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I was presented with the following question:

Evaluate: $$I:=\int e^x \cosh(x) \: dx$$

So we let $u=\cosh(x)$ and $v'=e^x$, therefore $u'=\sinh(x)$ and $v=e^x$.

Therefore $I:=e^x \cosh(x)-\int e^x \sinh(x)=e^x \cosh(x) - (e^x \sinh(x) - \int e^x \cosh(x)\: dx$

Thus: $I = e^x \cosh(x) - e^x \sinh(x) + I \implies 0 = e^x(\cosh(x)-\sinh(x))=e^x e^{-x} = 1$

How is this valid?

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I would just use the definition of $\cosh(x)$, that is, $$\cosh(x)=\frac{e^x+e^{-x}}{2}.$$ Then it's easy to integrate, and you don't even need to use parts. –  Clayton Feb 12 '13 at 10:04
Your manipulations are (useless and) correct since $I=I+e^x(\cosh x-\sinh x)+C$. –  Did Feb 12 '13 at 10:05
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1 Answer

$$e^x\cosh x=\frac{1}{2}e^x(e^x+e^{-x})=\frac{1}{2}(e^{2x}+1)\Longrightarrow$$

$$\int e^x\cosh x\,dx=\frac{1}{2}\int(e^{2x}+1)dx=\frac{1}{4}e^{2x}+\frac{x}{2}+C=\frac{1}{4}\left(e^{2x}+2x\right)+C$$

As you did by parts doesn't work since $\,e^x(\cosh x-\sinh x)=e^x(e^{-x})=1\,$...

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Without the 2 at the end. –  Did Feb 12 '13 at 10:16
...and before the end, too. Thanks. –  DonAntonio Feb 12 '13 at 10:22
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