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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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    $\begingroup$ 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. $\endgroup$ – Clayton Feb 12 '13 at 10:04
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    $\begingroup$ Your manipulations are (useless and) correct since $I=I+e^x(\cosh x-\sinh x)+C$. $\endgroup$ – Did Feb 12 '13 at 10:05
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I wasn't sure if it was worth answering a few month old question but seeing as this was one of the first results that came on Google when searching about it I thought it might be helpful for anyone else looking as well. I'm only a student but If I'm not mistaken, following on from:

$$I:=∫e^x \ \cosh \ x\ dx,$$ therefore: $$I =∫e^x \ \cosh \ x\ dx \ = \ e^x\ \cosh\ x\ -\ e^x\ \sinh\ x \ - ∫e^x\ \cosh\ x\ dx$$ $$I=∫e^x \ \cosh \ x\ dx \ = \ e^x\ \cosh\ x\ -\ e^x\ \sinh\ x \ - I$$ $$2I =2∫e^x\ \cosh\ x\ dx = e^x\ \cosh\ x - e^x\ \sinh\ x + c, $$ so: $$I =∫e^x \cosh\ x \ dx = {e^x \ \cosh\ x - e^x \ \sinh\ x + c\over 2}. $$

I think the problem was that if you added $I:=∫e^x\ \cosh\ x\ dx$ to right side, you needed to add it to the left as well. On another note, I apologize for the lack of structure in the answer and formatting. Never had to use Mathjax before. Wish the basic tutorial was improved and made more user friendly.

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  • $\begingroup$ This is the way I would have proceeded. Keep up the good work! $\endgroup$ – Dmoreno Jun 22 '14 at 17:11
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    $\begingroup$ @Dmoreno This won't work because $\cosh''=\cosh$ and $(-1)(-1)=1$ in the integration by parts. Indeed, integration by parts yields $I=I+ e^x(\cosh(x)-\sinh(x))+C$ as Did already commented. Note that $\cos(x)-\sinh(x)=e^{-x}$. This works on $e^{-x}\cos(x)$ because $\cos''=-\cos$. $\endgroup$ – Ian Mateus Jun 22 '14 at 17:30
  • $\begingroup$ The third line should read $$I=e^x\cosh(x)-e^x\sinh(x) \color{Red}{+}I$$ $\endgroup$ – Ian Mateus Jun 22 '14 at 17:35
  • $\begingroup$ You are right @Ian. $\endgroup$ – Dmoreno Jun 22 '14 at 17:38
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$$e^x\cosh x=\frac{1}{2}e^x(e^x+e^{-x})=\frac{1}{2}(e^{2x}+1)\Longrightarrow$$

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

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

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    $\begingroup$ Without the 2 at the end. $\endgroup$ – Did Feb 12 '13 at 10:16
  • $\begingroup$ ...and before the end, too. Thanks. $\endgroup$ – DonAntonio Feb 12 '13 at 10:22
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Given

$$ \int \exp(x) \cosh(x) dx $$

Write this as

$$ \int \exp(x) \cosh(ax) dx $$

then

$$ \begin{eqnarray} \int \exp(x) \cosh(ax) dx &=& \exp(x) \cosh(ax) - a \int \exp(x) \sinh(ax) dx\\ &=& \exp(x) \cosh(ax) - a \exp(x) \sinh(ax) + a^2 \int \exp(x) \cosh(ax) dx \end{eqnarray} $$

thus

$$ \begin{eqnarray} \int \exp(x) \cosh(ax) dx &=& \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)} { (1 - a)(1+a) } \end{eqnarray} $$

whence

$$ \begin{eqnarray} \int \exp(x) \cosh(x) dx &=& \lim_{a \rightarrow 1} \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)}{ (1 - a) (1 + a) } \end{eqnarray} $$


The last part - taking the limit gives

$$ \begin{eqnarray} \int \exp(x) \cosh(x) dx &=& \lim_{a \rightarrow 1} \frac{ \exp(x) \cosh(ax) - a \exp(x) \sinh(ax)}{ (1 - a) (1 + a) }\\ &=& \lim_{a \rightarrow 1} \frac{x \exp(x) [ \sinh(ax) - a \cosh(ax)] - \exp(x) \sinh(ax)}{-2}\\ &=& \frac{1}{2} \exp(x)\sinh(x) + \frac{1}{2} x \end{eqnarray} $$

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