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1












$begingroup$


So I have to integrate $$frac{sin^n x}{sin^n x + cos^n x}$$ and am coding this in Mathematica with



 (((Sin^n)[x])/(((Sin^n)[x]) + ((Cos^n)[x]))) 


with the bounds $0$ and $pi/2,$ where $n$ takes on various integer values.



I programmed the problem so that $n=1$ then $n=2$, etc...but every time I try to get the output, I only get back the integration symbol. For example, if I program $n=2$ and then do the integration command- the output is



 (((Sin^2)[x])/(((Sin^2)[x]) + ((Cos^2)[x]))), 


but does not solve it. Anyone know how to help or fix this??










share|improve this question









New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
    $endgroup$
    – enano9314
    4 hours ago












  • $begingroup$
    Related: math.stackexchange.com/questions/82489/…
    $endgroup$
    – Michael E2
    1 hour ago
















1












$begingroup$


So I have to integrate $$frac{sin^n x}{sin^n x + cos^n x}$$ and am coding this in Mathematica with



 (((Sin^n)[x])/(((Sin^n)[x]) + ((Cos^n)[x]))) 


with the bounds $0$ and $pi/2,$ where $n$ takes on various integer values.



I programmed the problem so that $n=1$ then $n=2$, etc...but every time I try to get the output, I only get back the integration symbol. For example, if I program $n=2$ and then do the integration command- the output is



 (((Sin^2)[x])/(((Sin^2)[x]) + ((Cos^2)[x]))), 


but does not solve it. Anyone know how to help or fix this??










share|improve this question









New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
    $endgroup$
    – enano9314
    4 hours ago












  • $begingroup$
    Related: math.stackexchange.com/questions/82489/…
    $endgroup$
    – Michael E2
    1 hour ago














1












1








1





$begingroup$


So I have to integrate $$frac{sin^n x}{sin^n x + cos^n x}$$ and am coding this in Mathematica with



 (((Sin^n)[x])/(((Sin^n)[x]) + ((Cos^n)[x]))) 


with the bounds $0$ and $pi/2,$ where $n$ takes on various integer values.



I programmed the problem so that $n=1$ then $n=2$, etc...but every time I try to get the output, I only get back the integration symbol. For example, if I program $n=2$ and then do the integration command- the output is



 (((Sin^2)[x])/(((Sin^2)[x]) + ((Cos^2)[x]))), 


but does not solve it. Anyone know how to help or fix this??










share|improve this question









New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




So I have to integrate $$frac{sin^n x}{sin^n x + cos^n x}$$ and am coding this in Mathematica with



 (((Sin^n)[x])/(((Sin^n)[x]) + ((Cos^n)[x]))) 


with the bounds $0$ and $pi/2,$ where $n$ takes on various integer values.



I programmed the problem so that $n=1$ then $n=2$, etc...but every time I try to get the output, I only get back the integration symbol. For example, if I program $n=2$ and then do the integration command- the output is



 (((Sin^2)[x])/(((Sin^2)[x]) + ((Cos^2)[x]))), 


but does not solve it. Anyone know how to help or fix this??







calculus-and-analysis






share|improve this question









New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 1 hour ago









Michael E2

151k12203482




151k12203482






New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









EmmaEmma

62




62




New contributor




Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Emma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
    $endgroup$
    – enano9314
    4 hours ago












  • $begingroup$
    Related: math.stackexchange.com/questions/82489/…
    $endgroup$
    – Michael E2
    1 hour ago














  • 1




    $begingroup$
    Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
    $endgroup$
    – enano9314
    4 hours ago












  • $begingroup$
    Related: math.stackexchange.com/questions/82489/…
    $endgroup$
    – Michael E2
    1 hour ago








1




1




$begingroup$
Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
$endgroup$
– enano9314
4 hours ago






$begingroup$
Hello! It seems that you haven't included any code. Can you please include that here in the post? Based on your textual description my guess is that you are using (Sin^2)[x] when that syntax is incorrect, you should instead write it as Sin[x]^2
$endgroup$
– enano9314
4 hours ago














$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
1 hour ago




$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
1 hour ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

This works for me:



Table[Integrate[Sin[x]^n/(Sin[x]^n+Cos[x]^n),{x,0,Pi/2}],{n,1,5}]


And it gives the output {Pi/4,Pi/4,Pi/4,Pi/4,Pi/4}.



