- #1
twoflower
- 368
- 0
Hi,
I have to find this one:
[tex]
\int \frac{dx}{\sqrt{1-e^{2x}}}
[/tex]
Is this right approach?
[tex]
\int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}
[/tex]
Substitution:
[tex]
t = \sqrt{1-e^{2x}}
[/tex]
[tex]
dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx
[/tex]
[tex]
e^{2x} = 1 - t^2\\
[/tex]
[tex]
\int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}} = - \int \frac{dt}{1-t^2} = - \int \frac{1-t}{1-t^2}dt - \int \frac{t}{1-t^2}dt
[/tex]
Substitution:
[tex]
y = 1 - t^2
[/tex]
[tex]
dy = -2t dt
[/tex]
[tex]
z = 1 + t
[/tex]
[tex]
dz = dt
[/tex]
[tex]
... = - \int \frac{dz}{z} - \frac{1}{2} \int \frac{dy}{y} = - \ln \left(1 + \sqrt{1-e^{2x}} \right) - \frac{1}{2} \ln \left(e^{2x} \right) + C
[/tex]
I'm afraid that the first substitution is not ok, but could someone please give me more detailed answer?
Thank you.
I have to find this one:
[tex]
\int \frac{dx}{\sqrt{1-e^{2x}}}
[/tex]
Is this right approach?
[tex]
\int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}
[/tex]
Substitution:
[tex]
t = \sqrt{1-e^{2x}}
[/tex]
[tex]
dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx
[/tex]
[tex]
e^{2x} = 1 - t^2\\
[/tex]
[tex]
\int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}} = - \int \frac{dt}{1-t^2} = - \int \frac{1-t}{1-t^2}dt - \int \frac{t}{1-t^2}dt
[/tex]
Substitution:
[tex]
y = 1 - t^2
[/tex]
[tex]
dy = -2t dt
[/tex]
[tex]
z = 1 + t
[/tex]
[tex]
dz = dt
[/tex]
[tex]
... = - \int \frac{dz}{z} - \frac{1}{2} \int \frac{dy}{y} = - \ln \left(1 + \sqrt{1-e^{2x}} \right) - \frac{1}{2} \ln \left(e^{2x} \right) + C
[/tex]
I'm afraid that the first substitution is not ok, but could someone please give me more detailed answer?
Thank you.