Finding the reason behind the value of the integral.
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I was just trying to find $$int_{0}^{pi / 2}frac{sin{9x}}{sin{x}},dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
Edit: It can also be noted that $$int_{a{pi}}^{bpi }frac{sin{9x}}{sin{x}},dx =(b-a){pi}$$ where $a,b$ are integers.
definite-integrals
asked Apr 2 at 14:23
JasmineJasmine
393213
$endgroup$
add a comment |
$begingroup$
I was just trying to find $$int_{0}^{pi / 2}frac{sin{9x}}{sin{x}},dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
Edit: It can also be noted that $$int_{a{pi}}^{bpi }frac{sin{9x}}{sin{x}},dx =(b-a){pi}$$ where $a,b$ are integers.
definite-integrals
asked Apr 2 at 14:23
JasmineJasmine
393213
$endgroup$
12
$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
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– Semiclassical
Apr 2 at 14:46
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I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
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– David K
Apr 2 at 21:12
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@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34
add a comment |
$begingroup$
I was just trying to find $$int_{0}^{pi / 2}frac{sin{9x}}{sin{x}},dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
Edit: It can also be noted that $$int_{a{pi}}^{bpi }frac{sin{9x}}{sin{x}},dx =(b-a){pi}$$ where $a,b$ are integers.
definite-integrals
asked Apr 2 at 14:23
JasmineJasmine
393213
$endgroup$
I was just trying to find $$int_{0}^{pi / 2}frac{sin{9x}}{sin{x}},dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
Edit: It can also be noted that $$int_{a{pi}}^{bpi }frac{sin{9x}}{sin{x}},dx =(b-a){pi}$$ where $a,b$ are integers.
definite-integrals
definite-integrals
asked Apr 2 at 14:23
JasmineJasmine
393213
asked Apr 2 at 14:23
JasmineJasmine
393213
edited Apr 2 at 21:32
Jasmine
asked Apr 2 at 14:23
JasmineJasmine
393213
asked Apr 2 at 14:23
JasmineJasmine
393213
asked Apr 2 at 14:23
JasmineJasmine
393213
393213
12
$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
Apr 2 at 14:46
$begingroup$
I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
$endgroup$
– David K
Apr 2 at 21:12
$begingroup$
@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34
add a comment |
12
$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
Apr 2 at 14:46
$begingroup$
I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
$endgroup$
– David K
Apr 2 at 21:12
$begingroup$
@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34
12
12
$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
Apr 2 at 14:46
$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
Apr 2 at 14:46
$begingroup$
I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
$endgroup$
– David K
Apr 2 at 21:12
$begingroup$
I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
$endgroup$
– David K
Apr 2 at 21:12
$begingroup$
@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34
$begingroup$
@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
Consider $I(n)=int_{0}^{pi/2} frac{sin(nx)}{sin x} dx$
$$I(2m+1)-I(2m-1)=int_{0}^{pi/2} frac{sin(2m+1)x-sin(2m-1)x}{sin{x}} dx=int_{0}^{pi/2} frac{2sin(x)cos(2mx)}{sin{x}} dx$$
$$implies 2int_{0}^{pi/2} cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=frac{pi}{2}$.
Edit
(As OP has changed the question a bit)
Now consider$I(n)=int_{api}^{bpi} frac{sin(nx)}{sin x} dx$
From(1) $$implies 2int_{api}^{bpi} cos(2mx)dx=2bigg[frac{sin(2mx)}{2m}bigg]_{api}^{bpi}$$
$$implies I(2m+1)-I(2m-1)=frac{1}{n} bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided ${a,b} in mathbb{Z} $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$
Hence
$$ bbox[5px,border:2px solid blue]
{int_{api}^{bpi} frac{sin(nx)}{sin x} dx=(b-a)pi
}
$$
When n is odd.
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
$endgroup$
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
Hint
Consider $I(n)=int_{0}^{pi/2} frac{sin(nx)}{sin x} dx$
$$I(2m+1)-I(2m-1)=int_{0}^{pi/2} frac{sin(2m+1)x-sin(2m-1)x}{sin{x}} dx=int_{0}^{pi/2} frac{2sin(x)cos(2mx)}{sin{x}} dx$$
$$implies 2int_{0}^{pi/2} cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=frac{pi}{2}$.
