# Math Help - Simple Integral Question (I think it's simple)

1. ## Simple Integral Question (I think it's simple)

I am supposed to know this by heart and I do:
http://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))

However, on a specific problem I'm having, I need to know to get the following:
http://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)

My question is:
Is there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?

Any input would be greatly appreciated!

2. Originally Posted by s3a
I am supposed to know this by heart and I do:
http://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))

However, on a specific problem I'm having, I need to know to get the following:
http://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)

My question is:
Is there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?

Any input would be greatly appreciated!
I'll give you a hint. See if you can figure it out yourself. Use the formula you memorized to make a general formula for the derivative of $tan^{-1}(\frac {x}{a})$

3. Eek, its better to just use LaTex (if you know how) and link the problem.

However this isn't a "simple" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:

$x = 6tan(\theta)$
$dx=6sec^{2}(\theta)d\theta$

Thus:

$\int \frac{6sec^{2}(\theta)d\theta}{(6tan{\theta})^{2}+ 36} \Rightarrow \int \frac{6sec^{2}(\theta)d\theta}{36(tan^{2}\theta+1) }$

From here we simplify:
$\int \frac{sec^{2}(\theta)d\theta}{6sec^{2}(\theta)} \Rightarrow \int \frac{1}{6}d\theta$

This integral is easy to evaluate:

$\int \frac{1}{6}d\theta \Rightarrow \frac{1}{6}\theta+C$

However, we need our final answer in terms of x, not theta. So, we go back to our original substitution:

$x = 6tan(\theta)$

. . .and solve for theta. Do not "memorize" formulas or "tricks" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.

4. I learned it like this:

$\int \frac{1}{x^2+36} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\frac{x^2}{36}+1} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\left(\frac{x}{6}\right)^2+1} \, dx$

(Then let $u=x/6 ~\implies~ du=dx/6$.)

$=~ \frac{1}{36} \int \frac{6}{u^2+1} \, du$

$=~ \frac{1}{6} \int \frac{1}{u^2+1} \, du$

You can probably complete it from there.

5. Originally Posted by ANDS!
Eek, its better to just use LaTex (if you know how) and link the problem.

However this isn't a "simple" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:

$x = 6tan(\theta)$
$dx=6sec^{2}(\theta)d\theta$

Thus:

$\int \frac{6sec^{2}(\theta)d\theta}{(6tan{\theta})^{2}+ 36} \Rightarrow \int \frac6{sec^{2}(\theta)d\theta}{36(tan^{2}\theta+1) }$

From here we simplify:
$\int \frac{sec^{2}(\theta)d\theta}{6sec^{2}(\theta)} \Rightarrow \int \frac{1}{6}d\theta$

This integral is easy to evaluate:

$\int \frac{1}{6}d\theta \Rightarrow \frac{1}{6}\theta+C$

However, we need our final answer in terms of x, not theta. So, we go back to our original substitution:

$x = 6tan(\theta)$

. . .and solve for theta. Do not "memorize" formulas or "tricks" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.
However, there is a MUCH simpler way to do this:

$\frac{d}{dx} tan^{-1}x = \frac {1}{x^2+1}$
$\frac{d}{dx} tan^{-1}(\frac {x}{a}) = \frac {1}{(\frac{x}{a})^2+1}.\frac{1}{a} = \frac {a}{x^2+a^2}
$

Notice how the question is in similar format:
$\int \frac{1}{x^2+36}dx = \frac{1}{6} \int \frac{6}{x^2+36}dx = \frac{1}{6} tan^{-1}\frac{x}{6}+C
$

Hope this helps, s3a

6. Originally Posted by drumist
I learned it like this:

$\int \frac{1}{x^2+36} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\frac{x^2}{36}+1} \, dx$

$=~ \frac{1}{36} \int \frac{1}{\left(\frac{x}{6}\right)^2+1} \, dx$

(Then let $u=x/6 ~\implies~ du=dx/6$.)

$=~ \frac{1}{36} \int \frac{6}{u^2+1} \, du$

$=~ \frac{1}{6} \int \frac{1}{u^2+1} \, du$

You can probably complete it from there.
Sorry, I was typing when you posted; your way works well, too.

And s3u: if you're going to take a timed test like AP Calc AB, memorizing $\frac{d}{dx} tan^{-1}(\frac {x}{a}) = \frac {a}{x^2+a^2}$ could save a lot of time. Of course, you don't need to memorize it, I'm sure, because you saw where that came from, you could re-figure that out in a few seconds.

7. All of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.

8. Originally Posted by ANDS!
All of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.
I actually prefer the method you used, but the one I used is more useful in a test perspective, to save a lot of time. And, yes, they are all, in essence, the same method; that's the beautiful consistency of Math.