Question #41c4f

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Question #41c4f

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Answer 1 · 2017-11-30T07:23:43

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

Explanation:

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$\int \frac{d x}{\sqrt{e^{2 x} - 1}}$ $intdx/sqrt(e^(2x)-1)$

We will solve this by Trigonometric Substitution:

In the right triangle below:

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If $c = e^{x}$ $c=e^x$ and $a = 1$ $a=1$ we can use the Pythagoras' formula to solve for the length of side $b$ $b$

$c^{2} = a^{2} + b^{2}$ $c^2=a^2+b^2$

$b^{2} = c^{2} - a^{2}$ $b^2=c^2-a^2$

$b = \sqrt{c^{2} - a^{2}} = \sqrt{e^{2 x} - 1}$ $b=sqrt(c^2-a^2)=sqrt(e^(2x)-1$

$sec θ = \frac{c}{a} = e^{x}$ $sectheta=c/a=e^x$

$tan θ = \frac{b}{a} = \sqrt{e^{2 x} - 1}$ $tantheta=b/a=sqrt(e^(2x)-1$

$sec θ tan θ d (θ) = e^{x} d x$ $secthetatanthetad(theta)=e^xdx$

$d x = \frac{sec θ tan θ d (θ)}{e^{x}} = \frac{sec θ tan θ d (θ)}{sec θ} = tan θ d (θ)$ $dx=(secthetatanthetad(theta))/e^x=(secthetatanthetad(theta))/sectheta=tanthetad(theta)$

We have all the pieces to substitute:

$\int \frac{d x}{\sqrt{e^{2 x} - 1}} = \int (\frac{1}{\sqrt{e^{2 x} - 1}}) (d x) = \int \frac{1}{tan θ} \cdot tan θ d (θ) = \int d (θ) = θ + C$ $intdx/sqrt(e^(2x)-1)=int(1/sqrt(e^(2x)-1))(dx)=int1/tantheta*tanthetad(theta)=intd(theta)=theta+C$

Now we can substitute back:

From $tan θ = \sqrt{e^{2 x} - 1}$ $tantheta=sqrt(e^(2x)-1$ we have:

$θ = arctan \sqrt{e^{2 x} - 1}$ $theta=arctansqrt(e^(2x)-1$

Therefore:

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

Answer 2 · 2017-11-30T15:24:51

$\frac{d x}{\sqrt{e^{2 x} - 1}} = a r c sec (e^{x}) + C$ $(dx)/sqrt(e^(2x)-1)=arcsec(e^x)+C$

Explanation:

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

= $\frac{e^{x} \cdot d x}{e^{x} \cdot \sqrt{e^{2 x} - 1}}$ $(e^x*dx)/[e^x*sqrt(e^(2x)-1)]$

After using $e^{x} = sec u$ $e^x=secu$ and $e^{x} \cdot d x = sec u \cdot tan u$ $e^x*dx=secu*tanu$ substitution, this integral became,

$\frac{sec u \cdot tan u \cdot d u}{sec u \cdot \sqrt{{(sec u)}^{2} - 1}}$ $(secu*tanu*du)/[secu*sqrt((secu)^2-1)]$

= $\frac{tan u \cdot d u}{\sqrt{{(tan u)}^{2}}}$ $(tanu*du)/sqrt((tanu)^2)$

= $\frac{tan u \cdot d u}{tan u}$ $(tanu*du)/tanu$

= $d u$ $du$

= $u + C$ $u+C$

After using $e^{x} = sec u$ $e^x=secu$ and $u = a r c sec (e^{x})$ $u=arcsec(e^x)$ inverse transforms, I found

$\frac{d x}{\sqrt{e^{2 x} - 1}} = a r c sec (e^{x}) + C$ $(dx)/sqrt(e^(2x)-1)=arcsec(e^x)+C$

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