# integral understanding question

• May 25th 2013, 02:15 PM
ryu1
integral understanding question
Hello,

$\displaystyle f(x) = \frac{1}{a^2x^2+b^2}$

I know the correct answer but I don't understand EXACTLY why this is incorrect:

$\displaystyle \int\frac{1}{a^2x^2+b^2} dx = \int\frac{1}{(ax)^2+b^2} dx$
$\displaystyle = \frac{1}{b} \arctan(\frac{ax}{b}) +C$

I understand it has something to do with the integral "in regards to x" (because we write the dx) there, but could you explain that to me a little better please?
Thanks a lot!

Thanks to Wolfram I avoided sending a wrong answer in this on my homework.(really powerful tool, too bad we can't use the internet on tests :( (but we can use it in real life though...(Headbang)))
• May 25th 2013, 03:31 PM
Shakarri
Re: integral understanding question
The variable has to be on its own in the equation, otherwise you are integrating a variable ax with respect to dx and they are different.
You can do a change of variables, let y=ax
$\displaystyle dx=\frac{dy}{a}$

Then the integral
$\displaystyle \int\frac{1}{(ax)^2+b^2} dx$

becomes
$\displaystyle \int\frac{1}{y^2+b^2}\cdot \frac{dy}{a}$

$\displaystyle =\frac{1}{a}\cdot \frac{1}{b} \arctan(\frac{y}{b}) +C$

Presumably in the derivation of the solution to the integral, x2 is not multiplied by a constant and the proof cannot be done the same way when x2 is multiplied by a constant
• May 25th 2013, 03:53 PM
Plato
Re: integral understanding question
Quote:

Originally Posted by ryu1
$\displaystyle f(x) = \frac{1}{a^2x^2+b^2}$
I know the correct answer but I don't understand EXACTLY why this is incorrect:
$\displaystyle \int\frac{1}{a^2x^2+b^2} dx = \int\frac{1}{(ax)^2+b^2} dx$
$\displaystyle = \frac{1}{b} \arctan(\frac{ax}{b}) +C$
I understand it has something to do with the integral "in regards to x" (because we write the dx) there, but could you explain that to me a little better please?

To understand, do the derivative of that answer. Carefully do ALL the algebra to simply that derivative.
That process will show you how to reverse engineer the problem.

Quote:

Originally Posted by ryu1
Thanks to Wolfram I avoided sending a wrong answer in this on my homework.(really powerful tool, too bad we can't use the internet

Frankly, I see absolutely no reason that you should not be allowed to use Wolframalpha on a test.
If you were asked to find $\displaystyle \sqrt{13}$ on a test would you not be allowed to use a calculator?
Tell me, what is the difference? This webpage sums up exactly what I think of the matter.

BTW. There is a cell phone app and tablet app for Wolfframalpha.
• May 25th 2013, 03:54 PM
Soroban
Re: integral understanding question
Hello, ryu1!

Quote:

I had this problem: .$\displaystyle \int \frac{dx}{a^2x^2+b^2}$

I know the correct answer, but I don't understand EXACTLY why this is incorrect:

$\displaystyle \int\frac{dx}{a^2x^2+b^2} \:=\: \int\frac{dx}{(ax)^2+b^2} \:=\:\frac{1}{b} \arctan(\frac{ax}{b}) +C$

If you know the correct answer, you can see what you are missing, can't you?

Okay, back to basics.

Formula: .$\displaystyle \int\frac{du}{u^2+b^2} \;=\;\frac{1}{b}\arctan\left(\frac{u}{b}\right) + C$

We have: .$\displaystyle \int\frac{dx}{(ax)^2 + b^2}$

That is: .$\displaystyle u \,=\,ax \quad\Rightarrow\quad du \,=\,a\,dx \quad\Rightarrow\quad dx \,=\,\tfrac{1}{a}du$

Substitute: .$\displaystyle \int\frac{\frac{1}{a}du}{u^2+b^2} \;=\;\frac{1}{a}\int\frac{du}{u^2+b^2}$

. . . . . . . . $\displaystyle =\;\frac{1}{a}\cdot\frac{1}{b}\arctan\left(\frac{u }{b}\right)+C \;=\;\frac{1}{ab}\arctan\left(\frac{u}{b}\right)+C$

Back-substitute: .$\displaystyle \frac{1}{ab}\arctan\left(\frac{ax}{b}\right)+C$
• May 25th 2013, 04:26 PM
ryu1
Re: integral understanding question
Quote:

Originally Posted by Soroban
Hello, ryu1!

If you know the correct answer, you can see what you are missing, can't you?

Okay, back to basics.

Formula: .$\displaystyle \int\frac{du}{u^2+b^2} \;=\;\frac{1}{b}\arctan\left(\frac{u}{b}\right) + C$

We have: .$\displaystyle \int\frac{dx}{(ax)^2 + b^2}$

That is: .$\displaystyle u \,=\,ax \quad\Rightarrow\quad du \,=\,a\,dx \quad\Rightarrow\quad dx \,=\,\tfrac{1}{a}du$

Substitute: .$\displaystyle \int\frac{\frac{1}{a}du}{u^2+b^2} \;=\;\frac{1}{a}\int\frac{du}{u^2+b^2}$

. . . . . . . . $\displaystyle =\;\frac{1}{a}\cdot\frac{1}{b}\arctan\left(\frac{u }{b}\right)+C \;=\;\frac{1}{ab}\arctan\left(\frac{u}{b}\right)+C$

Back-substitute: .$\displaystyle \frac{1}{ab}\arctan\left(\frac{ax}{b}\right)+C$

There is some saying that I can't remember but it's something like an animal seeing the reflection of the moon in the lake, trying to touch it but can't really reach it.
(or maybe something else, point is, even if I can see it, I can't always understand it, unfortunately, but then I can ask someone and then he tell me, if he is nice, then I think about what he said a little, and accept or not accept his explanation, if Plato explains it then I will most likely accept it :)

I see what you did there, I missed the part with the du = a dx , because I was just blindly using the formula wrong...

Quote:

Originally Posted by Plato
Frankly, I see absolutely no reason that you should not be allowed to use Wolframalpha on a test.
If you were asked to find $\displaystyle \sqrt{13}$ on a test would you not be allowed to use a calculator?
Tell me, what is the difference? This webpage sums up exactly what I think of the matter.

BTW. There is a cell phone app and tablet app for Wolfframalpha.

I guess they think it has something to do with like-cheating, like here, I had to do this problem, and I would make a mistake, but thanks to that tool I "didnt"... but on the contrary, graphing calculator is allowed (ti-83 also has some apps for it to help with many calculations)
But then again, most of the problems they give us don't require complex computations, it requires mostly understanding and making good use of mathematical theorems etc.

Also wolfram has a step by step tool that makes it an "almost" no brainer, which I think is bad, but it's really good and useful :)
As for the app, my cellphone doesn't have always on internet connection (data plan too expensive).

P.S.
I learned to use Latex there :)