# Solve dx/y = dy/x

1. Aug 17, 2014

### ibysaiyan

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

Hi I am trying to solve an indefinite integral of the form 1/ sqrt[ (x^2 +1)] dx..

2. Relevant equations

3. The attempt at a solution

Different ways of solving it are posted on the link below. I would like to know the following result :

dx/y = dy/x = d (x+y)/ (x+y) ???

http://math.stackexchange.com/questions/610733/how-to-integrate-int-dx-over-sqrt1-x2?lq=1

Thanks.

2. Aug 17, 2014

### Zondrina

If $1 = 1^2$, then denominator has the form $\sqrt{x^2 + a^2}$.

Make the substitution $x = a \tan(\theta)$ and use the identity $\sec^2(x) - \tan^2(x) = 1$.

3. Aug 17, 2014

### ibysaiyan

You have misunderstood my post. I want to know why $dx/y = dy/x = \int d (x+y) / (x+y) = ln (x+y) + C$

Thanks.

4. Aug 17, 2014

### HallsofIvy

Staff Emeritus
Probably the reason he misunderstood is that the equation you have is NOT correct. In fact it doesn't even makes sense- the first two parts are differentials, the last two are not. You need to separate it into two different equations. These are two completely different problems.

If dx/y= dy/x then, multiplying both sides by xy, xdx= ydy. Integrate both sides to find y as a function of x.

The other parts says that $\int d(x+y)/(x+ y)= ln(x+ y)+ C$. To see that that is true, let u= x+ y and use the fact that $\int du/u= ln(u)+ C$

But the two have nothing to do with each other.

5. Aug 17, 2014

### ibysaiyan

This is why I am stumped.
I am referring to the bottom most solution posted on the following link by juantheron (user): http://math.stackexchange.com/questions/610733/how-to-integrate-int-dx-over-sqrt1-x2?lq=1

6. Aug 17, 2014

### Zondrina

Observing the integral:

$\int \frac{d(x+y)}{(x+ y)} = \int \frac{1}{(x+ y)} d(x+y) = ln(x+ y)+ C$

Where $(x + y)$ is being treated as the independent. Back subbing for $y = \sqrt{1 + x^2}$ will give you the answer.

Does this alleviate the problem or is the concern something else entirely?

7. Aug 18, 2014

### Dipen

Hey there,

This can be done using the technique of trigonometric integration. Please try it out and let me know if you have anymore difficulty.
Thanks

8. Aug 18, 2014

### vanhees71

Another (more convenient?) way is to substitute $x=\sinh u$. As turns out, this is in fact a basic integral, involving hyperbolic functions.

9. Aug 18, 2014

### ibysaiyan

Guys thanks for the responses.

My issue is still not resolved. Let me rephrase the question.

I want to know how one can show that $\int dx/y = \int d(x+y) / (x+y)$

10. Aug 18, 2014

### HallsofIvy

Staff Emeritus
You can't. Unless there is some additional relation between x and y you haven't told us, such as "x= y", that equation is not true. What is true is that $\int dy/y= \int d(x+ y)/(x+ y)$. That equation is true but is confusing notation because we are using y to represent two different "dummy" variables. I would prefer to write it in the equivalent form $\int du/u= \int d(x+ y)/(x+ y)$ which is clearly true with u= x+ y.

11. Aug 18, 2014

### ibysaiyan

I was under the impression that maybe there is a way of showing the above relation, but now it all makes sense.

Thanks for the clarification.

12. Aug 18, 2014

### Curious3141

It IS correct.

It is a simple consequence of the fact that $\frac ab = \frac cd \implies \frac ab = \frac cd = \frac{a+c}{b+d}$ for nonzero $a,b,c,d$. This is trivial to prove.

Juantheron (on MathSE) is merely replacing some of the variables with infinitesimals to come up with a rather interesting way to evaluate the integral. He starts by defining $y^2 = x^2 + 1$, which then leads to $ydy = xdx$. That allows him to go to $\frac{dx}{y} = \frac{dy}{x}= \frac{dx + dy}{x + y} = \frac{d(x+y)}{x+y}$, and he uses that to evaluate the original integral.

Some may balk at this "fast and loose" use of the Leibniz notation, but it is sound.

Last edited: Aug 18, 2014