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I am at a loss for what to do.

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx$$ I tried to make $u = x^6 + 9 $, $du = 6x^5$

$$\frac{1}{5}\int_{-\infty}^\infty \frac {1}{x^3u}du$$

I can rewrite $x$ in a complex manner but I do not think that actually helps me.

I tried to do some algebra magic be rewriting $x^6$ as $x^3 \cdot x^3$ but I made no real progress like that. I know I can make it $(x^3 + 3)^2 - 6x$ but that doesn't seem to do any good.

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    It would be nice to have $x^2$ be more or less the derivative of $u$, so you might want to try $u=x^3$.2012-06-07
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    @Phira Is there a rule or process to follow to see that?2012-06-08
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    @Jordan It should be natural in the hierarchy of your mathematical toolkit you've learned in calc 1 and 2. If you can use u-substitution over integration by parts, partial fractions, trig substitutions, etc, you should. When using u-substitution, you're always looking for the $du$ to be part of integrand already - it makes things easy.2012-06-08
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    I was simply trying to help you understand the substitution made for your problem, not give you life advice.2012-06-08
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    Telling me it is something I should already know isn't useful, obvsiously I do not know this.2012-06-08

5 Answers 5

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This is supplementary to Peter and Michael's answers (and Phira's comment).

Here is what a plot of $\tan x$ looks like between $x=-\pi/2$ and $x=+\pi/2$:

$\hskip 2in$ tan

We have a couple one-side limits from the above, the first from the right and second from the left:

$$\lim_{x\to-\pi/2^+}\tan x=-\infty,\qquad \lim_{x\to+\pi/2^-}\tan x=+\infty. \tag{$\circ$}$$

This isn't terribly difficult to see visually by drawing right triangles with $\theta\approx\pm\pi/2$. Alternatively, we could find the left/right limits of $\sin$ and $\cos$ separately at $\pm\pi/2$ and reason the above limits.

Symmetrically, here is what $\arctan x$ looks like on the real line:

$\hskip 2in$ arctan

Above is the graphical depiction of why the limits $(\circ)$ are reversible. That is,

$$\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2},\qquad \lim_{x\to+\infty}\arctan x=+\frac{\pi}{2}. \tag{$\bullet$}$$

This is what allows you to finish off the computation method given by the substitution $u=x^3$. In my opinion this is the standard method that would be intended for calculus students, and the limits given in $(\circ),(\bullet)$ were intended to be in your personal "arsenal" a priori for this problem.

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    I am still confused on how I am suppose to calculuate, without the use of graphics, the limit, I don't need to graph every function I find the limit of, so how do I do this one?2012-06-08
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    Like I said, I think it was *intended* you know this limit *a priori* - it was not supposed to be new information for you, but an old fact to bring forth. I don't know if your class or textbook or whatnot actually covered it, though. Note that triangle edge ratios are a reasonable *definition* of the trigonometric functions, so evaluating the limits by reasoning with them would be reasonable too. Otherwise, as I alluded, one can use $\sin\to\pm1$ and $\cos\to0$ as $x\to\pm\pi/2$ (respectively) to get $(\circ)$, and then $(\circ)\implies(\bullet)$.2012-06-08
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    The last implication, if you really want something formal, can be done with limiting substitution. Namely, if $x\to\pm\infty$, we can replace it with $x=\tan\theta$ as $\theta\to \pm\pi/2^{\mp}$ respectively inside the limits of $(\bullet)$.2012-06-08
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    But I have arctan, that isn't really related in any way to sin or cos is it?2012-06-08
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    $\tan=\sin/\cos$2012-06-08
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    But arctan isn't acrsin/arccos2012-06-08
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    Correct: arcsin and arccos are irrelevant. We evaluate the limits for $\tan$ in $(\circ)$ and then use $(\circ)\implies(\bullet)$.2012-06-08
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    I do not understnad the dot arrow dot part.2012-06-08
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    I'm saying the two limits in my post tagged with $(\circ)$ logically imply the two limits tagged with $(\bullet)$ in my post. I just gave you the sin/cos sketch of why the limits in $(\circ)$ are true, and we can derive the limits in $(\bullet)$ from those in $(\circ)$ using my comment: replace e.g. $x\to+\infty$ with $\theta\to+\pi/2^-$ and $x$ with $\tan\theta$. This is substitution.2012-06-08
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    But the limits in the pictures are different, one apporaches infinity the other pi/22012-06-08
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    The two limits in the first picture correspond to $\tan$ and $(\circ)$, the two limits in the second picture correspond to $\arctan$ and $(\bullet)$.2012-06-08
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    Oh so you are saying that I need to have memorized the limits of arctan?2012-06-08
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    Precisely that, @Jordan. You will find that having many basic trig properties memorized will greatly aid you in a calculus class.2012-06-08
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    True, it would be helpful from this point forward to have the tan and arctan limits memorized - and if at some point you can't remember, the graphs can be invoked as visual aids. Plus I think this problem intended for you to have the limits memorized, so there is probably no need to submit a proof of the limits for your homework. However, you asked how the limits could be proved, so I gave a terse sketch as a response to this curiosity.2012-06-08
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    I was assuming that there was some sort of formula I could use like finding a derivative or something similar.2012-06-08
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You are given

