How to calculate the following integrals? [closed]
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How the calculate the following integrals? Therein $D$ is a constant.
$$(1);;int_{0}^{2pi}frac{1}{1-Dcdotcostheta} dtheta$$
and
$$(2);;int_{0}^{2pi}frac{1}{1+Dcdotcostheta} dtheta$$
calculus integration
asked Dec 11 '18 at 11:16
JLiuJLiu
396
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closed as off-topic by Nosrati, RRL, user10354138, Paramanand Singh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, user10354138, Paramanand Singh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 4 more comments
$begingroup$
How the calculate the following integrals? Therein $D$ is a constant.
$$(1);;int_{0}^{2pi}frac{1}{1-Dcdotcostheta} dtheta$$
and
$$(2);;int_{0}^{2pi}frac{1}{1+Dcdotcostheta} dtheta$$
calculus integration
asked Dec 11 '18 at 11:16
JLiuJLiu
396
$endgroup$
closed as off-topic by Nosrati, RRL, user10354138, Paramanand Singh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, user10354138, Paramanand Singh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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I suppose $dtheta$ is in the numerator?
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– Bernard
Dec 11 '18 at 11:17
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Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
1
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What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
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– Zacky
Dec 11 '18 at 11:20
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Aren't those two integrals the very same one...?
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– DonAntonio
Dec 11 '18 at 11:20
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@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24
|
show 4 more comments
$begingroup$
How the calculate the following integrals? Therein $D$ is a constant.
$$(1);;int_{0}^{2pi}frac{1}{1-Dcdotcostheta} dtheta$$
and
$$(2);;int_{0}^{2pi}frac{1}{1+Dcdotcostheta} dtheta$$
calculus integration
asked Dec 11 '18 at 11:16
JLiuJLiu
396
$endgroup$
How the calculate the following integrals? Therein $D$ is a constant.
$$(1);;int_{0}^{2pi}frac{1}{1-Dcdotcostheta} dtheta$$
and
$$(2);;int_{0}^{2pi}frac{1}{1+Dcdotcostheta} dtheta$$
calculus integration
calculus integration
asked Dec 11 '18 at 11:16
JLiuJLiu
396
asked Dec 11 '18 at 11:16
JLiuJLiu
396
edited Dec 11 '18 at 11:21
JLiu
asked Dec 11 '18 at 11:16
JLiuJLiu
396
asked Dec 11 '18 at 11:16
JLiuJLiu
396
asked Dec 11 '18 at 11:16
JLiuJLiu
396
396
closed as off-topic by Nosrati, RRL, user10354138, Paramanand Singh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, user10354138, Paramanand Singh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, RRL, user10354138, Paramanand Singh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, user10354138, Paramanand Singh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I suppose $dtheta$ is in the numerator?
$endgroup$
– Bernard
Dec 11 '18 at 11:17
$begingroup$
Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
1
$begingroup$
What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
$endgroup$
– Zacky
Dec 11 '18 at 11:20
$begingroup$
Aren't those two integrals the very same one...?
$endgroup$
– DonAntonio
Dec 11 '18 at 11:20
$begingroup$
@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24
|
show 4 more comments
$begingroup$
I suppose $dtheta$ is in the numerator?
$endgroup$
– Bernard
Dec 11 '18 at 11:17
$begingroup$
Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
1
$begingroup$
What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
$endgroup$
– Zacky
Dec 11 '18 at 11:20
$begingroup$
Aren't those two integrals the very same one...?
$endgroup$
– DonAntonio
Dec 11 '18 at 11:20
$begingroup$
@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24
$begingroup$
I suppose $dtheta$ is in the numerator?
$endgroup$
– Bernard
Dec 11 '18 at 11:17
$begingroup$
I suppose $dtheta$ is in the numerator?
$endgroup$
– Bernard
Dec 11 '18 at 11:17
$begingroup$
Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
$begingroup$
Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
1
1
$begingroup$
What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
$endgroup$
– Zacky
Dec 11 '18 at 11:20
$begingroup$
What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
$endgroup$
– Zacky
Dec 11 '18 at 11:20
$begingroup$
Aren't those two integrals the very same one...?
$endgroup$
– DonAntonio
Dec 11 '18 at 11:20
$begingroup$
Aren't those two integrals the very same one...?
