# Solve integral:`int (dx)/sqrt(1-e^(2x))`

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Step 1: integration by substitution

`u = 2x -gt du=2dx -gt dx=1/2du`

`= 1/2int(du)/(sqrt(1-e^u))`

Step 2: integration by substitution

`v = e^u -gt dv=e^udu -gt du=(dv)/v`

`= 1/2int(dv)/(vsqrt(1-v))`

Step 3: integration by substitution

`w = 1-v -gt v = 1-w`

`dw = -dv`

`=1/2int(dw)/((-1+w)sqrt(w))`

Step 4: integration by substitution

`z = sqrt(w) -gt w = z^2`

`dz = (dw)/(2sqrt(w)) -gt dw = 2zdz`

`= 1/2(2)int(zdz)/((-1+z^2)z)`

`= int(dz)/(-1+z^2)`

`= -int(dx)/(1-z^2)`

`= -tanh^-1(z) + C`

Final step: Resubstitute, w, v, and u:

`z=sqrt(w)=sqrt(1-v)=sqrt(1-e^u)=sqrt(1-e^2x)`

`int(dx)/(sqrt(1-e^(2x)))=-tanh^-1(sqrt(1-e^(2x)))+C`

**Sources:**

`f(x)= -tanh^-1(sqrt(1-e^(2x)) + C` is not an incorrect solution.

The derivative of tanh^-1(x) is `1/(1-x^2)` .

The solution that you have proposed is just tanh^-1 expressed on the complex plane.

**Sources:**

integration by substitution:

**`u=e^x`**

**`du=e^x.dx`**

**`dx=(du)/e^x=(du)/u`**

**` `****then, **

** **** **

`=- sec h^-1(u) +C`

` `

`=-sec h^-1 (e^x) +C`

i.e,

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Ya'r right, the problem is now to link it with my solution:

`int 1/sqrt(1-e^(2x)) dx` settings: `e^x= sin z` `e^xdx=cosz dz`

`dx=cot z dz`

`int 1/sqrt(1-e^(2x)) dx=` `int dz/sinz =` `ln[tan (z/2)]+C`

Now: `sin z=2sin(z/2)cos(z/2)= (2tan(z/2))/(1+tan^2(z/2))`

Settings `tan (z/2)=u`

`sin z +u^2 sinz-2u=0`

`u=(2+-sqrt(4-4sin^2z))/(2sinz )=(1+-sqrt(1-sin^2z))/sinz` `=(1+-sqrt(1-e^(2x)))/(e^(2x))`

So:

`int (dx)/sqrt(1-e^(2x))=ln(1+sqrt(1-e^(2x)))-2x +C`

But it doesn't look like right!

WROOOOOOONG: Cmhaske:

`-int dz/(1-z^2)=-int dz/((1+z)(1-z))=-1/2int (1/(1+z)+1/(1-z)) dz=`

`=-1/2 log((1+z)/(1-z))+c=1/2 log ((1-z)/(1+z)) +c=`

`=ln (sqrt((1-z)/(1+z))) +c`

indeed:

`d/dz log(sqrt((1-z)/(1+z)))=` `sqrt((1+z)/(1-z)) xx (-1/2)(sqrt(1+z)/sqrt(1-z)+ sqrt(1-z)/sqrt(1+z))/(1+z)` `=sqrt((1+z)/(1-z)) xx (-1/2) 2/(sqrt(1-z^2)(1+z))=` `-1/sqrt(1-z)xx1/((1+z)sqrt(1-z))=-1/(1-z^2)`