Finding the reason behind the value of the integral.












16












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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.)
    $endgroup$
    – Semiclassical
    Apr 2 at 14:46








  • 1




    $begingroup$
    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
















16












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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.)
    $endgroup$
    – Semiclassical
    Apr 2 at 14:46








  • 1




    $begingroup$
    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














16












16








16


9



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 2 at 21:32







Jasmine

















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








  • 1




    $begingroup$
    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




    $begingroup$
    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








  • 1




    $begingroup$
    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






1




1




$begingroup$
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










1 Answer
1






active

oldest

votes


















14












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    Apr 3 at 5:26














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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









14












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    Apr 3 at 5:26


















14












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    Apr 3 at 5:26
















14












14








14





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 3 at 5:22

























answered Apr 2 at 14:52









NewBornMATHNewBornMATH

539111




539111












  • $begingroup$
    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
    Apr 3 at 5:26




















  • $begingroup$
    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
    Apr 3 at 5:26


















$begingroup$
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
Apr 3 at 5:26






$begingroup$
Hope it helps else Let me know how i can improve my answer.
$endgroup$
– NewBornMATH
Apr 3 at 5:26




















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