Xjyuk

Why are there no cargo aircraft with "flying wing" design?

Did MS DOS itself ever use blinking text?

How to down pick a chord with skipped strings?

How to answer "Have you ever been terminated?"

How could we fake a moon landing now?

Circuit to "zoom in" on mV fluctuations of a DC signal?

When a candle burns, why does the top of wick glow if bottom of flame is hottest?

Why didn't Eitri join the fight?

representation of vector and matrix in latex

Fantasy story; one type of magic grows in power with use, but the more powerful they are, they more they are drawn to travel to their source

How do I make this wiring inside cabinet safer? (Pic)

If my PI received research grants from a company to be able to pay my postdoc salary, did I have a potential conflict interest too?

Maximum summed powersets with non-adjacent items

Do wooden building fires get hotter than 600°C?

Do square wave exist?

Extracting terms with certain heads in a function

What are the out-of-universe reasons for the references to Toby Maguire-era Spider-Man in ITSV

What's the meaning of "fortified infraction restraint"?

How does the math work when buying airline miles?

2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?

Significance of Cersei's obsession with elephants?

What causes the direction of lightning flashes?

When the Haste spell ends on a creature, do attackers have advantage against that creature?

How to compare two different files line by line in unix?

Integration Help

1

$begingroup$

So I have to integrate $$fracsin^n xsin^n x + cos^n x$$ 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

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.

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

1

$begingroup$

So I have to integrate $$fracsin^n xsin^n x + cos^n x$$ 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

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.

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

So I have to integrate $$fracsin^n xsin^n x + cos^n x$$ 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

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.

$endgroup$

 (((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

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.

share|improve this question

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

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.

edited 1 hour ago

Michael E2

151k12203482

edited 1 hour ago

Michael E2

151k12203482

asked 4 hours ago

EmmaEmma

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.

asked 4 hours ago

EmmaEmma

62

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 4 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
  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 3 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 1 hour ago

Michael E2Michael E2

151k12203482

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

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

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195438%2fintegration-help%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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 4 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
  4 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 4 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
  4 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 4 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 4 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 3 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 3 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 3 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 3 hours ago

m_goldberg

88.8k873200

answered 4 hours ago

amator2357amator2357

1437

edited 3 hours ago

m_goldberg

88.8k873200

edited 3 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

Michael E2Michael E2

151k12203482

answered 1 hour ago

Michael E2Michael E2

151k12203482

answered 1 hour ago

Michael E2Michael E2

151k12203482