$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
    $$(sin(sin(x)))'=cos(x)cos(sin(x)).$$






    share|cite|improve this answer



























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      $$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$






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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        active

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        up vote
        1
        down vote













        Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.



        Then we have the integral $intcos(u)du$.






        share|cite|improve this answer


























          up vote
          1
          down vote













          Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.



          Then we have the integral $intcos(u)du$.






          share|cite|improve this answer
























            up vote
            1
            down vote










            up vote
            1
            down vote









            Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.



            Then we have the integral $intcos(u)du$.






            share|cite|improve this answer














            Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.



            Then we have the integral $intcos(u)du$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 25 '14 at 13:52

























            answered May 5 '14 at 21:24









            beep-boop

            7,87452856




            7,87452856




















                up vote
                0
                down vote













                Hint:



                $$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
                $$(sin(sin(x)))'=cos(x)cos(sin(x)).$$






                share|cite|improve this answer
























                  up vote
                  0
                  down vote













                  Hint:



                  $$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
                  $$(sin(sin(x)))'=cos(x)cos(sin(x)).$$






                  share|cite|improve this answer






















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Hint:



                    $$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
                    $$(sin(sin(x)))'=cos(x)cos(sin(x)).$$






                    share|cite|improve this answer












                    Hint:



                    $$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
                    $$(sin(sin(x)))'=cos(x)cos(sin(x)).$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jul 4 '16 at 11:55









                    Yves Daoust

                    112k665206




                    112k665206




















                        up vote
                        0
                        down vote













                        $$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          $$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$






                            share|cite|improve this answer












                            $$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 10 at 19:11









                            Vincent Law

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