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx$$

Note first that the function in question is even. For any even function, we have that

$$\int_{-a}^{a}f(x) dx = 2\int_0^a f(x) dx$$. This can be generalized for improper integrals. So we have that

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx=2\int_0^\infty \frac {x^2}{x^6 + 9}dx$$

Now we can make an appropriate change of variables. Note that $x^6 = (x^3)^2$. We let $x^3=u$, then we get $3x^2 dx=du$. This works, because the expression $x^2 dx$ appears in the integral. Now we write the integral in terms of $u$ - remember we have to replace every instance of $x$ in the integral with $u$.

Note that when $x \to \infty$, $u \to \infty$, and $x \to 0$, then $u \to 0$. We get

$$\int_{-\infty}^\infty \frac {x^2}{x^6 + 9}dx=\frac 2 3\int_0^\infty \frac {du}{u^2 + 9}$$

I belive you can evaluate this integral now.

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    I end up with a limit of $1\9 arctan(x^3 /3$ and I have no clue how to evalutate that.2012-06-07
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    Can you find $\ell =\lim\limits_{x \to \infty} \arctan x$?2012-06-07
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    No I am not sure how to do that.2012-06-07
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    Are you aware that $\lim\limits_{x \to \pi/ 2^+}\tan x = +\infty$?2012-06-07
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    No I do not remember that, so it is "undefined" on the interal and I need to make 2 limits out of this expression?2012-06-07
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    Well, the integral you have is called an improper integral, so for the $0$ I guess you see that $\arctan 0=0$. Then, for the upper limit, you have to take $x \to +\infty$. As a general result, the $\arctan$ function has the property that $x \to \pm \infty$, then $\arctan x \to \pm \pi/2$. That follows from the way we define the inverse tangent, and how the tangent behaves for $\pi /2^-$ and $-\pi /2^+$. Errata: in my last comment, it should read $x \to \pi/2^-$. Maybe you should re-read your trig notes.2012-06-07
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    I can't figure out how to do the limit as x approaches pi over 2.2012-06-08
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    @Jordan Then do re-read your trig notes. The tangent tends to $+\infty$ as $x\to \pi/2^-$. In general, it has vertical asymptotes at any multiple of $\pi /2$.2012-06-08
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$$u = x^3, \quad du = 3x^2\,dx$$ Then $$ \int \frac{x^2\,dx}{x^6 + 9} = \int\frac{du/3}{u^2+9}. $$ You get an arctangent.

Be sure to learn this: The expression $x^2\,dx$ begs for letting $u$ be a thrid-degree polynomial, because $x^2$ is the derivative of a third-degree polyonomial.

Later addendum: $$ \int\frac{du/3}{u^2+9} = \int\frac{du/3}{9\left(\frac{u^2}{9}+1\right)} = \frac 1 9 \int \frac{du/3}{(u/3)^2 + 1} = \frac 1 9 \int \frac{dw}{w^2 + 1} $$ $$ = \frac 1 9 \arctan w + C = \frac 1 9 \arctan \frac u 3 + C = \frac 1 9 \arctan \frac{x^3}{3} + C. $$

Recall from trigonometry that $\arctan v\to\pi/2$ as $v\to\infty$ and $\arctan v\to-\pi/2$ as $v\to-\infty$. And $x^3$ approaches $\infty$ as $x\to\infty$ and similarly $x^3\to-\infty$ as $x\to-\infty$. Bottom line: $\pi/9$.

