# Evaluating integral

1. Sep 24, 2013

### Saitama

1. The problem statement, all variables and given/known data
If $y(x-y)^2=x$, then find
$$\int \frac{dx}{x-3y}$$

2. Relevant equations

3. The attempt at a solution
I tried expanding the given equation. That gives me a cubic in y. I found no way to solve the cubic equation. I plugged the equation in Wolfram Alpha to see if it gives any nice solution but nope, no luck there too. Now I think that this is some sort of trick question but I don't really know where to begin.

Any help is appreciated. Thanks!

2. Sep 24, 2013

### ehild

What about changing to the variable y/x=t? Just a guess...

ehild

3. Sep 24, 2013

### Saitama

I should have really attempted the problem properly before posting it here. I just woke up and gave it a try. I substituted x-y=t. The integral changes to dx/(3t-2x) and the equation changes to $(x-t)t^2=x$. I differentiated the equation wrt x and after some simplification, I got
$$\frac{dx}{3t-2x}=\frac{dt}{t^2-1}$$

Now its a very easy integral to solve. Thanks for the effort ehild.

4. Sep 24, 2013

### jackmell

I think that's wrong. Sides, once you got t, then what? How about checking it with the integral:

$$\int_1^2 \frac{1}{x-3y}dx,\quad y(x-y)^2=x$$

While we're at it, what is the geometric interpretation of that integral anyway?

5. Sep 24, 2013

### Saitama

I don't see what's wrong here. I differentiated both the sides of the equation wrt x. Did I do some mistake while simplifying the expression?

6. Sep 25, 2013

### Saitama

Just realised that the final equation after differentiating I get is wrong. It is actually
$$\frac{dx}{3t-2x}=\frac{t \, dt}{t^2-1}$$

7. Sep 25, 2013

### jackmell

To check it, I solved the definite integral above and obtained:

$$\int_1^2 \frac{1}{x-3y} dx=\int_a^b \frac{t}{t^2-1}dt$$

where a=1-lowroot and b=2-highroot with the roots being the real roots of the equations $y(1-y)^2=1$ and $y(2-y)^2=2$

which agrees numerically with direct substitution of y(x) from the solution of $y(x-y)^2=x$.

Always a good idea to check them numerically if we're trying to intercept an earth-bound meteor.

8. Sep 25, 2013

### juantheron

Given ::$y\cdot(x-y)^2 = x$

Now Diff. both side w. r . to $x$ , We Get

$\displaystyle y\cdot 2(x-y)\left\{1-\frac{dy}{dx}\right\}+(x-y)^2 \cdot \frac{dy}{dx} = 1$

$\displaystyle 2y\cdot(x-y)\left\{1-\frac{dy}{dx}\right\}+\frac{dy}{dx}\cdot (x-y)^2-(x-y)^2 = 1-(x-y)^2$

$\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(2xy-y^2-x^2+2xy-y^2\right) = 1-(x-y)^2$

$\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(-x^2+4xy-3y^2\right) = 1-(x-y)^2$

$\displaystyle \left\{\frac{dy}{dx}-1 \right\}\left(x^2-4xy+3y^2\right) = 1 - (x-y)^2$

$\displaystyle \left\{\frac{dy}{dx}-1 \right\}.\left(x-y\right).\left(x-3y\right) = 1-(x-y)^2$

$\displaystyle \left(dy - dx\right).(x-y).(x-3y) = \left\{1-(x-y)^2\right\}dx$

$\displaystyle \frac{dx}{(x-3y)} = \frac{\left(dy - dx\right)\cdot(x-y)}{1-(x-y)^2}$

Now Integrate both side , We Get

$\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{\left(dx - dy\right)\cdot(x-y)}{(x-y)^2 -1}$

Let $(x-y) = t\Leftrightarrow d(x-y) = dt\Leftrightarrow \left(dx - dy\right) = dt$

So $\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{t}{t^2-1}dt$

So again Let $\displaystyle t^2 - 1 = u\Leftrightarrow tdt = \frac{1}{2}du$

So $\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \int\frac{1}{u}du = \frac{1}{2}\cdot \ln \left|u\right|+C$

So $\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \ln \left|(x-y)^2 - 1\right|+C$