# Homework Help: Evaluating a Complicated Integral; why is this method incorrect?

1. Feb 20, 2013

### AmagicalFishy

1. The problem statement, all variables and given/known data
Solve (k and Q are constants):
$$\frac{dP}{dt} = kP(Q-P)$$

2. Relevant equations
$$∫\frac{1}{x^2 + a^2} = \frac{1}{a} arctan(\frac{x}{a})$$
Where "a" is a constant.

3. The attempt at a solution
Ok, none of this seems relevant to evaluating a complicated integral, but that differential problem is what gave rise to my question. I know how to solve it (and solve the associated integral using partial fractions), but I don't know why my uh... adventurous method of solving the integral is wrong. This is about solving an integral, not the ODE. :)

It's frequently the case that I understand one way of solving a problem, but don't understand why solving it another way is incorrect. If I can't differentiate between the correct and incorrect ways, then I'm not getting anywhere.

$$\frac{dP}{dt} = kP(Q-P) \\ \frac{dP}{dt} = k(PQ-P^2) \\ \frac{dP}{dt} + \frac{kQ^2}{4}= k(PQ-P^2 +\frac{Q^2}{4}) \text{—Completing the Square}\\ \frac{dP}{dt} = k(P-\frac{Q}{2})^2 - \frac{kQ^2}{4} \\ \frac{1}{k(P-\frac{Q}{2})^2 - \frac{kQ^2}{4}} dP = dt \\ \frac{1}{\sqrt(k)^2(P-\frac{Q}{2})^2 - \frac{\sqrt{k}^2Q^2}{4}} dP = dt \\ (1) = \frac{1}{[\sqrt(k)(P-\frac{Q}{2})]^2 - [\frac{\sqrt{k}Q}{2}]^2} dP = dt \\ \text{Let:} \\ \sqrt(k)(P-\frac{Q}{2}) = x \\ - [\frac{\sqrt{k}Q}{2}] = a\\ \frac{dx}{\sqrt(k)-\frac{Q}{2}} = dP \\ \\ \text{Then:} \\ (1) = \frac{1}{\sqrt(k)-\frac{Q}{2}}\frac{1}{x^2 + a^2} dx = dt \\ \frac{1}{\sqrt(k)-\frac{Q}{2}}∫\frac{1}{x^2 + a^2} dx = ∫dt$$

When you integrate both sides, you use the above arctangent equation and replace "x" and "a" respectively. In the above substitution, "a" is a constant and "x" is the variable.

Algebraically, this all seems correct to me. I'm not worried about the answer to the differential equation—I'd like to know why evaluating that integral is incorrect. This is what I used to check whether or not this fruits the correct answer. If it had, Wolfram would say, simply, "True".

I think it has something to do with the substitution of "a" having a negative, where a sign issue invalidates everything, but I'd like to confirm that here.

Thanks for looking it through, anyone who does.

2. Feb 20, 2013

### haruspex

Except, that does not complete the square. You have a sign wrong for that.

3. Feb 20, 2013

### AmagicalFishy

...

... mother f#&@

Thanks, dude. I just... ugh. :l