$endgroup$
– DonAntonio
Dec 11 '18 at 11:20
$begingroup$
@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24
$begingroup$
@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24
|
show 4 more comments
1 Answer
1
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oldest
votes
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$$I_2=int_0^{2pi}frac{dtheta}{1+Dcos(theta)}=int_0^{pi}frac{dtheta}{1+Dcos(theta)}+int_{pi}^{2pi}frac{dtheta}{1+Dcos(theta)}$$ Let $theta=tan(frac{x}{2})$
$$I_2=int_0^{infty}frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}+int_{-infty}^0frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}=2int_{-infty}^{infty}frac{dx}{1+x^2+D-Dx^2}=2int_{-infty}^infty frac{dx}{(1-D)x^2+(1+D)}=frac{2}{1-D}int_{-infty}^{infty}frac{dx}{x^2+frac{1+D}{1-D}}=frac{2pi}{sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
$endgroup$
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$I_2=int_0^{2pi}frac{dtheta}{1+Dcos(theta)}=int_0^{pi}frac{dtheta}{1+Dcos(theta)}+int_{pi}^{2pi}frac{dtheta}{1+Dcos(theta)}$$ Let $theta=tan(frac{x}{2})$
$$I_2=int_0^{infty}frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}+int_{-infty}^0frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}=2int_{-infty}^{infty}frac{dx}{1+x^2+D-Dx^2}=2int_{-infty}^infty frac{dx}{(1-D)x^2+(1+D)}=frac{2}{1-D}int_{-infty}^{infty}frac{dx}{x^2+frac{1+D}{1-D}}=frac{2pi}{sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
$endgroup$
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
add a comment |
$begingroup$
$$I_2=int_0^{2pi}frac{dtheta}{1+Dcos(theta)}=int_0^{pi}frac{dtheta}{1+Dcos(theta)}+int_{pi}^{2pi}frac{dtheta}{1+Dcos(theta)}$$ Let $theta=tan(frac{x}{2})$
$$I_2=int_0^{infty}frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}+int_{-infty}^0frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}=2int_{-infty}^{infty}frac{dx}{1+x^2+D-Dx^2}=2int_{-infty}^infty frac{dx}{(1-D)x^2+(1+D)}=frac{2}{1-D}int_{-infty}^{infty}frac{dx}{x^2+frac{1+D}{1-D}}=frac{2pi}{sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
$endgroup$
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
add a comment |
$begingroup$
$$I_2=int_0^{2pi}frac{dtheta}{1+Dcos(theta)}=int_0^{pi}frac{dtheta}{1+Dcos(theta)}+int_{pi}^{2pi}frac{dtheta}{1+Dcos(theta)}$$ Let $theta=tan(frac{x}{2})$
$$I_2=int_0^{infty}frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}+int_{-infty}^0frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}=2int_{-infty}^{infty}frac{dx}{1+x^2+D-Dx^2}=2int_{-infty}^infty frac{dx}{(1-D)x^2+(1+D)}=frac{2}{1-D}int_{-infty}^{infty}frac{dx}{x^2+frac{1+D}{1-D}}=frac{2pi}{sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
$endgroup$
$$I_2=int_0^{2pi}frac{dtheta}{1+Dcos(theta)}=int_0^{pi}frac{dtheta}{1+Dcos(theta)}+int_{pi}^{2pi}frac{dtheta}{1+Dcos(theta)}$$ Let $theta=tan(frac{x}{2})$
$$I_2=int_0^{infty}frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}+int_{-infty}^0frac{1}{1+Dfrac{1-x^2}{1+x^2}}frac{2dx}{1+x^2}=2int_{-infty}^{infty}frac{dx}{1+x^2+D-Dx^2}=2int_{-infty}^infty frac{dx}{(1-D)x^2+(1+D)}=frac{2}{1-D}int_{-infty}^{infty}frac{dx}{x^2+frac{1+D}{1-D}}=frac{2pi}{sqrt{1-D^2}}$$ Hopefully all my computation is correct, the other integral can be done in the same way. Just use Weierstrass Substitution.
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
answered Dec 11 '18 at 15:37
aledenaleden
2,362511
2,362511
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
add a comment |
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
$begingroup$
Many thanks for your contributions!
$endgroup$
– JLiu
Dec 12 '18 at 2:17
add a comment |
$begingroup$
I suppose $dtheta$ is in the numerator?
$endgroup$
– Bernard
Dec 11 '18 at 11:17
$begingroup$
Thanks for your checking, $dtheta$ is not in the numerator, I re-edit the problem
$endgroup$
– JLiu
Dec 11 '18 at 11:19
1
$begingroup$
What is the difference between $(1)$ and $(2)$? Also have you tried to use this: en.wikipedia.org/wiki/Tangent_half-angle_substitution ? After splitting the interval from $0, pi$ to $pi , 2pi$ cause $tan frac{x}{2}$ is undefined there.
$endgroup$
– Zacky
Dec 11 '18 at 11:20
$begingroup$
Aren't those two integrals the very same one...?
$endgroup$
– DonAntonio
Dec 11 '18 at 11:20
$begingroup$
@ClaudeLeibovici Salut Claude, mon ami!
$endgroup$
– DonAntonio
Dec 11 '18 at 11:24