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Evaluate $$\int\limits_2^4\frac{\sqrt{x^2-4}}{x^2}\mathrm dx$$

My working:

$x=2\sec\theta\quad\Rightarrow\quad\theta=\arccos\left(\frac{2}{x}\right)$

$dx=2\sec\theta\tan\theta d\theta$

$I=\int\frac{\sqrt{4\sec^2\theta-4}}{4\sec^2\theta}2\sec\theta\tan\theta d\theta=\int\frac{\tan^2\theta}{\sec\theta}d\theta=\int\frac{\sin^2\theta}{\cos\theta}d\theta=\int\sec\theta d\theta-\int\cos\theta d\theta\\=\ln|\sec\theta+\tan\theta|-\sin\theta+C\\=\left.\ln\left|\frac{x}{2}+\frac{\sqrt{1-(2/x)^2}}{2/x}\right|-\sqrt{1-\left(\frac{2}{x}\right)^2}\right]_2^4$

EDIT

$=\ln\left|\frac{4+\sqrt{12}}{2}\right|-\sqrt{1-\frac{1}{4}}-\ln\left|\frac{2+\sqrt{0}}{2}\right|-\sqrt{1-1}\\=\ln|2+\sqrt{3}|-\frac{\sqrt{3}}{2}\qquad\blacksquare$

Ali Caglayan
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ahorn
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    Why don't you take the limits along? It makes the substitution easier. I typed it in my TI and it seems to be correct IF that -1 at the end is ommitted – imranfat Aug 19 '14 at 19:04
  • I did this in a test 2 hours ago and it was running in my mind until I wrote this down. In the process I found my initial error. Thanks for checking my work! I agree, taking the limits along would have been easier. It's irritating how much effort I have to go to to figure out where I went wrong on one little integral. – ahorn Aug 19 '14 at 19:21

1 Answers1

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Yes, you are completely right, except the $-1$ term. You have:

\begin{align} & \left.\ln\left|\frac{x}{2}+\frac{\sqrt{1-(2/x)^2}}{2/x}\right|-\sqrt{1-\left(\frac{2}{x}\right)^2}\right|_2^4 = \\ = & \ln\left|\frac{4+\sqrt{12}}{2}\right|-\sqrt{1-\frac{1}{4}}-\ln\left|\frac{2+\sqrt{0}}{2}\right|-\color{red}{\sqrt{1-0}} \end{align}

You divide $2$ by $2$ and get's $0$ under the square root sign, it should be $\sqrt{1-1}$, and the result:

$$ \ln|2+\sqrt{3}|-\frac{\sqrt{3}}{2}\blacksquare $$

m0nhawk
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