enter image description here






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Recommend that you add a Plot to make it easier to understand why the result is a constant.
    $endgroup$
    – Bob Hanlon
    4 hours ago










  • $begingroup$
    Good suggestion. Editing.
    $endgroup$
    – Kevin Ausman
    3 hours ago



















2












$begingroup$

Perhaps you're writing your function in the wrong format Emma. The following works fine:



n = 2;
Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n),{x, 0, π/2}]



π/4







share|improve this answer











$endgroup$













  • $begingroup$
    I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
    $endgroup$
    – Kevin Ausman
    4 hours ago






  • 1




    $begingroup$
    Yes, just realized that, thanks for pointing it out.
    $endgroup$
    – amator2357
    4 hours ago



















1












$begingroup$

A common trick (see this Math.SE post:



ClearAll[symmetrizeIntegrate];
SetAttributes[symmetrizeIntegrate, HoldAll];
symmetrizeIntegrate[Integrate[f_, {x_, a_, b_}, opts___]] :=
Integrate[(f + (f /. x -> a + b - x))/2, {x, a, b}, opts]

symmetrizeIntegrate[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, [Pi]/2}]]
(* [Pi]/4 *)





share|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    This works for me:



    Table[Integrate[Sin[x]^n/(Sin[x]^n+Cos[x]^n),{x,0,Pi/2}],{n,1,5}]


    And it gives the output {Pi/4,Pi/4,Pi/4,Pi/4,Pi/4}.



    enter image description here






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Recommend that you add a Plot to make it easier to understand why the result is a constant.
      $endgroup$
      – Bob Hanlon
      4 hours ago










    • $begingroup$
      Good suggestion. Editing.
      $endgroup$
      – Kevin Ausman
      3 hours ago
















    3












    $begingroup$

    This works for me:



    Table[Integrate[Sin[x]^n/(Sin[x]^n+Cos[x]^n),{x,0,Pi/2}],{n,1,5}]


    And it gives the output {Pi/4,Pi/4,Pi/4,Pi/4,Pi/4}.



    enter image description here






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Recommend that you add a Plot to make it easier to understand why the result is a constant.
      $endgroup$
      – Bob Hanlon
      4 hours ago










    • $begingroup$
      Good suggestion. Editing.
      $endgroup$
      – Kevin Ausman
      3 hours ago














    3












    3








    3





    $begingroup$

    This works for me:



    Table[Integrate[Sin[x]^n/(Sin[x]^n+Cos[x]^n),{x,0,Pi/2}],{n,1,5}]


    And it gives the output {Pi/4,Pi/4,Pi/4,Pi/4,Pi/4}.



    enter image description here






    share|improve this answer











    $endgroup$



    This works for me:



    Table[Integrate[Sin[x]^n/(Sin[x]^n+Cos[x]^n),{x,0,Pi/2}],{n,1,5}]


    And it gives the output {Pi/4,Pi/4,Pi/4,Pi/4,Pi/4}.



    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 3 hours ago

























    answered 4 hours ago









    Kevin AusmanKevin Ausman

    32417




    32417








    • 1




      $begingroup$
      Recommend that you add a Plot to make it easier to understand why the result is a constant.
      $endgroup$
      – Bob Hanlon
      4 hours ago










    • $begingroup$
      Good suggestion. Editing.
      $endgroup$
      – Kevin Ausman
      3 hours ago














    • 1




      $begingroup$
      Recommend that you add a Plot to make it easier to understand why the result is a constant.
      $endgroup$
      – Bob Hanlon
      4 hours ago










    • $begingroup$
      Good suggestion. Editing.
      $endgroup$
      – Kevin Ausman
      3 hours ago








    1




    1




    $begingroup$
    Recommend that you add a Plot to make it easier to understand why the result is a constant.
    $endgroup$
    – Bob Hanlon
    4 hours ago




    $begingroup$
    Recommend that you add a Plot to make it easier to understand why the result is a constant.
    $endgroup$
    – Bob Hanlon
    4 hours ago












    $begingroup$
    Good suggestion. Editing.
    $endgroup$
    – Kevin Ausman
    3 hours ago




    $begingroup$
    Good suggestion. Editing.
    $endgroup$
    – Kevin Ausman
    3 hours ago











    2












    $begingroup$

    Perhaps you're writing your function in the wrong format Emma. The following works fine:



    n = 2;
    Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n),{x, 0, π/2}]



