Definite integral of $cos(x^2)$

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Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?







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  • 3




    The obvious answer is change the order of integration.
    – Doug M
    Jul 25 at 22:01










  • Ah, thank you, that explains it!
    – BMac
    Jul 25 at 22:06






  • 2




    What was the whole integral?
    – Henry Lee
    Jul 26 at 16:48














up vote
0
down vote

favorite












Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?







share|cite|improve this question

















  • 3




    The obvious answer is change the order of integration.
    – Doug M
    Jul 25 at 22:01










  • Ah, thank you, that explains it!
    – BMac
    Jul 25 at 22:06






  • 2




    What was the whole integral?
    – Henry Lee
    Jul 26 at 16:48












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?







share|cite|improve this question













Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?









share|cite|improve this question












share|cite|improve this question




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edited Jul 25 at 22:00









Bernard

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110k635103









asked Jul 25 at 21:59









BMac

24517




24517







  • 3




    The obvious answer is change the order of integration.
    – Doug M
    Jul 25 at 22:01










  • Ah, thank you, that explains it!
    – BMac
    Jul 25 at 22:06






  • 2




    What was the whole integral?
    – Henry Lee
    Jul 26 at 16:48












  • 3




    The obvious answer is change the order of integration.
    – Doug M
    Jul 25 at 22:01










  • Ah, thank you, that explains it!
    – BMac
    Jul 25 at 22:06






  • 2




    What was the whole integral?
    – Henry Lee
    Jul 26 at 16:48







3




3




The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01




The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01












Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06




Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06




2




2




What was the whole integral?
– Henry Lee
Jul 26 at 16:48




What was the whole integral?
– Henry Lee
Jul 26 at 16:48










1 Answer
1






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1
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Well, I think that you're looking at:



$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$



Using:



$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$



We can write:



$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Well, I think that you're looking at:



    $$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$



    Using:



    $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$



    We can write:



    $$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
    $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
    $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
    $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
    $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$






    share|cite|improve this answer

























      up vote
      1
      down vote













      Well, I think that you're looking at:



      $$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$



      Using:



      $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$



      We can write:



      $$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
      $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
      $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
      $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
      $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Well, I think that you're looking at:



        $$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$



        Using:



        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$



        We can write:



        $$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$






        share|cite|improve this answer













        Well, I think that you're looking at:



        $$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$



        Using:



        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$



        We can write:



        $$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
        $$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$







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        answered Jul 28 at 11:50









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