Find the arclength from 0 to 1 on the function $y =\arcsin(e^{-x})$.

Question

I need to find the arclength from 0 to 1 on the function $y = \arcsin(e^{-x})$. I know that $$y' = \frac{-e^{-x}}{\sqrt{1-e^{-2x}}}$$ By applying arc length formula I get this nasty integral:

$$\int_{a}^{b} \sqrt{1 + (y')^2} \ dx=\int_{0}^{1}\sqrt{\frac{1-2e^{-2x}}{1-e^{-2x}}}dx$$

However I am unsure how to evaluate this. Putting it through software, I seem to get complex values which I am not sure are possible. Could someone offer an attempt?

Answer 1

hint

$$(y')^2=\frac{e^{-2x}}{1-e^{-2x}}$$

$$1+(y')^2=\frac{1}{1-e^{-2x}}$$

The length is

$$L=\int_a^b\frac{e^{-2x}dx}{e^{-2x}\sqrt{1-e^{-2x}}}$$

Now, make substitution $$u=\sqrt{1-e^{-2x}}$$

to get $$L=\int_A^B\frac{du}{1-u^2}$$

Now, use partial fraction decomposition.

Answer 2

You have committed a mistake. $$\int_{a}^{b} \sqrt{1 + (y')^2} \ dx=\int_{0}^{1}\sqrt{\frac{1+e^{-2x}-e^{-2x}}{1-e^{-2x}}}dx=\int_{0}^{1}\sqrt{\frac{1}{1-e^{-2x}}}dx=\int_{0}^{1}\frac{e^{x}}{\sqrt{e^{2x}-1}}dx$$ Can you finish it now?