I've been doing this question for a bit and I can't get my head around it.
I'm meant to evaluate $I = \int \frac1{\cosh(x)} dx$ by substitution, using $e^x$ as $u$.

So far I have,

$$\int \frac{2}{e^x+e^{-x}} dx$$

$u=e^x, du=e^xdx$

"$dx=du/e^x$" --Not sure if I can do this.

$$2\int \frac{\frac{1}{e^x}}{e^x+\frac{1}{e^x}} du\\ 2\int \frac{\frac{1}{u}}{u+\frac{1}{u}} du$$

And at this stage I get stuck. It looks kind of a function and its derivative I think? If I represent it as:

$$2\int u^{-1}(u+u^{-1})^{-1}$$

But not enough for me to be able to obviously recognise it and doctor it to use integration by substitution again. I think I'm overthinking the question and it's probably a lot simpler than this. Thanks heaps for any help.

  • $\begingroup$ The solution may be very similar to $\int\frac1{\cos(x)}~dx$ $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 23:39
  • $\begingroup$ $$\int\frac{\frac1u}{u+\frac1u}=\int\frac1{u^2+1}=\arctan(\dots$$ $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 23:41
  • $\begingroup$ As per the "Not sure if I can do this" step, consider rewriting it as $x=\ln(u)$ and $dx=\frac1udu$, which is equivalent to what you reached. $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 23:43
  • $\begingroup$ Thank you for your help! $\endgroup$ – ialmao Oct 16 '17 at 0:27

$$\int \frac{2}{e^x+e^{-x}} dx$$ Let $u=e^x\implies du=e^xdx=udx$. So, $$\int \frac{2}{e^x+e^{-x}} d=2\int\dfrac{\frac{1}{u}}{u+\frac{1}{u}}du=2\int\dfrac{1}{u^2+1}du=?$$ Hints: multiplied by $\frac{u}{u}$ in second to last step. If you don't recognize that final integral, try using the substitution $u=\tan(v)$ to finish (and recall the derivative of $\tan(.)$!).

  • $\begingroup$ Ahhh. This really helped, thanks! $\endgroup$ – ialmao Oct 16 '17 at 0:28

So we have $$\int \frac{1}{\cosh(x)}\,dx=\int\text{sech}{(x)}\,dx=\int \frac{2}{e^x+e^{-x}}\,dx$$ Now we substitute $x=\ln(u)$. $dx=\frac{1}{u}du$ $$\int \frac{2}{e^x+e^{-x}}\,dx=2\int\left(\frac{1}{u+u^{-1}}\right)\left(\frac{du}{u}\right)=2\int\frac{du}{u^2+1}$$ The last integral is a standard one; substitute $u=\tan(\theta)$. You should then get $$2\int\frac{du}{u^2+1}=2\arctan(u)+C$$ Remembering that $x=\ln(u)$ or equivalently that $u=e^x$ we get $$\int \text{sech}{(x)}\,dx=2\arctan(e^x)+C$$ Added bonus: From this, we get that $$\int_{-\infty}^{\infty}\text{sech}(x)\,dx=\lim_{b\to\infty}\left(2\arctan(e^x)\Big|_{-b}^{b}\right)=2\left(\frac{\pi}{2}\right)=\pi$$

  • $\begingroup$ I really see no reason for the "added bonus". $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 0:29
  • $\begingroup$ I just thought that it was a rather beautiful result that might interest ialmao $\endgroup$ – Zachary Oct 16 '17 at 12:28
  • $\begingroup$ I suppose it is beautiful, but I'll probably say that about every non-trivial integral that evaluates to $\pi$. $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 13:30

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