# Domain of antiderivative problem

#### hasek747

So I have the following problem:

$$\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx$$

I realize this can be computed a simple U-substitution (followed by rationalizing the function), however I wanted to get in some practice with trig substitutions.

$$\displaystyle \int \frac{4e^x}{(1-e^{2x})^3} dx =\int \frac{4e^x}{(1-(e^x)^2)^3} dx$$

Now I set the following:

$$\displaystyle e^x = sin(t) \rightarrow x = ln(sin(t)) \rightarrow dx = \frac{1}{sin(t)} * cos(t) dt$$

Substituting:

$$\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4\int \frac{sin(t)}{(1-sin(t)^2)^3} * \frac{1}{sin(t)} * cos *dt = 4\int \frac{1}{cos(t)^5}$$
$$\displaystyle 4\int sec(t)^5 dt = 4\int sec(t)^3 sec(t)^2 = sec(t)^3 tan(t) - \int 3 sec(t)^2 tan(t) dt$$

So we get:

$$\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4(sec(t)^3 tan - \frac{3}{2}tan(t)^2 + C$$

To express it in terms of X:

$$\displaystyle cos(t) = \sqrt{1-(e^x)^2}\rightarrow sec(t) = \frac{1}{\sqrt{1-(e^x)^2}} \rightarrow tan(t) = \frac{e^x}{\sqrt{1-(e^x)^2}}$$

Finally substituting back, the final answer becomes:

$$\displaystyle 4((\frac{1}{\sqrt{1-(e^x)^2}})^3 \frac{e^x}{\sqrt{1-(e^x)^2}} - \frac{3}{2}(\frac{e^x}{\sqrt{1-(e^x)^2}})^2) + C$$

My problem is that the domain of this function is only x < 0 (for real numbers). What should I make of that?

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#### skeeter

MHF Helper
Note the domain of the original integrand is all reals, $x \ne 0$

By doing the trig sub, $e^x = \sin{t}$, you've restricted the domain to $-\infty < x < 0$ from the very start ...

#### hasek747

So the only way this trig substitution would work here is if this were a definite integral problem with a range that falls somewhere between $$\displaystyle -\infty$$ and 0 ?

#### Plato

MHF Helper
So I have the following problem:
$$\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx$$
I realize this can be computed a simple U-substitution (followed by rationalizing the function), however I wanted to get in some practice with trig substitutions.
My problem is that the domain of this function is only x < 0 (for real numbers). What should I make of that?
If instead use $u=e^x$ and get $\dfrac{du}{(1-u^2)^3}$
Now use partial fractions, SEE HERE, to eliminate the domain issues.

#### hasek747

Thank you for the comment Plato. I'm aware of that, but the problem I'm having is that I'm not fully understanding how the domain changes with the various substitutions, so I'm trying to work the problem in ways that will expose me to domain-related issues. This is a general weakness of mine when it comes to integrals and I'm trying to improve it.

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#### Plato

MHF Helper
I'm not fully understanding how the domain changes with the various substitutions, so I'm trying to work the problem in ways that will expose me to domain-related issues.
That is the whole point of using $u=e^x$. That substitution does not change the domain.
Still $x\ne 0$ makes $(1-u)\ne 0$.

Do you understand how making a u-substitution in a definite integral changes the limits of integration?

#### hasek747

Do you understand how making a u-substitution in a definite integral changes the limits of integration?
I believe I do. However up until now I was always substituting back to the original integration term, so I wasn't actually calculating the new limits when making the substitution; so far it had been working out for me and this excercise over here is the first time I actually noticed an issue with the domain of the function can arise when substituting (it's not something that I recall was mentioned in any of the online materials I've been learning from).

So essentially whenever I'm about to make a substitution of any kind, I should first make sure that the substitution I am planning to use does change the range of the function? And if it does change the range, then I need to use a different method of integration?

#### hasek747

Hello,

Just bumping this in case anyone could answer the question in my last post. Would appreciate the help as I'm still not sure about this.

#### Plato

MHF Helper
Just bumping this in case anyone could answer the question in my last post.