How do you integrate $\int \frac{1}{\sqrt{e^{2 x} - 2 e^{x}}} d x$ $int 1/sqrt(e^(2x)-2e^x)dx$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{e^{2 x} - 2 e^{x}}} d x$ $int 1/sqrt(e^(2x)-2e^x)dx$ using trigonometric substitution?

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Answer 1 · 2017-02-01T23:30:07

$\int \frac{d x}{\sqrt{e^{2 x} - 2 e^{x}}} = \sqrt{1 - 2 e^{- x}} + C$ $int (dx)/sqrt(e^(2x)-2e^x) = sqrt(1-2e^(-x)) +C$

Explanation:

Complete the square under the root:

$\frac{1}{\sqrt{e^{2 x} - 2 e^{x}}} = \frac{1}{\sqrt{e^{2 x} - 2 e^{x} + 1 - 1}} = \frac{1}{\sqrt{{(e^{x} - 1)}^{2} - 1}}$ $1/sqrt(e^(2x)-2e^x) = 1/sqrt(e^(2x)-2e^x+1-1)=1/sqrt((e^x-1)^2-1)$

Now substitute:

$e^{x} - 1 = t$ $e^x-1 = t$

$x = ln (1 + t)$ $x = ln(1+t)$

$d x = \frac{d t}{1 + t}$ $dx = (dt)/(1+t)$

We have:

$\int \frac{d x}{\sqrt{e^{2 x} - 2 e^{x}}} = \int \frac{d x}{\sqrt{{(e^{x} - 1)}^{2} - 1}} = \int \frac{d t}{(1 + t) \sqrt{t^{2} - 1}}$ $int (dx)/sqrt(e^(2x)-2e^x) = int (dx)/sqrt((e^x-1)^2-1) = int (dt) / ((1+t)sqrt(t^2-1))$

Substitute again:

$t = sec u$ $t=secu$

$d t = sec u tan u d u$ $dt = secu tan u du$

to have:

$\int \frac{d t}{(1 + t) \sqrt{t^{2} - 1}} = \int \frac{sec u tan u d u}{(1 + sec u) \sqrt{{sec}^{2} u - 1}}$ $int (dt) / ((1+t)sqrt(t^2-1)) = int (sec u tan u du) / ((1+secu)sqrt(sec^2u-1))$

Using the identity:

${tan}^{2} u = {sec}^{2} u - 1$ $tan^2u = sec^2u-1$

this becomes:

$\int \frac{sec u tan u d u}{(1 + sec u) \sqrt{{tan}^{2} u}} = \int \frac{sec u}{1 + sec u} d u = \int \frac{1}{cos u} \frac{1}{1 + \frac{1}{cos u}} d u = \int \frac{d u}{1 + cos u}$ $int (sec u tan u du) / ((1+secu)sqrt(tan^2u)) = int secu/(1+secu)du = int 1/cosu 1/(1+1/cosu) du = int (du)/(1+cosu)$

Use now:

${cos}^{2} (\frac{u}{2}) = \frac{1 + cos u}{2}$ $cos^2(u/2) = (1+cosu)/2$

$\int \frac{d u}{1 + cos u} = \frac{1}{2} \int \frac{d u}{{cos}^{2} (\frac{u}{2})} = \int \frac{d (\frac{u}{2})}{{cos}^{2} (\frac{u}{2})} = tan (\frac{u}{2}) + C$ $int (du)/(1+cosu) = 1/2int (du)/cos^2(u/2) = int (d(u/2))/cos^2(u/2) = tan(u/2)+C$

To undo the substitutions, consider first the identity:

$tan (\frac{u}{2}) = \sqrt{\frac{1 - cos u}{1 + cos u}}$ $tan(u/2) = sqrt((1-cosu)/(1+cosu))$

and as $cos u = \frac{1}{sec u} = \frac{1}{t}$ $cosu = 1/secu = 1/t$ :

$tan (\frac{u}{2}) = \sqrt{\frac{1 - \frac{1}{t}}{1 + \frac{1}{t}}} = \sqrt{\frac{t - 1}{t + 1}} = \sqrt{\frac{e^{x} - 2}{e^{x}}} = \sqrt{1 - 2 e^{- x}}$ $tan(u/2) = sqrt ( ( 1-1/t)/(1+1/t)) = sqrt((t-1)/(t+1)) = sqrt ((e^x-2)/e^x) =sqrt(1-2e^(-x))$

so that eventually:

$\int \frac{d x}{\sqrt{e^{2 x} - 2 e^{x}}} = \sqrt{1 - 2 e^{- x}} + C$ $int (dx)/sqrt(e^(2x)-2e^x) = sqrt(1-2e^(-x)) +C$

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