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Integration Help

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??

calculus-and-analysis

share|improve this question

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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
  56 mins ago
  

  
  
  
  

add a comment | 

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??

calculus-and-analysis

share|improve this question

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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
  56 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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??

calculus-and-analysis

share|improve this question

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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.

$endgroup$

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

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

share|improve this question

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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.

edited 1 hour ago

Michael E2

151k12203482

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

61

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

61

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}.

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 46 mins ago

Michael E2Michael E2

151k12203482

$endgroup$

add a comment | 

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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}.

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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}.

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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}.

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

$endgroup$

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

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

answered 4 hours ago

Kevin AusmanKevin Ausman

32417

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

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

$endgroup$

n = 2;

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

share|improve this answer

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

edited 2 hours ago

m_goldberg

88.8k873200

edited 2 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

answered 4 hours ago

amator2357amator2357

1437

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

answered 46 mins ago

Michael E2Michael E2

151k12203482

$endgroup$

add a comment | 

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

answered 46 mins ago

Michael E2Michael E2

151k12203482

$endgroup$

add a comment | 

$begingroup$

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

answered 46 mins ago

Michael E2Michael E2

151k12203482

$endgroup$

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

answered 46 mins ago

Michael E2Michael E2

151k12203482

answered 46 mins ago

Michael E2Michael E2

151k12203482

answered 46 mins ago

Michael E2Michael E2

151k12203482