# Homework Help: Difficult Integration

1. Dec 4, 2014

### basty

1. The problem statement, all variables and given/known data

$(e^y + 1)^2 e^{-y} dx + (e^x + 1)^3 e^{-x} dy = 0$

2. Relevant equations

3. The attempt at a solution

$(e^y + 1)^2 e^{-y} dx + (e^x + 1)^3 e^{-x} dy = 0$
$(e^{2y} + 2 e^y + 1) e^{-y} dx + (e^{3x} + 3e^{2x} + 3e^x + 1) e^{-x} dy = 0$
$(e^{2y - y} + 2 e^{y - y} + e^{-y}) dx + (e^{3x - x} + 3e^{2x - x} + 3e^{x - x} + e^{-x}) dy = 0$
$(e^{y} + 2 e^{0} + e^{-y}) dx + (e^{2x} + 3e^{x} + 3e^{0} + e^{-x}) dy = 0$
$(e^{y} + 2 (1) + e^{-y}) dx + (e^{2x} + 3e^{x} + 3 (1) + e^{-x}) dy = 0$
$(e^{y} + 2 + e^{-y}) dx + (e^{2x} + 3e^{x} + 3 + e^{-x}) dy = 0$
$(e^{y} + e^{-y} + 2) dx + (e^{2x} + 3e^{x} + e^{-x} + 3) dy = 0$
$(e^{2x} + 3e^{x} + e^{-x} + 3) dy = - (e^{y} + e^{-y} + 2) dx$
$- \frac{1}{e^{y} + e^{-y} + 2} dy = \frac{1}{e^{2x} + 3e^{x} + e^{-x} + 3} dx$
$- \int \frac{1}{e^{y} + e^{-y} + 2} dy = \int \frac{1}{e^{2x} + 3e^{x} + e^{-x} + 3} dx$

It is a difficult integration. What next?

2. Dec 4, 2014

### ShayanJ

The only thing that I can see here is $\frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2$.

3. Dec 4, 2014

### haruspex

You mean sech(), right?

4. Dec 4, 2014

### basty

How do you get $\frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2$?

I know that $sech \ x = \frac{2}{e^x + e^{-x}}$.

Last edited: Dec 4, 2014
5. Dec 4, 2014

### ShayanJ

Sorry...Yeah, I meant the hyperbolic thing!

6. Dec 4, 2014

### basty

How do you get $\frac{1}{e^y+e^{-y}+2}=\frac 1 4 \sec^2 \frac y 2$?

I know that $sech \ y = \frac{2}{e^y + e^{-y}}$.

Last edited: Dec 4, 2014
7. Dec 4, 2014

### ShayanJ

$sech x=\frac{2}{e^x+e^{-x}} -^{x=\frac{y}{2}}\rightarrow sech \frac y 2=\frac{2}{e^{\frac y 2}+e^{-\frac y 2}} \rightarrow sech^2 \frac y 2=\frac{4}{e^{y}+e^{-y}+2}$

Last edited: Dec 4, 2014
8. Dec 4, 2014

### basty

Why $sech \ x = \frac{2}{e^x + e^{-x}} - x$?

And why it equal to $\frac{y}{2}$?

The correct one is $sech \ x = \frac{2}{e^x + e^{-x}}$, not $\frac{2}{e^x + e^{-x}} - x$.

What this -x variable mean?

9. Dec 4, 2014

### ShayanJ

Oh...I meant to be a little fancy but looks like it was a bad idea. I meant take $sech x=\frac{2}{e^x+e^{-x}}$ and then set $x=\frac y 2$.

10. Dec 4, 2014

### basty

You said $sech \ \frac{y}{2} = \frac{2}{e^{\frac{y}{2}} + e^{-\frac{y}{2}}}$
$= \frac{2}{e^{\frac{y}{2}} + \frac{1}{e^{\frac{y}{2}}}}$

$= \frac{2}{\frac{(e^{\frac{y}{2}})^2 + 1}{e^{\frac{y}{2}}}}$

$=\frac{2 e^{\frac{y}{2}}}{(e^{\frac{y}{2}})^2 + 1}$

$=\frac{2 e^{\frac{y}{2}}}{(e^{\frac{y}{2}}. \ e^{\frac{y}{2}}) + 1}$

$=\frac{2 e^{\frac{y}{2}}}{e^{\frac{y}{2} + \frac{y}{2}} + 1}$

$=\frac{2 e^{\frac{y}{2}}}{e^{\frac{y + y}{2}} + 1}$

$=\frac{2 e^{\frac{y}{2}}}{e^{\frac{2y}{2}} + 1}$

$=\frac{2 e^{\frac{y}{2}}}{e^{y} + 1}$

Please explain. I am really confuse.

