Solving Difficult Integrals: Step by Step Guide

[basty]

Homework Statement

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

Homework Equations

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?

 

[ShayanJ]

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

 

[haruspex]

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

You mean sech(), right?

 

[basty]

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

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

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

 

[ShayanJ]

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

 

[basty]

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

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

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

 

[ShayanJ]

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

 

[basty]

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

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?

 

[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 $x=\frac y 2$.

 

[basty]

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

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.

 

[epenguin]

When you write this

e^xdx/(e^x + 1)³ = - e^ydy/(e^y + 1)²

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

 

[vela]

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

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

 

[basty]

When you write this

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

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

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

 

[Mark44]

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

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

 

[basty]

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

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

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 $u = e^x + 1$?

 

[Mark44]

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

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 $u = e^x + 1$?

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

 

[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?

 

[STEMucator]

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

(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?

That is certainly not correct. What is:

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

 

[Mark44]

That is certainly not correct. What is:

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

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

 

[basty]

That is certainly not correct. What is:

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

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?

 

[Mark44]

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:
$- \frac{1}{2}(e^x + 1)^{-2} = (e^y + 1)^{-1} + c$

 

[basty]

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:
$- \frac{1}{2}(e^x + 1)^{-2} = (e^y + 1)^{-1} + c$

OK.

 

[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)