Integration question - use of substitution

[twoflower]

Hi,

I have to find this one:

<br /> \int \frac{dx}{\sqrt{1-e^{2x}}}<br />

Is this right approach?

<br /> \int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}<br />

Substitution:
<br /> t = \sqrt{1-e^{2x}}<br />

<br /> dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx<br />

<br /> e^{2x} = 1 - t^2\\<br />

<br /> \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 <br />

Substitution:
<br /> y = 1 - t^2<br />

<br /> dy = -2t dt<br />


<br /> z = 1 + t<br />

<br /> dz = dt<br />

<br /> ... = - \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<br />

I'm afraid that the first substitution is not ok, but could someone please give me more detailed answer?

Thank you.

 

[dextercioby]

Well,the last sign should be a "+" in the term with "ln of e^{2x}"...

\int \frac{dt}{1-t^{2}} can be don also using the substitution

t=\tanh u

Daniel.

 

[twoflower]

Thank you dextercioby,

could you tell me whether the use of the first substitution in my approach is ok?

 

[twoflower]

Btw we haven't learned hyperbolic functions.

 

[dextercioby]

I told u,everything is okay,except for the last sign.

Daniel.

 

[dextercioby]

It's okay.U could do it by simple fraction decomposition,but u said u needed to do it by substitution.

Daniel.

 

[twoflower]

Ok, thank you. But I have one doubt about my approach anyway

The theorem about substition says we can use the substitution in case we have something like this:

<br /> \int f(g(x)) g&#039;(x) dx<br />

But...I'm afraid this is not exactly my case. My integral just doesn't have this form...

 

[dextercioby]

Which integral...?

Daniel.

 

[twoflower]

This one:

<br /> \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}<br />

 

[dextercioby]

Well,look at it this way

x\rightarrow t(x)\rightarrow \frac{e^{2x}}{e^{2x}\sqrt{1-e^{2x}}}

and u see that f(x) is your initial function & g(x) is t(x)=sqrt(1-e^(2x)) ...

U needn't worry about the form of the functions.Just make the substitutions which would provide simpler forms for the integrals.

Daniel.

 

[twoflower]

Ok, so we're quite free to use substitutions if it helps us. I worried about whether my substitution meets the requirements of the theorem..

 

[dextercioby]

Since u haven't provided the intervals on which u wish to integrate that function,then u can do pretty much everything...

For example,you function can be integrated only on the domain in which

e^{2x}&lt;1​

,so that would set a condition on the variable u want to use as a substitution for "x"...

Daniel.