Integration of two exponential multiplied by each otherIntegrating $sin^2(x)$ using imaginary numbers.Need...
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Integration of two exponential multiplied by each other
Integrating $sin^2(x)$ using imaginary numbers.Need help with integration by partsIntegrate $int frac{ln(sin x)}{sin^2 x},mathrm dx.$Integral $int frac{sqrt{16-x^2}}{x} mathrm{d}x$Solid Angle IntegrationDouble Integral of an Exponential Function with an Absolute Value in the Numerator of the ExponentComplex integral with exponential and tangentShow the value of an integral using integration by parts.How to calculate $int xe^{1/x^2} dx$How to integrate $int 2xe^{x^2-y^2}cos(2xy)- 2ye^{x^2-y^2}sin(2xy) mathrm dy$?
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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
edited 2 hours ago
Thomas Shelby
3,6342525
asked 3 hours ago
articatarticat
163
$endgroup$
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
edited 2 hours ago
Thomas Shelby
3,6342525
asked 3 hours ago
articatarticat
163
$endgroup$
$begingroup$
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
1 hour ago
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
edited 2 hours ago
Thomas Shelby
3,6342525
asked 3 hours ago
articatarticat
163
$endgroup$
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
integration
edited 2 hours ago
Thomas Shelby
3,6342525
asked 3 hours ago
articatarticat
163
edited 2 hours ago
Thomas Shelby
3,6342525
asked 3 hours ago
articatarticat
163
edited 2 hours ago
Thomas Shelby
3,6342525
edited 2 hours ago
Thomas Shelby
3,6342525
edited 2 hours ago
Thomas Shelby
3,6342525
3,6342525
asked 3 hours ago
articatarticat
163
asked 3 hours ago
articatarticat
163
asked 3 hours ago
articatarticat
163
163
$begingroup$
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
1 hour ago
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
2 Answers
2
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 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
$endgroup$
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
$endgroup$
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 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
$endgroup$
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 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
$endgroup$
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 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
edited 3 hours ago
answered 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
answered 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
answered 3 hours ago
Thomas ShelbyThomas Shelby
3,6342525
3,6342525
add a comment |
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
$endgroup$
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
$endgroup$
add a comment |
$begingroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
$endgroup$
Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
answered 3 hours ago
Graham KempGraham Kemp
86.1k43478
86.1k43478
add a comment |
add a comment |
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$begingroup$
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
1 hour ago