Question #41c4f

2 Answers
Nov 30, 2017

intdx/sqrt(e^(2x)-1)=arctansqrt(e^(2x)-1)+C

Explanation:

.

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 and a=1 we can use the Pythagoras' formula to solve for the length of side b

c^2=a^2+b^2

b^2=c^2-a^2

b=sqrt(c^2-a^2)=sqrt(e^(2x)-1

sectheta=c/a=e^x

tantheta=b/a=sqrt(e^(2x)-1

secthetatanthetad(theta)=e^xdx

dx=(secthetatanthetad(theta))/e^x=(secthetatanthetad(theta))/sectheta=tanthetad(theta)

We have all the pieces to substitute:

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 tantheta=sqrt(e^(2x)-1 we have:

theta=arctansqrt(e^(2x)-1

Therefore:

intdx/sqrt(e^(2x)-1)=arctansqrt(e^(2x)-1)+C

Nov 30, 2017

(dx)/sqrt(e^(2x)-1)=arcsec(e^x)+C

Explanation:

(dx)/sqrt(e^(2x)-1)

=(e^x*dx)/[e^x*sqrt(e^(2x)-1)]

After using e^x=secu and e^x*dx=secu*tanu substitution, this integral became,

(secu*tanu*du)/[secu*sqrt((secu)^2-1)]

=(tanu*du)/sqrt((tanu)^2)

=(tanu*du)/tanu

=du

=u+C

After using e^x=secu and u=arcsec(e^x) inverse transforms, I found

(dx)/sqrt(e^(2x)-1)=arcsec(e^x)+C