Question

Question #41c4f

Answer 1

`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:

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`

Answer 2

`(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`