0
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$$\int \frac{2}{e^x+e^{-x}} dx$$ 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 $\tan(.)$!).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Ahhh. This really helped, thanks! $\endgroup$ – ialmao Oct 16 '17 at 0:28
1
$\begingroup$

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 $$\int_{-\infty}^{\infty}\text{sech}(x)\,dx=\lim_{b\to\infty}\left(2\arctan(e^x)\Big|_{-b}^{b}\right)=2\left(\frac{\pi}{2}\right)=\pi$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.