# Integration question - use of substitution

1. Mar 27, 2005

### twoflower

Hi,

I have to find this one:

$$\int \frac{dx}{\sqrt{1-e^{2x}}}$$

Is this right approach?

$$\int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}$$

Substitution:
$$t = \sqrt{1-e^{2x}}$$

$$dt = - \frac{e^{2x}}{\sqrt{1-e^{2x}}} dx$$

$$e^{2x} = 1 - t^2\\$$

$$\int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}} = - \int \frac{dt}{1-t^2} = - \int \frac{1-t}{1-t^2}dt - \int \frac{t}{1-t^2}dt$$

Substitution:
$$y = 1 - t^2$$

$$dy = -2t dt$$

$$z = 1 + t$$

$$dz = dt$$

$$... = - \int \frac{dz}{z} - \frac{1}{2} \int \frac{dy}{y} = - \ln \left(1 + \sqrt{1-e^{2x}} \right) - \frac{1}{2} \ln \left(e^{2x} \right) + C$$

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

Thank you.

2. Mar 27, 2005

### 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.

Last edited: Mar 27, 2005
3. Mar 27, 2005

### twoflower

Thank you dextercioby,

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

4. Mar 27, 2005

### twoflower

Btw we haven't learned hyperbolic functions.

5. Mar 27, 2005

### dextercioby

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

Daniel.

6. Mar 27, 2005

### dextercioby

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

Daniel.

7. Mar 27, 2005

### 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:

$$\int f(g(x)) g'(x) dx$$

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

8. Mar 27, 2005

### dextercioby

Which integral...?

Daniel.

9. Mar 27, 2005

### twoflower

This one:

$$\int \frac{e^{2x} dx}{e^{2x} \sqrt{1-e^{2x}}}$$

10. Mar 27, 2005

### 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.

11. Mar 27, 2005

### 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..

12. Mar 27, 2005

### 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

$$e^{2x}<1$$​

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

Daniel.