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$\displaystyle \int_0^8 \sqrt{x^4+4x^2}\,dx$.

Alright, so I thought I had this figured out. Here's what I did:

  1. I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$.
  2. I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta) (4\tan^2(\theta) + 4)}$.
  3. Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta) (\tan^2(\theta) + 1))}$
  4. Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$
  5. This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$.
  6. Now, the integral of this is $4\sec(\theta)$
  7. And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$...
  8. But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer

I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)

Any help is greatly appreciated!

PS, what's the syntax for doing sqrts and exponentials?

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By the way, you can right click on LaTex code and select "Show Source" to get the correct sytanx. –  JavaMan Jul 29 '11 at 16:23
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You started with $x$ in the integral (in the title), then switched to $\theta$; I rewrote it so your original integral agrees with the title, and $\theta$ is the new variable. –  Arturo Magidin Jul 29 '11 at 16:54
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@Silver: You changed the integrand, but you didn't change the $dx$... –  Arturo Magidin Jul 29 '11 at 17:04
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@Silver: I think I may have told you this before, but I really appreciate it when a poster shows their work up front, uses proper spelling, and responds. –  mixedmath Jul 29 '11 at 17:47
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If you put braces around the argument of sqrt the whole thing goes under the root sign. I did it on step 3. –  Ross Millikan Jul 29 '11 at 19:12
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3 Answers 3

up vote 13 down vote accepted

There are other ways of doing this integral, but let me try to fix your attempt, which is certainly a fine idea as far as it goes.

The main problem I spot with your development is that you forgot to change the $dx$ when you did the change of variable. (And you should be able to evaluate $\sec(\arctan a)$ as well; we'll get to that shortly).

So: you start with $$\int_0^8 \sqrt{x^4+4x^2}\,dx = \int_0^8 \sqrt{x^2(x^2+4)}\,dx.$$

Then you do the change of variable $x=2\tan(\theta)$. If you do this, then $$dx = 2\sec^2\theta\,d\theta;$$ when $x=0$, you want $\theta=0$, and when $x=8$ you want $\theta=\arctan(4)$ (you are correct there). So the integral actually becomes, after changing integrand, limits, and the $dx$: $$\begin{align*} \int_0^8\sqrt{x^2(x^2+4)}\,dx &= \int_0^{\arctan(4)}\sqrt{4\tan^2\theta(4\tan^2\theta+4)}2\sec^2\theta\,d\theta\\ &= \int_0^{\arctan(4)} \sqrt{16\tan^2\theta(\tan^2\theta+1)}2\sec^2\theta\,d\theta\\ &= \int_0^{\arctan(4)}8\sec^2\theta\sqrt{\tan^2\theta\sec^2\theta}\,d\theta\\ &= 8\int_0^{\arctan(4)}\sec^2\theta|\tan\theta\sec\theta|\,d\theta. \end{align*}$$ Now, on $[0,\arctan(4)]$, both tangent and secant are positive, so we can drop the absolute value signs (something else you were not careful with), and the integral becomes $$8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta.$$ Set $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, so we have $$\begin{align*} 8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta &= 8\int_{\sec(0)}^{\sec(\arctan(4))}u^2\,du\\ &= \frac{8}{3}u^3\Biggm|_{\sec(0)}^{\sec(\arctan(4))}\\ &=\frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right). \end{align*}$$

Now, $\sec(0) = 1$. What about $\sec(\arctan(4))$?

Say $\psi$ is an angle with $\tan(\psi)=4$. Take a right triangle with this angle; by scaling, we may assume the opposite side has length $4$ and the adjacent side has length $1$. Then the hypotenuse has length $\sqrt{17}$, so the cosine of $\psi$ is $\frac{1}{\sqrt{17}}$, hence the secant has value $\sqrt{17}$. So $\sec(\arctan(4)) = \sec(\psi) = \sqrt{17}$. Thus, the integral is: $$\begin{align*} \int_0^8\sqrt{x^4+4x^2}\,dx &= \frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right)\\ &=\frac{8}{3}\left( \sqrt{17}^3 - 1^3\right)\\ &= \frac{8}{3}\left(17\sqrt{17} - 1\right). \end{align*}$$

In summary: your mistake was that when you did the change of variable, you forgot to change the differential as well; and at the end you could have simplified $\sec(\arctan(4))$.

Of course, the better way of doing this is to factor out $x$ from the square root, and then recognize that you can do $$\int_0^8\sqrt{x^4+4x^2}\,dx = \int_0^8x\sqrt{x^2+4}\,dx$$ with the change of variable $u=x^2+4$, like DJC suggested. But I thought you might like to know where exactly your approach went wrong (the substitution), and whether it could be brought to a correct conclusion (it could).

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That was a very thorough explanation! Wow... thank you :) –  user13327 Jul 29 '11 at 17:28
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@Silver, there's more where that came from (literally)! –  The Chaz 2.0 Jul 29 '11 at 17:59
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If $x=2\tan\theta$, what is $\mbox{d}x$ in terms of $\theta$?

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oooooooooooooooooooooooh! dang –  user13327 Jul 29 '11 at 17:07
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Hint:

$$\begin{align} \int_0^8 \sqrt{x^4 + 4x^2}dx &= \int_0^8\sqrt{x^2(x^2 + 4)}dx \\ &= \int_0^8 |x| \sqrt{x^2 + 4}dx \\ &= \int_0^8 x \sqrt{x^2 + 4} dx \end{align}$$

Now try a $u$-substitution.

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Or just spot an antiderivative directly. –  Geoff Robinson Jul 29 '11 at 16:24
    
Alright, so that works and I got the right answer that way, but why didn't my trig substitution work? @Geoff: Well, it looks a lot like an arcsec.. if it were in the form of 1 / sqrt{x^4 + 4x^2}dx, then I'd know how to turn it into an arcsec, but... I don't know how from it's current state –  user13327 Jul 29 '11 at 16:33
    
^^^ I did, to 0 to arctan(4) from 0 to 8 –  user13327 Jul 29 '11 at 16:53
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@Silver: Maybe a $u$ substitution was the way to go if you can't spot an antiderivative right off, but you still seem to be trying to make it more complicated than it is. You're looking for an antiderivative for $x(x^{2}+4)^{\frac{1}{2}}$, and what's before the bracket looks an awful lot like the derivative of what's inside the bracket. –  Geoff Robinson Jul 29 '11 at 17:04
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@ Silver: PS: I think your mistake in your working was that you didn't remember the factor $\frac{dx}{d \theta}$. –  Geoff Robinson Jul 29 '11 at 17:09
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