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?