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

  1. #1
    s3a
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    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!
    Thanks in advance!
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by s3a View Post
    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!
    Thanks in advance!
    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})

    If you need anymore help, I'll be glad to help you through the whole answer.
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  3. #3
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    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.
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  4. #4
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    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.
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  5. #5
    Member mathemagister's Avatar
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    Quote Originally Posted by ANDS! View Post
    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}<br />

    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<br />

    Hope this helps, s3a
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  6. #6
    Member mathemagister's Avatar
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    Quote Originally Posted by drumist View Post
    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.
    Last edited by mathemagister; February 13th 2010 at 01:06 PM. Reason: LaTex
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  7. #7
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    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.
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  8. #8
    Member mathemagister's Avatar
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    Quote Originally Posted by ANDS! View Post
    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.
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