Evaluate $int_0^pi frac{sin^2 nx}{sin^2 x}dx$
$begingroup$
How do I evaluate this definite integral, I'm not even getting a slightest idea on how to approach this. Tried converting into cos using double angle property but that didn't help.
$$int_0^pi frac{sin^2 nx}{sin^2 x}dx$$
definite-integrals
edited May 20 '18 at 4:04
John Glenn
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
$endgroup$
add a comment |
$begingroup$
How do I evaluate this definite integral, I'm not even getting a slightest idea on how to approach this. Tried converting into cos using double angle property but that didn't help.
$$int_0^pi frac{sin^2 nx}{sin^2 x}dx$$
definite-integrals
edited May 20 '18 at 4:04
John Glenn
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
$endgroup$
add a comment |
$begingroup$
How do I evaluate this definite integral, I'm not even getting a slightest idea on how to approach this. Tried converting into cos using double angle property but that didn't help.
$$int_0^pi frac{sin^2 nx}{sin^2 x}dx$$
definite-integrals
edited May 20 '18 at 4:04
John Glenn
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
$endgroup$
How do I evaluate this definite integral, I'm not even getting a slightest idea on how to approach this. Tried converting into cos using double angle property but that didn't help.
$$int_0^pi frac{sin^2 nx}{sin^2 x}dx$$
definite-integrals
definite-integrals
edited May 20 '18 at 4:04
John Glenn
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
edited May 20 '18 at 4:04
John Glenn
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
edited May 20 '18 at 4:04
John Glenn
1,958524
edited May 20 '18 at 4:04
John Glenn
1,958524
edited May 20 '18 at 4:04
John Glenn
1,958524
1,958524
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
asked May 20 '18 at 2:24
Aryan VermaAryan Verma
111
111
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$frac{sin nx}{sin x}=frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=sum_{r=1}^ne^{i(n+1-2r)x}$$
$$frac{sin^2 nx}{sin^2 x}
=n+sum_{(r,s)in A}e^{i(2n+2-2r-2s)x}
=n+sum_{(r,s)in A}cos(2n+2-2r-2s)x
$$
where $A$ is the set of $(r,s)$ with $r$, $sin{1,2,ldots,n}$
with $r+sne n+1$. The details of this are not important
as $int_0^pi cos mx,dx=0$ for nonzero integers $m$. Therefore
$$int_0^pifrac{sin^2 nx}{sin^2 x},dx=npi.$$
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
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add a comment |
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Let $$I_{n}=int_{0}^{pi}frac{sin^2nx}{sin^2x} dx$$
$$I_{n+1}-2I_{n}+I_{n-1}=0$$
Thus $I_{1},I_{2},I_{3},...$ are in A.P
$I_{1}=pi$
$I_{2}=2pi$
So, $I_{n}=npi$
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
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add a comment |
$begingroup$
Hint:
De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.
Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
$endgroup$
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
$begingroup$
$$frac{sin nx}{sin x}=frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=sum_{r=1}^ne^{i(n+1-2r)x}$$
$$frac{sin^2 nx}{sin^2 x}
=n+sum_{(r,s)in A}e^{i(2n+2-2r-2s)x}
=n+sum_{(r,s)in A}cos(2n+2-2r-2s)x
$$
where $A$ is the set of $(r,s)$ with $r$, $sin{1,2,ldots,n}$
with $r+sne n+1$. The details of this are not important
as $int_0^pi cos mx,dx=0$ for nonzero integers $m$. Therefore
$$int_0^pifrac{sin^2 nx}{sin^2 x},dx=npi.$$
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
add a comment |
$begingroup$
$$frac{sin nx}{sin x}=frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=sum_{r=1}^ne^{i(n+1-2r)x}$$
$$frac{sin^2 nx}{sin^2 x}
=n+sum_{(r,s)in A}e^{i(2n+2-2r-2s)x}
=n+sum_{(r,s)in A}cos(2n+2-2r-2s)x
$$
where $A$ is the set of $(r,s)$ with $r$, $sin{1,2,ldots,n}$
with $r+sne n+1$. The details of this are not important
as $int_0^pi cos mx,dx=0$ for nonzero integers $m$. Therefore
$$int_0^pifrac{sin^2 nx}{sin^2 x},dx=npi.$$
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
add a comment |
$begingroup$
$$frac{sin nx}{sin x}=frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=sum_{r=1}^ne^{i(n+1-2r)x}$$
$$frac{sin^2 nx}{sin^2 x}
=n+sum_{(r,s)in A}e^{i(2n+2-2r-2s)x}
=n+sum_{(r,s)in A}cos(2n+2-2r-2s)x
$$
where $A$ is the set of $(r,s)$ with $r$, $sin{1,2,ldots,n}$
with $r+sne n+1$. The details of this are not important
as $int_0^pi cos mx,dx=0$ for nonzero integers $m$. Therefore
$$int_0^pifrac{sin^2 nx}{sin^2 x},dx=npi.$$
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
$$frac{sin nx}{sin x}=frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=sum_{r=1}^ne^{i(n+1-2r)x}$$
$$frac{sin^2 nx}{sin^2 x}
=n+sum_{(r,s)in A}e^{i(2n+2-2r-2s)x}
=n+sum_{(r,s)in A}cos(2n+2-2r-2s)x
$$
where $A$ is the set of $(r,s)$ with $r$, $sin{1,2,ldots,n}$
with $r+sne n+1$. The details of this are not important
as $int_0^pi cos mx,dx=0$ for nonzero integers $m$. Therefore
$$int_0^pifrac{sin^2 nx}{sin^2 x},dx=npi.$$
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered May 20 '18 at 5:14
Lord Shark the UnknownLord Shark the Unknown
107k1161133
107k1161133
add a comment |
add a comment |
$begingroup$
Let $$I_{n}=int_{0}^{pi}frac{sin^2nx}{sin^2x} dx$$
$$I_{n+1}-2I_{n}+I_{n-1}=0$$
Thus $I_{1},I_{2},I_{3},...$ are in A.P
$I_{1}=pi$
$I_{2}=2pi$
So, $I_{n}=npi$
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
$endgroup$
add a comment |
$begingroup$
Let $$I_{n}=int_{0}^{pi}frac{sin^2nx}{sin^2x} dx$$
$$I_{n+1}-2I_{n}+I_{n-1}=0$$
Thus $I_{1},I_{2},I_{3},...$ are in A.P
$I_{1}=pi$
$I_{2}=2pi$
So, $I_{n}=npi$
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
$endgroup$
add a comment |
$begingroup$
Let $$I_{n}=int_{0}^{pi}frac{sin^2nx}{sin^2x} dx$$
$$I_{n+1}-2I_{n}+I_{n-1}=0$$
Thus $I_{1},I_{2},I_{3},...$ are in A.P
$I_{1}=pi$
$I_{2}=2pi$
So, $I_{n}=npi$
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
$endgroup$
Let $$I_{n}=int_{0}^{pi}frac{sin^2nx}{sin^2x} dx$$
$$I_{n+1}-2I_{n}+I_{n-1}=0$$
Thus $I_{1},I_{2},I_{3},...$ are in A.P
$I_{1}=pi$
$I_{2}=2pi$
So, $I_{n}=npi$
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
answered Dec 30 '18 at 15:49
MaverickMaverick
2,101621
2,101621
add a comment |
add a comment |
$begingroup$
Hint:
De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.
Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
$endgroup$
add a comment |
$begingroup$
Hint:
De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.
Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
$endgroup$
add a comment |
$begingroup$
Hint:
De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.
Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
$endgroup$
Hint:
De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.
Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
answered May 20 '18 at 3:36
DzoooksDzoooks
905416
905416
add a comment |
add a comment |
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