Edit
(As OP has changed the question a bit)
Now consider$I(n)=int_{api}^{bpi} frac{sin(nx)}{sin x} dx$
From(1) $$implies 2int_{api}^{bpi} cos(2mx)dx=2bigg[frac{sin(2mx)}{2m}bigg]_{api}^{bpi}$$
$$implies I(2m+1)-I(2m-1)=frac{1}{n} bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided ${a,b} in mathbb{Z} $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$
Hence
$$ bbox[5px,border:2px solid blue]
{int_{api}^{bpi} frac{sin(nx)}{sin x} dx=(b-a)pi
}
$$
When n is odd.
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
$endgroup$
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
add a comment |
$begingroup$
Hint
Consider $I(n)=int_{0}^{pi/2} frac{sin(nx)}{sin x} dx$
$$I(2m+1)-I(2m-1)=int_{0}^{pi/2} frac{sin(2m+1)x-sin(2m-1)x}{sin{x}} dx=int_{0}^{pi/2} frac{2sin(x)cos(2mx)}{sin{x}} dx$$
$$implies 2int_{0}^{pi/2} cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=frac{pi}{2}$.
Edit
(As OP has changed the question a bit)
Now consider$I(n)=int_{api}^{bpi} frac{sin(nx)}{sin x} dx$
From(1) $$implies 2int_{api}^{bpi} cos(2mx)dx=2bigg[frac{sin(2mx)}{2m}bigg]_{api}^{bpi}$$
$$implies I(2m+1)-I(2m-1)=frac{1}{n} bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided ${a,b} in mathbb{Z} $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$
Hence
$$ bbox[5px,border:2px solid blue]
{int_{api}^{bpi} frac{sin(nx)}{sin x} dx=(b-a)pi
}
$$
When n is odd.
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
$endgroup$
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
add a comment |
$begingroup$
Hint
Consider $I(n)=int_{0}^{pi/2} frac{sin(nx)}{sin x} dx$
$$I(2m+1)-I(2m-1)=int_{0}^{pi/2} frac{sin(2m+1)x-sin(2m-1)x}{sin{x}} dx=int_{0}^{pi/2} frac{2sin(x)cos(2mx)}{sin{x}} dx$$
$$implies 2int_{0}^{pi/2} cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=frac{pi}{2}$.
Edit
(As OP has changed the question a bit)
Now consider$I(n)=int_{api}^{bpi} frac{sin(nx)}{sin x} dx$
From(1) $$implies 2int_{api}^{bpi} cos(2mx)dx=2bigg[frac{sin(2mx)}{2m}bigg]_{api}^{bpi}$$
$$implies I(2m+1)-I(2m-1)=frac{1}{n} bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided ${a,b} in mathbb{Z} $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$
Hence
$$ bbox[5px,border:2px solid blue]
{int_{api}^{bpi} frac{sin(nx)}{sin x} dx=(b-a)pi
}
$$
When n is odd.
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
$endgroup$
Hint
Consider $I(n)=int_{0}^{pi/2} frac{sin(nx)}{sin x} dx$
$$I(2m+1)-I(2m-1)=int_{0}^{pi/2} frac{sin(2m+1)x-sin(2m-1)x}{sin{x}} dx=int_{0}^{pi/2} frac{2sin(x)cos(2mx)}{sin{x}} dx$$
$$implies 2int_{0}^{pi/2} cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=frac{pi}{2}$.
Edit
(As OP has changed the question a bit)
Now consider$I(n)=int_{api}^{bpi} frac{sin(nx)}{sin x} dx$
From(1) $$implies 2int_{api}^{bpi} cos(2mx)dx=2bigg[frac{sin(2mx)}{2m}bigg]_{api}^{bpi}$$
$$implies I(2m+1)-I(2m-1)=frac{1}{n} bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided ${a,b} in mathbb{Z} $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$
Hence
$$ bbox[5px,border:2px solid blue]
{int_{api}^{bpi} frac{sin(nx)}{sin x} dx=(b-a)pi
}
$$
When n is odd.
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
edited 2 days ago
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
answered Apr 2 at 14:52
NewBornMATHNewBornMATH
520111
520111
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
add a comment |
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
Apr 2 at 20:52
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
2 days ago
add a comment |
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$begingroup$
The following identity seems like it may help:$$frac{sin((n+1/2)theta}{sin(theta/2)}=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
Apr 2 at 14:46
$begingroup$
I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
$endgroup$
– David K
Apr 2 at 21:12
$begingroup$
@David I checked for few other rational a and b and they satisfied but you are right.
$endgroup$
– Jasmine
Apr 2 at 21:34