Mixing
Find upper boundary of an absolute value of a integral
0
$begingroup$
I can't prove that
$mid int_0^{pi}frac{Rcosx}{R^2cos^2x+1}dxmid leq int_0^pifrac{R}{R^2-1}dx$.
Could you help me with a hint?
calculus definite-integrals
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
$endgroup$
add a comment |
0
$begingroup$
I can't prove that
$mid int_0^{pi}frac{Rcosx}{R^2cos^2x+1}dxmid leq int_0^pifrac{R}{R^2-1}dx$.
Could you help me with a hint?
calculus definite-integrals
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
$endgroup$
$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
1
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00
add a comment |
0
0
0
$begingroup$
I can't prove that
$mid int_0^{pi}frac{Rcosx}{R^2cos^2x+1}dxmid leq int_0^pifrac{R}{R^2-1}dx$.
Could you help me with a hint?
calculus definite-integrals
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
$endgroup$
I can't prove that
$mid int_0^{pi}frac{Rcosx}{R^2cos^2x+1}dxmid leq int_0^pifrac{R}{R^2-1}dx$.
Could you help me with a hint?
calculus definite-integrals
calculus definite-integrals
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
edited Feb 2 at 20:17
Novak Djokovic
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
asked Feb 2 at 12:30
Novak DjokovicNovak Djokovic
45038
45038
$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
1
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00
add a comment |
$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
1
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00
$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
1
1
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00
add a comment |
1 Answer
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$begingroup$
You asked for a hint, so here it is:
Make the substitution $u=x-frac{pi}{2}$ and then use the fact that $sin(u)$ is an odd function.
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
1
$begingroup$
You asked for a hint, so here it is:
Make the substitution $u=x-frac{pi}{2}$ and then use the fact that $sin(u)$ is an odd function.
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
add a comment |
1
$begingroup$
You asked for a hint, so here it is:
Make the substitution $u=x-frac{pi}{2}$ and then use the fact that $sin(u)$ is an odd function.
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
add a comment |
1
1
1
$begingroup$
You asked for a hint, so here it is:
Make the substitution $u=x-frac{pi}{2}$ and then use the fact that $sin(u)$ is an odd function.
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
You asked for a hint, so here it is:
Make the substitution $u=x-frac{pi}{2}$ and then use the fact that $sin(u)$ is an odd function.
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Feb 3 at 0:25
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
add a comment |
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
Note that I hesitated to answer your question because (1) you gave no evidence that you had tried to solve it yourself and (2) your history shows that when other made serious attempts to answer your past questions, you neither selected nor upvoted their answers. I suspect this is why no one had responded to your question hours after you asked it.
$endgroup$
– John Wayland Bales
Feb 3 at 0:32
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
$begingroup$
I am not supposed to solve the integrals, but to use inequalities instead. And, that is what I can't do. So, substitution you've suggested won't work.Thank you very much! I will do my best to up vote answers.
$endgroup$
– Novak Djokovic
Feb 3 at 17:05
add a comment |
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$begingroup$
Did you at least evaluate either integral? If so, you should edit your question to show what you found.
$endgroup$
– John Wayland Bales
Feb 3 at 0:11
$begingroup$
I am not sure if you have a typo in your question. As far as I can tell, the right hand side integral is equal to $pifrac{R}{R^2-1}$ which is less than zero if $R<-1$ or $0<R<1$. In those cases, you can't prove the result because the result is false. If $R>1$, I suggest you graph $cos(x)/(R^2 cos^2 x +1)$ for $R=2$ or $R=3$. I think that will give you a hint about the value of the left hand side.
$endgroup$
– irchans
Feb 3 at 0:54
$begingroup$
I was not supposed to evaluate integrals. Only to show inequality,
$endgroup$
– Novak Djokovic
Feb 3 at 17:04
1
$begingroup$
Since, as pointed out by @irchans the right side can be negative, making the inequality false, I suspect the expression is supposed to be $dfrac{R}{R^2+1}$ rather than $dfrac{R}{R^2-1}$. In that case the inequality is true because the integral on the left is identically zero.
$endgroup$
– John Wayland Bales
Feb 3 at 22:05
$begingroup$
I was thinking the same, it doesn't work for $R>1$ too.
$endgroup$
– Novak Djokovic
Feb 4 at 18:00