$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx$ [closed]
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I am having confusion on how to go about integrating this integral:
$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$
I attempted by using integration by parts but that didn't work.
integration exponential-function indefinite-integrals
edited Feb 26 at 7:55
user21820
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
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closed as off-topic by user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會 Feb 26 at 12:11
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." – user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am having confusion on how to go about integrating this integral:
$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$
I attempted by using integration by parts but that didn't work.
integration exponential-function indefinite-integrals
edited Feb 26 at 7:55
user21820
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
$endgroup$
closed as off-topic by user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會 Feb 26 at 12:11
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." – user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
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– Gus
Feb 26 at 3:42
add a comment |
$begingroup$
I am having confusion on how to go about integrating this integral:
$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$
I attempted by using integration by parts but that didn't work.
integration exponential-function indefinite-integrals
edited Feb 26 at 7:55
user21820
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
$endgroup$
I am having confusion on how to go about integrating this integral:
$$int[exp( jcdotphicdot x)cdot exp(jcdot kcdot xcdot sin theta)] mathrm dx.$$
I attempted by using integration by parts but that didn't work.
integration exponential-function indefinite-integrals
integration exponential-function indefinite-integrals
edited Feb 26 at 7:55
user21820
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
edited Feb 26 at 7:55
user21820
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
edited Feb 26 at 7:55
user21820
39.3k543154
edited Feb 26 at 7:55
user21820
39.3k543154
edited Feb 26 at 7:55
user21820
39.3k543154
39.3k543154
asked Feb 26 at 2:00
articatarticat
314
asked Feb 26 at 2:00
articatarticat
314
asked Feb 26 at 2:00
articatarticat
314
314
closed as off-topic by user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會 Feb 26 at 12:11
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." – user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會 Feb 26 at 12:11
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." – user21820, mrtaurho, José Carlos Santos, Delta-u, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
Feb 26 at 3:42
add a comment |
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
Feb 26 at 3:42
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
Feb 26 at 3:42
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
Feb 26 at 3:42
add a comment |
2 Answers
2
active
oldest
votes
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Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
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add a comment |
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Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
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2
$begingroup$
exponentiation rules - yes it does
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– qwr
Feb 26 at 7:17
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
$endgroup$
add a comment |
$begingroup$
Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
$endgroup$
add a comment |
$begingroup$
Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
$endgroup$
Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
edited Feb 26 at 2:09
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
answered Feb 26 at 2:05
Thomas ShelbyThomas Shelby
3,7492525
3,7492525
add a comment |
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
$endgroup$
2
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
$endgroup$
2
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
$endgroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
answered Feb 26 at 2:04
Graham KempGraham Kemp
86.3k43478
86.3k43478
2
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
add a comment |
2
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
2
2
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
$begingroup$
exponentiation rules - yes it does
$endgroup$
– qwr
Feb 26 at 7:17
add a comment |
$begingroup$
This is a bit off-topic, but you might have encountered problems, with integrands like exp(x) * sin(x) where you needed to do two rounds of integration by parts. Using complex exponentials lets you avoid integration by parts in those problems by putting everything in one exponential.
$endgroup$
– Gus
Feb 26 at 3:42