$ int e^-sin t cos t~ dt - int e^-sin t cos t sin t~ dt $

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$ int e^-sin t cos t~ dt - int e^-sin t cos t sin t~ dt $



According to the answer,



It substituted $x=sin t$



$ int e^-x~ dx- int e^-x(x)~ dx $



Why did the answer remove $cos t$ ?






integration indefinite-integrals







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  • SImply because $dx=cos(t),dt$
    – Claude Leibovici
    Aug 30 at 3:58










  • @ClaudeLeibovici is there a way to prove this ?
    – user185692
    Aug 30 at 4:06










  • It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
    – farruhota
    Aug 30 at 4:30














up vote
0
down vote

favorite












$ int e^-sin t cos t~ dt - int e^-sin t cos t sin t~ dt $



According to the answer,



It substituted $x=sin t$



$ int e^-x~ dx- int e^-x(x)~ dx $



Why did the answer remove $cos t$ ?






integration indefinite-integrals







share|cite|improve this question























  • SImply because $dx=cos(t),dt$
    – Claude Leibovici
    Aug 30 at 3:58










  • @ClaudeLeibovici is there a way to prove this ?
    – user185692
    Aug 30 at 4:06










  • It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
    – farruhota
    Aug 30 at 4:30












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$ int e^-sin t cos t~ dt - int e^-sin t cos t sin t~ dt $



According to the answer,



It substituted $x=sin t$



$ int e^-x~ dx- int e^-x(x)~ dx $



Why did the answer remove $cos t$ ?






integration indefinite-integrals







share|cite|improve this question















$ int e^-sin t cos t~ dt - int e^-sin t cos t sin t~ dt $



According to the answer,



It substituted $x=sin t$



$ int e^-x~ dx- int e^-x(x)~ dx $



Why did the answer remove $cos t$ ?






integration indefinite-integrals




integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 30 at 5:34









tarit goswami

1,156219




1,156219










asked Aug 30 at 3:37









user185692

1536




1536











  • SImply because $dx=cos(t),dt$
    – Claude Leibovici
    Aug 30 at 3:58










  • @ClaudeLeibovici is there a way to prove this ?
    – user185692
    Aug 30 at 4:06










  • It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
    – farruhota
    Aug 30 at 4:30
















  • SImply because $dx=cos(t),dt$
    – Claude Leibovici
    Aug 30 at 3:58










  • @ClaudeLeibovici is there a way to prove this ?
    – user185692
    Aug 30 at 4:06










  • It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
    – farruhota
    Aug 30 at 4:30















SImply because $dx=cos(t),dt$
– Claude Leibovici
Aug 30 at 3:58




SImply because $dx=cos(t),dt$
– Claude Leibovici
Aug 30 at 3:58












@ClaudeLeibovici is there a way to prove this ?
– user185692
Aug 30 at 4:06




@ClaudeLeibovici is there a way to prove this ?
– user185692
Aug 30 at 4:06












It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
– farruhota
Aug 30 at 4:30




It is $int e^-sin t cos t dt - int e^-sin t cos t sin t dt=int e^-sin t dsin t - int e^-sin t sin t d sin t$
– farruhota
Aug 30 at 4:30










2 Answers
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$$I=int e^-sin t cos t dt - int e^-sin t cos t sin t dt$$



You are substituting $x=sin t$
Differentiate both sides w.r.t $t$



Thus $$fracdxdt=cos t$$
$$dx=cos(t) dt$$
Thus $$I=int e^-sin t cdot dx - int e^-sin t sin t cdot dx$$
$$I=int e^-x cdot dx - int e^-x cdot x cdot dx$$






share|cite|improve this answer



























    up vote
    1
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    [beginarraylint e^ - sin t cos tdt - int e^ - sin t sin tcos tdt\There,are,two,parts,of,this,question.\Use,the,formulaint U^n du = fracu^n + 1n + 1 + c,to,solve,the,sec ond,part.\You,donot,need,to,solve,the,first,part,it,will,cancel,when,you,mathoprm int egrate,the,sec ond,part.endarray]






    share|cite|improve this answer






















    • Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
      – Kevin
      Aug 30 at 14:39






    • 1




      Thanks, it is perfect.
      – Cipher
      Aug 30 at 23:06










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    $$I=int e^-sin t cos t dt - int e^-sin t cos t sin t dt$$



    You are substituting $x=sin t$
    Differentiate both sides w.r.t $t$



    Thus $$fracdxdt=cos t$$
    $$dx=cos(t) dt$$
    Thus $$I=int e^-sin t cdot dx - int e^-sin t sin t cdot dx$$
    $$I=int e^-x cdot dx - int e^-x cdot x cdot dx$$






