Finding arclength of $f(x) = e^{-x}$ from $0 \le x \le t$.

Question

Essentially, this Q&A is about evaluating the integral $$L = \int_{0}^{t}{\sqrt{1+e^{-2x}}} \mathrm{d}x, \quad t > 0$$ where $L$ is the arclength as stated in the title and employs the formula $\displaystyle{L_{\text{Arc}} = \int_{a}^{b}{\sqrt{1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2}}} \space \mathrm{d}x$

This is a Q&A, with my working posted below. Please feel welcome to share other methods, especially, if they are simpler compared to mine.

Answer 1

My working for the integral as follows:

$$L = \int_{0}^{t}{\sqrt{1+e^{-2x}}} \space \mathrm{dx} = \int_{0}^{t}{\sqrt{1+e^{-2x}}} \space \mathrm{dx} \\= \int_{0}^{t}{\left(-\dfrac{1}{2}e^{2x}\right)\left(-2e^{-2x}\right)\sqrt{1+e^{-2x}}} \space \mathrm{dx} \quad \xrightarrow{\large{u \space = \space 1 + e^{-2x} \\ du \space = \space-2e^{-2x}}} \quad \dfrac{-1}{2}\int_{2}^{1+e^{-2t}}{\dfrac{\sqrt{u}}{u - 1}} \space \mathrm{du} + \\ = \dfrac{-1}{2}\int_{2}^{1+e^{-2t}}{\dfrac{\sqrt{u}}{u - 1} \cdot \dfrac{\sqrt{u}}{\sqrt{u}}} \space \mathrm{du} = \dfrac{-1}{2}\int_{2}^{1+e^{-2t}}{2 \cdot \dfrac{\big(\sqrt{u}\big)^2}{\big(\sqrt{u}\big)^2 - 1}} \space \mathrm{d\left[\sqrt{u}\right]} = -\Large\int_{2}^{1+e^{-2t}}\small{\left(\dfrac{1}{\big(\sqrt{u} - 1 \big)\big(\sqrt{u} + 1 \big)} + 1\right)} \space \mathrm{d\left[\sqrt{u}\right]} \\ \quad \\= \left. \left[ \dfrac{-1}{2}\ln{\bigg{|} \dfrac{\sqrt{u} - 1}{\sqrt{u} + 1} \bigg{|}} - \sqrt{u} \right] \space \right|_{\;2}^{\;1 + e^{-2t}} \\ = \dfrac{-1}{2}\ln{\bigg{|} \dfrac{\sqrt{1 + e^{-2t}} - 1}{\sqrt{1 + e^{-2t}} + 1} \bigg{|}} - \sqrt{1 + e^{-2t}} + \dfrac{1}{2} \ln{\left( 3 - 2\sqrt{2} \right)} + \sqrt{2}$$

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I considered trig substitution prior to method above: $$L = \int_{0}^{t}{\sqrt{1+e^{-2x}}} \space \mathrm{dx} = \int{e^{-t}\sqrt{e^{2t}+1}} \space \mathrm{dt} + \sqrt{2} = -\int{\sqrt{\dfrac{1}{v^2}+1}} \space \mathrm{dv} + \sqrt{2} \\ \quad \xrightarrow{\large{v \space = \space \cot{y} \\ dv \space = \space -\csc^2{y}}} \quad \int{\sec{y} \cdot \csc^2{y}} \space \mathrm{dy}$$ However, with the numerous substitutions that will be tedious to reverse later, and the ugly algebra that I encountered after, I am unable to continue.

Answer 2

Quite the same method but simpler written.

$y=\sqrt{1+e^{-2x}}\longrightarrow\mathrm dx=\frac y{1-y^2}\mathrm dy,$

$$\begin{align}\int\sqrt{1+e^{-2x}}\,\mathrm dx&=\int\frac{y^2}{1-y^2}\,\mathrm dy\\ &=\int\left(-1+\frac12\left(\frac1{1-y}+\frac1{1+y}\right)\right)\,\mathrm dy\\ &=\left[-y+\frac12\ln\frac{y+1}{y-1}\right]\\ &=\left[-\sqrt{1+e^{-2x}}+\frac12\ln\frac{(1+\sqrt{1+e^{-2x}})^2}{e^{-2x}}\right]\\ &=\left[-\sqrt{1+e^{-2x}}+\ln(e^x+\sqrt{e^{2x}+1})\right]\\ \end{align}$$ hence $$\begin{align}\int_0^t\sqrt{1+e^{-2x}}\,\mathrm dx&=\left[-\sqrt{1+e^{-2x}}+\ln(e^x+\sqrt{e^{2x}+1})\right]_0^t\\ &=-\sqrt{1+e^{-2t}}+\sqrt2+\ln\frac{e^t+\sqrt{e^{2t}+1}}{1+\sqrt2}. \end{align}$$ One can also note that $\ln(u+\sqrt{u^2+1})=\sinh^{-1}(u).$

Answer 3

Consider $$\int \sqrt{1+e^u} du $$ note that if you let $u=-2x$, you have the original integral. Now set $e^u=t$, therefore $du=dt/t$: $$\int \sqrt{1+e^u} du=\int \frac{1}{t}\sqrt{1+t}\,dt $$ Now perform the substitution $1+t=z^2$: $$\int \frac{1}{t}\sqrt{1+t}\,dt =\int\frac{2z^2}{z^2-1}dz $$ and the last integral can be solved easily.