twoflower
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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.