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$$\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}$$

Hello,

I have to substitute with $$t = \cosh (x)$$

I just don´t know what to do with the e-function.

Thanks.

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  • $\begingroup$ you have to substitute or you would like to ? $\endgroup$ – MrYouMath Jun 1 '16 at 16:58
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    $\begingroup$ Note that e^x = cosh(x) + sinh(x) $\endgroup$ – florence Jun 1 '16 at 16:58
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    $\begingroup$ I don´t want to, it is given. $\endgroup$ – user295683 Jun 1 '16 at 17:01
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As OP said "I have to ...".

$$\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}=\int_{0}^{1}{ \frac{ \cosh(x)+\sinh(x)}{ \cosh(x)} \,dx}=\int_{0}^{1}{1+ \frac{\sinh(x)}{ \cosh(x)} \,dx}$$ $$=\int_{0}^{1}{1 \,dx}+\int_{0}^{1}{\frac{\sinh(x)}{ \cosh(x)} \,dx}$$

First integral is trivial. Now substitute $u=\cosh(x)$ and $du=\sinh(x)dx$ in the second integral.

EDIT: You also have to substitute the boundary values $u_{lower}=\cosh(0)$ and $u_{upper}=\cosh(1)$.

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    $\begingroup$ Thank you very much for your work, but also for other´s work. $\endgroup$ – user295683 Jun 1 '16 at 17:07
  • $\begingroup$ Nice! The only answer that actually answers the OP's question. $\endgroup$ – Thomas Jun 1 '16 at 17:08
  • $\begingroup$ One more question, I have to substitute the borders too, how do I substitute 1 and 0 now? $\endgroup$ – user295683 Jun 1 '16 at 17:11
  • $\begingroup$ Yes you have to substitute the borders. $\endgroup$ – MrYouMath Jun 1 '16 at 17:16
  • $\begingroup$ So I use cosh (1) and cosh (0)? $\endgroup$ – user295683 Jun 1 '16 at 17:20
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$$\int_{0}^{1}\frac{e^x}{\cosh x}\,dx = \int_{0}^{1}\frac{2e^{2x}}{e^{2x}+1}\,dx = \left.\log(1+e^{2x})\right|_{0}^{1}=\color{red}{\log\left(\frac{e^2+1}{2}\right)}. $$

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$$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ and then get the value of $e^x$ in terms of t.

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Note that $\cosh(x) = \frac{e^x + e^{-x}}{2}$, so $$\frac{e^{x}}{\cosh(x)} = \frac{2 e^{x}}{e^{x}+e^{-x}} = \frac{2}{1+e^{-2x}}$$

Now subsitute $u=e^{-x}$ (which gives $du = -e^{-x}dx = -u dx$) to find: $$\int \frac{2}{1+e^{-2x}} dx = \int \frac{-2}{u(1+u^2)} du$$

The rest is straightforward.


The remainder of the work can be completed using partial fraction decomposition:

$$\frac{1}{u(1+u^2)} = \frac{A}{u} + \frac{Bu+C}{1+u^2}$$

Which gives: $$1=A(1+u^2) + (Bu+C)u$$

Setting $u=0$ tells us $A=1$. Then $$1=1+u^2 + Bu^2 + Cu$$ $$0=(1+B)u^2 + Cu$$ and we see that $B=-1$ and $C=0$.

Therefore $$\frac{1}{u(1+u^2)} = \frac{1}{u} - \frac{u}{1+u^2}.$$

Now we have $$\int \frac{-2}{u(1+u^2)} du = -2 \int \frac{1}{u} -\frac{u}{1+u^2} du$$

$$= -2 \left( \ln(u) - \int \frac{u}{1+u^2} du \right)$$

Finish by using another substitution $w=1+u^2$. This gives $dw = 2u du$ and $$= -2 \left( \ln(u) - \frac{1}{2}\int \frac{1}{w} dw \right) = -2 \left( \ln(u) - \frac{\ln(1+u^2)}{2}\right)$$

Finally, replace $u$ by $e^{-x}$ to find:

$$\int \frac{2}{1+e^{-2x}} dx = 2x + \ln(1+e^{-2x}) +C.$$

Then plug in the bounds.

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  • $\begingroup$ Could you elaborate on how to solve it? Thanks. $\endgroup$ – user295683 Jun 1 '16 at 18:59
  • $\begingroup$ Great, thanks for this. $\endgroup$ – user295683 Jun 1 '16 at 21:14

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