11. Dec 4, 2014

### epenguin

When you write this

exdx/(ex + 1)3 = - eydy/(ey + 1)2

isn't each side easily recognised as derivative of something?

12. Dec 4, 2014

### vela

Staff Emeritus
Or if you don't recognize the derivatives, try the obvious substitutions.

13. Dec 5, 2014

### basty

I still don't get it. Give me more clue please.

14. Dec 5, 2014

### Staff: Mentor

Vela gave you a clue in post #12. Try it.

15. Dec 5, 2014

### basty

$\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}$

I shall integrate the left side (the x variable) first. If using the substitution method, which one should be substituted?
Let $u = e^x$ or $u = e^x + 1$?

16. Dec 5, 2014

### Staff: Mentor

Try them both. You should see that one is slightly better than the other.

17. Dec 5, 2014

### basty

$\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}$

(1)
Let $u = e^x + 1$ then $\frac{du}{dx} = e^x$ or $du = e^x dx$

$\int \frac{e^x dx}{(e^x + 1)^3}$
$= \int \frac{1}{u^3}du$
$= \ln |u^3| + c$
$= \ln (e^x + 1)^3 + c$

(2)
Let $u = e^y + 1$ then $\frac{du}{dy} = e^y$ or $du = e^y dy$

$\int \frac{-e^y dy}{(e^y + 1)^2}$
$= - \int \frac{e^y dy}{(e^y + 1)^2}$
$= - \int \frac{1}{u^2}du$
$= - \ln |u^2| + c$
$= - \ln (e^y + 1)^2 + c$

From (1) and (2), I get:

$\int \frac{e^x dx}{(e^x + 1)^3} = \int \frac{-e^y dy}{(e^y + 1)^2}$
$\ln (e^x + 1)^3 + c = - \ln (e^y + 1)^2 + c$

Is it correct?

18. Dec 5, 2014

### Zondrina

That is certainly not correct. What is:

$$\int u^{-3} \space du$$

19. Dec 5, 2014

### Staff: Mentor

Good catch, Zondrina. I totally missed that he was using this integration "rule" (in quotes because it's not an integration rule):
$\int \frac{dx}{f(x)} = ln|f(x)| + C$

20. Dec 5, 2014

### basty

Thank you for the correction.

Let I correct the mistake:

$\frac{e^x dx}{(e^x + 1)^3} = \frac{-e^y dy}{(e^y + 1)^2}$

(1)
Let $u = e^x + 1$ then $\frac{du}{dx} = e^x$ or $du = e^x dx$

$\int \frac{e^x dx}{(e^x + 1)^3}$
$= \int \frac{1}{u^3}du$
$= \int u^{-3} du$
$= \frac{1}{-3+1}u^{-3+1} + c$
$= \frac{1}{-2}u^{-2} + c$
$= - \frac{1}{2}(e^x + 1)^{-2} + c$

(2)
Let $u = e^y + 1$ then $\frac{du}{dy} = e^y$ or $du = e^y dy$

$\int \frac{-e^y dy}{(e^y + 1)^2}$
$= - \int \frac{e^y dy}{(e^y + 1)^2}$
$= - \int \frac{1}{u^2}du$
$= - \int u^{-2} du$
$= - \frac{1}{-2+1} u^{-2+1} + c$
$= - \frac{1}{-1} u^{-1} + c$
$= u^{-1} + c$
$= (e^y + 1)^{-1} + c$

From (1) and (2), I get:

$\int \frac{e^x dx}{(e^x + 1)^3} = \int \frac{-e^y dy}{(e^y + 1)^2}$
$- \frac{1}{2}(e^x + 1)^{-2} + c = (e^y + 1)^{-1} + c$

Is it correct?

21. Dec 5, 2014

### Staff: Mentor

That looks better. In the last line you have this:
$- \frac{1}{2}(e^x + 1)^{-2} + c = (e^y + 1)^{-1} + c$
You shouldn't use the same constant on both sides.
$- \frac{1}{2}(e^x + 1)^{-2} + c_1 = (e^y + 1)^{-1} + c_2$
Or you can combine both constants into a third constant, like so:
$- \frac{1}{2}(e^x + 1)^{-2} = (e^y + 1)^{-1} + c$

22. Dec 5, 2014

### basty

OK.

23. Dec 5, 2014

### epenguin

It's on right lines at least, but not quite correct is you have written the same integration constant on both sides which is tantamount to an integration constant of 0, so that is conceptually not quite correct. You only need one integration constant - I hope clear why.

Apart from that you don't ever need to ask us whether an integration is correct - you can tell us rather because you can always check correctness by differentiation.

Also (Polya Principle) in doing that you might notice how you might have arrived at the result a bit faster, ideally, though getting there somehow is the most important.

(posting overlap)