Solve integral:`int (dx)/sqrt(1-e^(2x))`
5 Answers
crmhaske | College Teacher | (Level 3) Associate Educator
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`
crmhaske | College Teacher | (Level 3) Associate Educator
`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.
anuroopash9 | eNotes Newbie
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,
``
` `
` `
` `
` `
` `
` `
` `
` `
` `` `
` `
` `
` `
` `
` `
` `
oldnick | (Level 1) Valedictorian
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!
oldnick | (Level 1) Valedictorian
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)`