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

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.

• 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. – Lutz Lehmann Nov 19 '16 at 23:37
• @LutzL, Thank you for your comment. How did you integrate exp(x)*sinh(ax) by parts? – Frank Nov 19 '16 at 23:42
• Just follow the answer of johannesvalks in you linked question. But it is really a much too complicated way to get this simple integral. – Lutz Lehmann Nov 20 '16 at 0:20

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

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

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

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

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.

• The reason I want to use integration by parts is for educational enrichment. Thank you. – Frank Nov 20 '16 at 0:36