Question

Provide all details and eveluate the integral dx/e^x square root of 1-e^_2x

Answer 1

Given integral is, $I = \int \frac{dx}{e^x\sqrt{1-e^{-2x}}}$

$I = \int \frac{dx}{e^x\sqrt{1-e^{-2x}}} = \int \frac{e^{-x}dx}{\sqrt{1-e^{-2x}}}$

Let $e^{-x} = t \implies e^{-x}dx = -dt$

$I= -\int \frac{dt}{\sqrt{1-t^2}}$

Using integration formula, $\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}(x) + C$

Then,

$I= -\int \frac{dt}{\sqrt{1-t^2}} =- sin^{-1}(t) + C$

Putting value of t,

$I = -sin^{-1}(e^{-x}) + C$