# 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?

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