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I have been tasked to find $\int{\text{sech}(x)dx}$ using both hyperbolic and trig substitutions, for the trig substitution method I did the following. $$I=\int{\frac{2e^x}{e^{2x}+1}dx} $$ $$\text{Let} \space u=e^x \implies dx=\frac{1}{e^x}du $$ Then applying the first substitution and using the trig substitution of $u=\tan(t)$: $$\therefore I=\int\frac{2u(\frac{1}{u})}{u^2+1}du \iff \int\frac{2}{u^2+1}du$$ $$\text{Let}\space u=\tan(t) \implies du=\sec^2(t)dt$$ And simplifying: $$\therefore I=2\int{\frac{\sec^2(t)}{tan^2(t)+1}dt \iff 2\int{1dt}}$$ $$I=2t$$ And finally resubstituting variables to bring it back to in terms of $x$: $$\because t=\arctan(u) , \space u=e^x$$ $$\therefore I=2\arctan(e^x) + c$$

Which checks out on wolfram alpha, however for hyperbolic substitutions I have tried using $u=\text{sinh}(t)$ which just returns the original integral back:

$$\text{Let} \space u=\text{sinh}(t) \iff du=\text{cosh}(t)dt$$ $$\therefore I=2\int{\frac{\text{cosh}(t)}{\text{sinh}^2(t)+1}dt} \iff 2\int{\frac{1}{\text{cosh}(t)}dt}$$

I have also tried using the substitution of $u=\text{csch}(t)$ which also led back to the original integral, too my knowledge there happen to be no other useful hyperbolic substitutions to carry out on this integral.

Have I made a mistake in my integration or am I missing some other useful substitution which may be carried out here?

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  • $\begingroup$ Additionally, I am well aware I could have used the identity $\int{\frac{1}{u^2+1}}=\arctan(u) + c$ however I was told by my professor to not do so $\endgroup$ Dec 10, 2020 at 12:17
  • $\begingroup$ Does this answer your question? $\endgroup$
    – Toby Mak
    Dec 10, 2020 at 12:25
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    $\begingroup$ @TobyMak I did read through that post but the majority consensus was that its not doable without involving complex constants $\endgroup$ Dec 10, 2020 at 12:32
  • $\begingroup$ Incidentally, it has a name. $\endgroup$
    – J.G.
    Dec 10, 2020 at 15:00

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Note that for any $ x\in\mathbb{R} $, $ \cosh{x}=\frac{\cosh{\left(2\times\frac{x}{2}\right)}}{1}=\frac{\cosh^{2}{\left(\frac{x}{2}\right)}+\sinh^{2}{\left(\frac{x}{2}\right)}}{\cosh^{2}{\left(\frac{x}{2}\right)}-\sinh^{2}{\left(\frac{x}{2}\right)}}=\frac{1+\tanh^{2}{\left(\frac{x}{2}\right)}}{1-\tan^{2}{\left(\frac{x}{2}\right)}} $.

Substituting $ \left\lbrace\begin{matrix}y=\tanh{\left(\frac{x}{2}\right)}\\ \mathrm{d}x=\frac{2\,\mathrm{d}y}{1-y^{2}}\ \ \ \ \ \ \ \ \ \ \end{matrix}\right. $, we get : \begin{aligned}\int{\frac{\mathrm{d}x}{\cosh{x}}}&=2\int{\frac{\mathrm{d}y}{1+y^{2}}}\\&=2\arctan{y}+C\\&=2\arctan{\left(\tanh{\left(\frac{x}{2}\right)}\right)}+C\end{aligned}

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