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$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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 asSin[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
New contributor
$endgroup$
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
calculus-and-analysis
New contributor
New contributor
edited 1 hour ago
Michael E2
151k12203482
151k12203482
New contributor
asked 4 hours ago
EmmaEmma
61
61
New contributor
New contributor
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 asSin[x]^2
$endgroup$
– enano9314
4 hours ago
$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
56 mins ago
add a comment |
1
$begingroup$
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 asSin[x]^2
$endgroup$
– enano9314
4 hours ago
$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
56 mins 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
56 mins ago
$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
56 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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}.
$endgroup$
1
$begingroup$
Recommend that you add aPlot
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 |
$begingroup$
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
$endgroup$
$begingroup$
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 |
$begingroup$
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 *)
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}.
$endgroup$
1
$begingroup$
Recommend that you add aPlot
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}.
$endgroup$
1
$begingroup$
Recommend that you add aPlot
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}.
$endgroup$
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}.
edited 3 hours ago
answered 4 hours ago
Kevin AusmanKevin Ausman
32417
32417
1
$begingroup$
Recommend that you add aPlot
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 |
1
$begingroup$
Recommend that you add aPlot
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
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
$endgroup$
$begingroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
edited 2 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
add a comment |
$begingroup$
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
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 *)
$endgroup$
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 *)
$endgroup$
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 *)
$endgroup$
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 *)
answered 46 mins ago
Michael E2Michael E2
151k12203482
151k12203482
add a comment |
add a comment |
Emma is a new contributor. Be nice, and check out our Code of Conduct.
Emma is a new contributor. Be nice, and check out our Code of Conduct.
Emma is a new contributor. Be nice, and check out our Code of Conduct.
Emma is a new contributor. Be nice, and check out our Code of Conduct.
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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 asSin[x]^2
$endgroup$
– enano9314
4 hours ago
$begingroup$
Related: math.stackexchange.com/questions/82489/…
$endgroup$
– Michael E2
56 mins ago