Integrate The Function E^x[sinh X] | Math Forums

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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Integrate the function e^x[sinh x]
  • Thread starter ChloeG
  • Start date Feb 6, 2011
  • Tags exsinh function integrate
C

ChloeG

Joined Feb 2011 49 Posts | 0+ I am trying to integrate e^x[sinh x] dx integrating by parts e^x[sinh x] dx=e^x[cosh x] - integral e^x[cosh x] dx =e^x[cosh x] - e^x[sinh x] + integral e^x[sinh x] dx This does not make sense. Should I begin by substituting the definition of sinh x in the original integral? Is the reason it did not make sense the constant of integration or is e^x an improper function? MarkFL

MarkFL

Joined Jul 2010 12K Posts | 548+ St. Augustine, FL., U.S.A.'s oldest city I think I would use the definition \sinh x\equiv\frac{1}{2}\(e^x - e^{-x}\) to write the integral as: \frac{1}{2}\int e^{2x}-1\,dx=\frac{1}{2}\(\frac{1}{2}e^{2x}-x\)+C=\frac{1}{4}\(e^{2x}-2x\)+C
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ChloeG

Joined Feb 2011 49 Posts | 0+ Thank you. :) skipjack

skipjack

Joined Dec 2006 25K Posts | 4K+
ChloeG said: This does not make sense. Click to expand...
It makes sense, but merely reminds you that e^x(cosh x - sinh x) = e^x(e^(-x)) = 1 (a constant). Using your method, ?e^(ax)sinh(x)dx = (e^{ax}cosh(x) - ae^{ax}sinh(x))/(1 - a²) + C unless a = ±1. As a ? 1, by de l'Hôpital's rule, this tends to e^x(sinh(x) - x(cosh(x) - sinh(x)))/2 + C = (e^x sinh(x) - x)/2 + C. This result's derivative = (e^x cosh(x) + e^x sinh(x) - 1)/2                                = (e^x (e^(-x) + sinh(x)) + e^x sinh(x) - 1)/2                                = e^x sinh(x).
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ChloeG

Joined Feb 2011 49 Posts | 0+ Thank you. B

bah

Joined Oct 2014 1 Posts | 0+ the gambia e^xsinhx Login or Register / Reply

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