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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$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Thomas Shelby

3,6342525




3,6342525










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


















  • $begingroup$
    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










2 Answers
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Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      4












      $begingroup$

      Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$






          share|cite|improve this answer











          $endgroup$



          Recall that $e^acdot e^b=e^{a+b} $. So we can write $$intexp( jcdotphicdot x)cdot exp(jcdot kcdot xcdotsin theta) mathrm dx=intexpleft(( jcdot phi+jcdot kcdotsin theta)xright) mathrm dx=dfrac1 { jcdotphi+jcdot kcdotsin theta}expleft(( jcdot phi+jcdot kcdotsin theta)xright).$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Thomas ShelbyThomas Shelby

          3,6342525




          3,6342525























              2












              $begingroup$

              Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint: exponentiation rules. $$exp(acdot x)cdotexp(bcdot x)=exp((a+b)cdot x)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Graham KempGraham Kemp

                  86.1k43478




                  86.1k43478






























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