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    I am not sure what to do next.2012-06-07
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    I am not sure what to do next. I am not sure if this method works, I need to have atleast one number to work with and this just leaves two infinity values which are useless.2012-06-07
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    @Jordan $$\begin{eqnarray*} \left. \frac{1}{9}\arctan \frac{x^{3}}{3}\right\vert _{-\infty }^{+\infty } &=&\lim_{x\rightarrow +\infty }\frac{1}{9}\arctan \frac{x^{3}}{3} \\ &&-\lim_{x\rightarrow -\infty }\frac{1}{9}\arctan \frac{x^{3}}{3} \end{eqnarray*}$$2012-06-08
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    @AméricoTavares What about the undefined value and pi/2? It seems like I get an infinity over infinity value.2012-06-08
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    @Jordan $\lim_{x\rightarrow +\infty }\frac{1}{9}\arctan \frac{x^{3}}{3}$ and $\lim_{x\rightarrow -\infty }\frac{1}{9}\arctan \frac{x^{3}}{3}$ are finite, they are not undefined.2012-06-08
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    Is there some trick I use to calculate these values?2012-06-08
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    @Jordan No! You need to understand how $\tan x$ behaves as $x\to \pi/2$ from left and as $x\to -\pi/2$ from right. And what that implies to the behaviour of $\arctan x$. Note here you have $x^3/3$ instead of $x$.2012-06-08
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    @AméricoTavares But I am not working with tanx I am working with $\frac{1}{9} arctanx^3/3$2012-06-08
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    @Jordan The integral is given in terms of $\frac 1 9 \arctan \frac {x^3}{ 3}+C$2012-06-08
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    @AméricoTavares I do not know what that looks like.2012-06-08
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    @Jordan I could give you the answer but I think that you should have a look at your trigonometry book. How is the graph of $\arctan x$?2012-06-08
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    @AméricoTavares I do not have a trig book but according to wolfram the graph of arctan is symetric about the origin and the function we are working with is mirrored around the y axis.2012-06-08
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    @Jordan And what are the asymptotes as $x\to \infty$ and as $x\to -\infty$?2012-06-08
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    @AméricoTavares I am not sure how to find that with arctan.2012-06-08
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    @Jordan We cannot continue discussions here as you know. See http://en.wikipedia.org/wiki/File:Arctangent_Arccotangent.svg2012-06-08
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    @Jordan : Draw a right triangle and look at the sides suitably labeled "opposite" and "adjacent". Make the length of the "adjacent" side $1$. Then tangent $=$ opp/adj $=$ opp/$1$ $=$ opp. As the angle approaches a right angle, i.e. $\pi/2$ radians, the "opposite" side goes to $\infty$. Hence $\lim\limits_{v\to\infty}\arctan v=\pi/2$.2012-06-08
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    ....and more generally, recalling trigonometry will help with many of the things you've posted questions about.2012-06-08
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This method will evaluate the integral, however not in an basic way using substitution, as I assume you want. I just think some will find this method interesting.

Letting $$f(z)=\frac {z^2}{z^6 + 9}$$

Integrating in the positive sense around the contour formed by a semicircle of radius $R$ (called $C_R$) on the complex plane, we have

$$\int_{C_R}f(z)\, dz=\int_{-R}^R f(z)\, dz+\int_\text{Arc} f(z)\, dz$$

As $R \to \infty$, $\int_\text{Arc} f(z)\, dz=0$, thus

$$\lim_{R \to \infty} \int_{C_R}f(z)\, dz=\int_{-\infty}^\infty f(z)\, dz = 2 \pi i\sum \text{Residues of f(z) in }\lim_{R \to \infty}C_R$$

The poles, $z_1$, $z_2$ and $z_3$ of $f(z)$ are (with de Moivre's theorem)

$$z_1=3^{1/3}\exp(\frac{i \pi}{6})$$ $$z_2=3^{1/3}i$$ $$z_3=3^{1/3}\exp(\frac{5 i \pi}{6})$$

The residues of the poles, $b_1$, $b_2$ and $b_3$ respectively, are (I used L'Hopital's rule and the limit definition of the residue, but an alternative may be possible)

$$b_1=-\frac{i}{18}$$ $$b_2=\frac{i}{18}$$ $$b_3=-\frac{i}{18}$$

So finally,

$$\int_{-\infty}^\infty f(z)\, dz=2\pi i (-\frac{i}{18}+\frac{i}{18}-\frac{i}{18})=-\frac{\pi i^2}{9}=\frac{\pi}{9}$$

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    So you need about 6 years of college math to understand this answer?2012-06-08
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    @Jordan I'm not in any university, my friend, nor have I ever been. I have only an interest in mathematics. Anyone who wants to learn complex integration will be able to understand my answer. I don't expect you to understand this, I am merely contributing another interesting method to evaluate the integral.2012-06-08
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    Plus, I object to posting a complete answer to a homework question within one hour!2012-06-08
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In this situation you can either adopt real analysis methods (as shown very elegantly by the posters prior to me), or you can use the residue theorem, a complex analysis technique.

We have an even integrand of a real definite integral from the negative infinity to positive infinity, define $f(z)=\frac{z^2}{z^6+9}$ and define $C$ to be a semicircle on the upper half of the complex plane with a radius large enough to enclose any singularities of $f$.

We say $\int_C f(z) \, dz = \text{answer} = 2\pi i \operatorname{Res}_f$

Solving $z^6=-9$ gives us 6 imaginary solutions, 3 of which are on the upper plane, they are: $\alpha=9^{1/6}\exp(i \pi /6)$, $\beta=9^{1/6} \, i$, $\gamma=9^{1/6} \exp(5i \pi /6)$

Using the residue theorem, we differentiate the denominator of f and evaluate this new function at alpha, beta and gamma and multiply by $2\pi i$ to find the integral:

We evaluate $\pi i z^{-3}$ at alpha. beta and gamma and sum up our results to obtain $\frac{\pi i}{3} 9^{-1/2} (-i)= \frac{\pi i}{9}$

My answer is rough (written on an iPad), but as this is homework perhaps seeing a different method may help you understand the question!

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    I just posted a complex analysis solution here: http://math.stackexchange.com/a/155394/27624 . You may have missed it.2012-06-08
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    Ah it is very unfortunate we posted the same solution, I began working on my mine around 20 minutes prior to posting it (using a tablet) and did not refresh the page! Either way, your solution explains everything step by step2012-06-08