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Integration Help

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2

$begingroup$

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??

---

Update: Even with the syntax fixed, Mathematica does not solve it, with or without assumptions:

Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), x, 0, Pi/2,
 Assumptions -> n > 0 && n [Element] Integers]

calculus-and-analysis

share|improve this question

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Apr 17 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/questions/82489/…
  $endgroup$
  – Michael E2
  Apr 18 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Was it closed simply because of a syntax error by the OP? The integral with fixed syntax can be evaluated at specific values for n, as shown in two of the answers, but it cannot be evaluated with a unspecified parameter n. (Of course we often get this kind of question, which reveals limitiations of Integrate.)
  $endgroup$
  – Michael E2
  Apr 18 at 13:47
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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??

---

Update: Even with the syntax fixed, Mathematica does not solve it, with or without assumptions:

Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), x, 0, Pi/2,
 Assumptions -> n > 0 && n [Element] Integers]

calculus-and-analysis

share|improve this question

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Apr 17 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/questions/82489/…
  $endgroup$
  – Michael E2
  Apr 18 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Was it closed simply because of a syntax error by the OP? The integral with fixed syntax can be evaluated at specific values for n, as shown in two of the answers, but it cannot be evaluated with a unspecified parameter n. (Of course we often get this kind of question, which reveals limitiations of Integrate.)
  $endgroup$
  – Michael E2
  Apr 18 at 13:47
  

  
  
  
  

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??

---

Update: Even with the syntax fixed, Mathematica does not solve it, with or without assumptions:

Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), x, 0, Pi/2,
 Assumptions -> n > 0 && n [Element] Integers]

calculus-and-analysis

share|improve this question

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

$endgroup$

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

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

Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), x, 0, Pi/2,
 Assumptions -> n > 0 && n [Element] Integers]

share|improve this question

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

share|improve this question

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

edited Apr 21 at 16:14

Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

asked Apr 17 at 20:43

KatieKatie

3344 bronze badges

3 Answers
3

active

oldest

votes

5

$begingroup$

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 Apr 17 at 21:40

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Recommend that you add a Plot to make it easier to understand why the result is a constant.
  $endgroup$
  – Bob Hanlon
  Apr 17 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good suggestion. Editing.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:38
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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 Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, just realized that, thanks for pointing it out.
  $endgroup$
  – amator2357
  Apr 17 at 21:09
  

  
  
  
  

add a comment
 | 

3

$begingroup$

A common trick (see this Math.SE post):

$$int_a^b f(x) ; dx
buildrel x = a+b-u over = -int_b^a f(a + b - u) ; du
= int_a^b f(a + b - x) ; dx, ,$$
so therefore
$$int_a^b f(x) ; dx = int_b^a f(x) + f(a + b - x) over 2 ; dx, .$$

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]]
(* π/4 *)

share|improve this answer

edited Apr 18 at 13:38

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

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5

$begingroup$

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 Apr 17 at 21:40

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Recommend that you add a Plot to make it easier to understand why the result is a constant.
  $endgroup$
  – Bob Hanlon
  Apr 17 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good suggestion. Editing.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:38
  

  
  
  
  

add a comment
 | 

5

$begingroup$

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 Apr 17 at 21:40

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Recommend that you add a Plot to make it easier to understand why the result is a constant.
  $endgroup$
  – Bob Hanlon
  Apr 17 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good suggestion. Editing.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:38
  

  
  
  
  

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 Apr 17 at 21:40

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

$endgroup$

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

share|improve this answer

edited Apr 17 at 21:40

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

answered Apr 17 at 21:03

Kevin AusmanKevin Ausman

1,24333 silver badges1616 bronze badges

3

$begingroup$

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 Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, just realized that, thanks for pointing it out.
  $endgroup$
  – amator2357
  Apr 17 at 21:09
  

  
  
  
  

add a comment
 | 

3

$begingroup$

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 Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you have the parentheses in the wrong place relative to the original question. Right idea for the solution, though.
  $endgroup$
  – Kevin Ausman
  Apr 17 at 21:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, just realized that, thanks for pointing it out.
  $endgroup$
  – amator2357
  Apr 17 at 21:09
  

  
  
  
  

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 Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

$endgroup$

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

share|improve this answer

edited Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

edited Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

edited Apr 17 at 22:26

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

answered Apr 17 at 20:57

amator2357amator2357

8201010 bronze badges

3

$begingroup$

A common trick (see this Math.SE post):

$$int_a^b f(x) ; dx
buildrel x = a+b-u over = -int_b^a f(a + b - u) ; du
= int_a^b f(a + b - x) ; dx, ,$$
so therefore
$$int_a^b f(x) ; dx = int_b^a f(x) + f(a + b - x) over 2 ; dx, .$$

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]]
(* π/4 *)

share|improve this answer

edited Apr 18 at 13:38

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

A common trick (see this Math.SE post):

$$int_a^b f(x) ; dx
buildrel x = a+b-u over = -int_b^a f(a + b - u) ; du
= int_a^b f(a + b - x) ; dx, ,$$
so therefore
$$int_a^b f(x) ; dx = int_b^a f(x) + f(a + b - x) over 2 ; dx, .$$

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]]
(* π/4 *)

share|improve this answer

edited Apr 18 at 13:38

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

A common trick (see this Math.SE post):

$$int_a^b f(x) ; dx
buildrel x = a+b-u over = -int_b^a f(a + b - u) ; du
= int_a^b f(a + b - x) ; dx, ,$$
so therefore
$$int_a^b f(x) ; dx = int_b^a f(x) + f(a + b - x) over 2 ; dx, .$$

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]]
(* π/4 *)

share|improve this answer

edited Apr 18 at 13:38

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

$endgroup$

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]]
(* π/4 *)

share|improve this answer

edited Apr 18 at 13:38

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges

answered Apr 18 at 0:26

Michael E2Michael E2

159k1313 gold badges219219 silver badges518518 bronze badges