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$$\int_0^{\infty}\frac{\sinh x}{1+\cosh^2x}dx$$

Here's what I've attempted:

Using the identity $1+\cosh^2x=\sinh^2x$ I got:

$$\int_0^{\infty}\frac{\sinh x}{\sinh^2x}dx=\int_0^{\infty}\frac1{\sinh x}dx$$

Is this right? However, where can I go from here?

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    $\begingroup$ why not just the substitution $\ \cosh x=t$ ? $\endgroup$ – Mosk Apr 30 '15 at 16:06
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    $\begingroup$ Or substitute $e^x = t$ and use partial fractions. $\endgroup$ – abnry Apr 30 '15 at 16:07
  • $\begingroup$ Is what I've done wrong? $\endgroup$ – RobChem Apr 30 '15 at 16:08
  • $\begingroup$ Generally $$\sinh^2(x) \neq 1 + \cosh^2(x).$$ $\endgroup$ – Chantry Cargill Apr 30 '15 at 16:13
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    $\begingroup$ The identity is $~1+\sinh^2x=\cosh^2x,~$ not the other way around. :-$)$ $\endgroup$ – Lucian Apr 30 '15 at 16:13
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The identity you are using is not correct. The correct one is

\begin{equation} \cosh^2 x -\sinh^2 x = 1 \end{equation}

Moreover this integration is a simple change of variables:

\begin{equation} \begin{aligned} &\int_0^{+\infty} \frac{\sinh x}{1 + \cosh^2 x} dx\\ =& \int_0^{+\infty} \frac{1}{1 + \cosh^2 x} d(\cosh x)\\ =& \int_1^{+\infty} \frac{1}{1+y^2} dy\\[0.5em] =&~ \tan^{-1} y~|_{+\infty} - \tan^{-1} y~|_{1} \\[0.5em] =&~ \frac{\pi}{2} - \frac{\pi}{4}\\[0.5em] =&~ \frac{\pi}{4} \end{aligned} \end{equation}

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Set $x=\text{arccosh(y)}$, $dx=\frac{1}{\sqrt{y^2-1}}dy$ ,

furthermore we know that $\sinh(\text{arccosh(y)})=\sqrt{y^2-1}$ .

Note that this part exactly cancels the stuff from the differential and therefore we get: $$ I=\int_1^{\infty}dy\frac{1}{1+y^2}=\text{arctan}(\infty)-\text{arctan}(1)=\frac{\pi}{4} $$

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