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      $$I=int e^-sin t cos t dt - int e^-sin t cos t sin t dt$$



      You are substituting $x=sin t$
      Differentiate both sides w.r.t $t$



      Thus $$fracdxdt=cos t$$
      $$dx=cos(t) dt$$
      Thus $$I=int e^-sin t cdot dx - int e^-sin t sin t cdot dx$$
      $$I=int e^-x cdot dx - int e^-x cdot x cdot dx$$






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        $$I=int e^-sin t cos t dt - int e^-sin t cos t sin t dt$$



        You are substituting $x=sin t$
        Differentiate both sides w.r.t $t$



        Thus $$fracdxdt=cos t$$
        $$dx=cos(t) dt$$
        Thus $$I=int e^-sin t cdot dx - int e^-sin t sin t cdot dx$$
        $$I=int e^-x cdot dx - int e^-x cdot x cdot dx$$






        share|cite|improve this answer












        $$I=int e^-sin t cos t dt - int e^-sin t cos t sin t dt$$



        You are substituting $x=sin t$
        Differentiate both sides w.r.t $t$



        Thus $$fracdxdt=cos t$$
        $$dx=cos(t) dt$$
        Thus $$I=int e^-sin t cdot dx - int e^-sin t sin t cdot dx$$
        $$I=int e^-x cdot dx - int e^-x cdot x cdot dx$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 30 at 4:22









        Deepesh Meena

        3,2492824




        3,2492824




















            up vote
            1
            down vote













            [beginarraylint e^ - sin t cos tdt - int e^ - sin t sin tcos tdt\There,are,two,parts,of,this,question.\Use,the,formulaint U^n du = fracu^n + 1n + 1 + c,to,solve,the,sec ond,part.\You,donot,need,to,solve,the,first,part,it,will,cancel,when,you,mathoprm int egrate,the,sec ond,part.endarray]






            share|cite|improve this answer






















            • Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
              – Kevin
              Aug 30 at 14:39






            • 1




              Thanks, it is perfect.
              – Cipher
              Aug 30 at 23:06














            up vote
            1
            down vote













            [beginarraylint e^ - sin t cos tdt - int e^ - sin t sin tcos tdt\There,are,two,parts,of,this,question.\Use,the,formulaint U^n du = fracu^n + 1n + 1 + c,to,solve,the,sec ond,part.\You,donot,need,to,solve,the,first,part,it,will,cancel,when,you,mathoprm int egrate,the,sec ond,part.endarray]






            share|cite|improve this answer






















            • Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
              – Kevin
              Aug 30 at 14:39






            • 1




              Thanks, it is perfect.
              – Cipher
              Aug 30 at 23:06












            up vote
            1
            down vote










            up vote
            1
            down vote









            [beginarraylint e^ - sin t cos tdt - int e^ - sin t sin tcos tdt\There,are,two,parts,of,this,question.\Use,the,formulaint U^n du = fracu^n + 1n + 1 + c,to,solve,the,sec ond,part.\You,donot,need,to,solve,the,first,part,it,will,cancel,when,you,mathoprm int egrate,the,sec ond,part.endarray]






            share|cite|improve this answer














            [beginarraylint e^ - sin t cos tdt - int e^ - sin t sin tcos tdt\There,are,two,parts,of,this,question.\Use,the,formulaint U^n du = fracu^n + 1n + 1 + c,to,solve,the,sec ond,part.\You,donot,need,to,solve,the,first,part,it,will,cancel,when,you,mathoprm int egrate,the,sec ond,part.endarray]







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 4 at 22:29

























            answered Aug 30 at 14:28









            Cipher

            113




            113











            • Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
              – Kevin
              Aug 30 at 14:39






            • 1




              Thanks, it is perfect.
              – Cipher
              Aug 30 at 23:06
















            • Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
              – Kevin
              Aug 30 at 14:39






            • 1




              Thanks, it is perfect.
              – Cipher
              Aug 30 at 23:06















            Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
            – Kevin
            Aug 30 at 14:39




            Hello Cipher, welcome to MathStackExchange! I have edited your post to included the mathjax syntax we use here. Please take a look and familiarise yourself with the usage of the typesetting language, and also check my edits to make sure I did not make a mistake in your intentions!
            – Kevin
            Aug 30 at 14:39




            1




            1




            Thanks, it is perfect.
            – Cipher
            Aug 30 at 23:06




            Thanks, it is perfect.
            – Cipher
            Aug 30 at 23:06

















             

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