1
$\begingroup$

Is it possible to integrate $e^{x}\ \sinh(x)$ using only integration by parts?

If so, how would it be done?

I read the following Stack Overflow article, Integrate $\int e^x \cosh(x) \: dx$ by parts, which contains an arithmetic mistake.

Any help is greatly appreciated.

$\endgroup$
  • $\begingroup$ You can follow the answer about modifying the function to integrate $e^x\sinh(ax)$ and perform the limit $a\to 1$ in the result. -- The direct approach will again lead to a tautological result. $\endgroup$ – Lutz Lehmann Nov 19 '16 at 23:37
  • $\begingroup$ @LutzL, Thank you for your comment. How did you integrate exp(x)*sinh(ax) by parts? $\endgroup$ – Frank Nov 19 '16 at 23:42
  • $\begingroup$ Just follow the answer of johannesvalks in you linked question. But it is really a much too complicated way to get this simple integral. $\endgroup$ – Lutz Lehmann Nov 20 '16 at 0:20
2
$\begingroup$

$\int e^x \sinh x = e^x \sinh x-\int e^x \cosh x=e^x \sinh x - \int e^x (\sinh x + e^{-x})=e^x \sinh x - \int e^x \sinh x - \int 1 = e^x \sinh x - x - \int e^x \sinh x $

Gives: $\int e^x \sinh x = \frac{e^x \sinh x-x}{2}$

Edit: I am using $\cosh x = \sinh x +e^{-x}$.

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

$e^x \sinh(x) = e^x \frac{e^x - e^{-x}}{2} = \frac{1}{2} \left( e^{2x} -1 \right)$. Why would you want to use integration by parts?

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Sure. Set $\mathrm{e}^x = \cos(-\mathrm{i}x) + \mathrm{i} \sin( -\mathrm{i} x)$ and then use the usual method for trigonometric functions.

Although Jon Warneke's answer is a much better way to attack this integral.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The reason I want to use integration by parts is for educational enrichment. Thank you. $\endgroup$ – Frank Nov 20 '16 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.