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$\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx$.

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?

  • 0
    After step 1, rewrite $\sqrt{x^2(x^2+4)}$ as $x\sqrt{x^2+4}$ and use regular substition.2012-06-08

3 Answers 3

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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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    @Silver, there's more where **that** came from (literally)!2011-07-29
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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.

  • 0
    ^^^ Definitely where I messed up!2011-07-29
3

If $x=2\tan\theta$, what is $\mbox{d}x$ in terms of $\theta$?

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    oooooooooooooooooooooooh! d$a$$n$g2011-07-29