    π/4







    share|improve this answer











    $endgroup$













    • $begingroup$
      I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
      $endgroup$
      – Kevin Ausman
      4 hours ago






    • 1




      $begingroup$
      Yes, just realized that, thanks for pointing it out.
      $endgroup$
      – amator2357
      4 hours ago
















    2












    $begingroup$

    Perhaps you're writing your function in the wrong format Emma. The following works fine:



    n = 2;
    Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n),{x, 0, π/2}]



    π/4







    share|improve this answer











    $endgroup$













    • $begingroup$
      I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
      $endgroup$
      – Kevin Ausman
      4 hours ago






    • 1




      $begingroup$
      Yes, just realized that, thanks for pointing it out.
      $endgroup$
      – amator2357
      4 hours ago














    2












    2








    2





    $begingroup$

    Perhaps you're writing your function in the wrong format Emma. The following works fine:



    n = 2;
    Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n),{x, 0, π/2}]



    π/4







    share|improve this answer











    $endgroup$



    Perhaps you're writing your function in the wrong format Emma. The following works fine:



    n = 2;
    Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n),{x, 0, π/2}]



    π/4








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 3 hours ago









    m_goldberg

    88.8k873200




    88.8k873200










    answered 4 hours ago









    amator2357amator2357

    1437




    1437












    • $begingroup$
      I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
      $endgroup$
      – Kevin Ausman
      4 hours ago






    • 1




      $begingroup$
      Yes, just realized that, thanks for pointing it out.
      $endgroup$
      – amator2357
      4 hours ago


















    • $begingroup$
      I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
      $endgroup$
      – Kevin Ausman
      4 hours ago






    • 1




      $begingroup$
      Yes, just realized that, thanks for pointing it out.
      $endgroup$
      – amator2357
      4 hours ago
















    $begingroup$
    I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
    $endgroup$
    – Kevin Ausman
    4 hours ago




    $begingroup$
    I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
    $endgroup$
    – Kevin Ausman
    4 hours ago




    1




    1




    $begingroup$
    Yes, just realized that, thanks for pointing it out.
    $endgroup$
    – amator2357
    4 hours ago




    $begingroup$
    Yes, just realized that, thanks for pointing it out.
    $endgroup$
    – amator2357
    4 hours ago











    1












    $begingroup$

    A common trick (see this Math.SE post:



    ClearAll[symmetrizeIntegrate];
    SetAttributes[symmetrizeIntegrate, HoldAll];
    symmetrizeIntegrate[Integrate[f_, {x_, a_, b_}, opts___]] :=
    Integrate[(f + (f /. x -> a + b - x))/2, {x, a, b}, opts]

    symmetrizeIntegrate[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, [Pi]/2}]]
    (* [Pi]/4 *)





    share|improve this answer









    $endgroup$


















      1












      $begingroup$

      A common trick (see this Math.SE post:



      ClearAll[symmetrizeIntegrate];
      SetAttributes[symmetrizeIntegrate, HoldAll];
      symmetrizeIntegrate[Integrate[f_, {x_, a_, b_}, opts___]] :=
      Integrate[(f + (f /. x -> a + b - x))/2, {x, a, b}, opts]

      symmetrizeIntegrate[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, [Pi]/2}]]
      (* [Pi]/4 *)





      share|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        A common trick (see this Math.SE post:



        ClearAll[symmetrizeIntegrate];
        SetAttributes[symmetrizeIntegrate, HoldAll];
        symmetrizeIntegrate[Integrate[f_, {x_, a_, b_}, opts___]] :=
        Integrate[(f + (f /. x -> a + b - x))/2, {x, a, b}, opts]

        symmetrizeIntegrate[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, [Pi]/2}]]
        (* [Pi]/4 *)





        share|improve this answer









        $endgroup$



        A common trick (see this Math.SE post:



        ClearAll[symmetrizeIntegrate];
        SetAttributes[symmetrizeIntegrate, HoldAll];
        symmetrizeIntegrate[Integrate[f_, {x_, a_, b_}, opts___]] :=
        Integrate[(f + (f /. x -> a + b - x))/2, {x, a, b}, opts]

        symmetrizeIntegrate[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, [Pi]/2}]]
        (* [Pi]/4 *)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Michael E2Michael E2

        151k12203482




        151k12203482






















            Emma is a new contributor. Be nice, and check out our Code of Conduct.










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