right arrow key to step forward through presentation.left arrow key to step backwards through presentation.b key to return to the beginning of the presentation.e key to go to the end of the presentation.We wish to compute \[\int\frac{\sqrt{x^2-4}}x\, dx\]
The substitution to use is \(x=2\sec\theta\).
This substitution depends on the Pythagorean identity \[ 1 + \tan^2\theta = \sec^2\theta \]
or \[ \sec^2\theta - 1 = \tan^2\theta \]
So \[ x^2 - 4 = (2\sec\theta)^2 - 4\]
\[ = 4\sec^2\theta - 4 = 4(\sec^2\theta-1) \]
\[ = 4\tan^2\theta \]
So \[ \sqrt{x^2-4} = \sqrt{(2\sec\theta)^2 - 4} \]
\[ = \sqrt{4\tan^2\theta} = 2\tan\theta \]
(ignoring the \(\pm\)).
Now \[ \frac{d}{d\theta}\sec\theta = \sec\theta\tan\theta \]
So we may write \[ d \sec\theta = \sec\theta\tan\theta \, d\theta. \]
So we can change variables in the integral:
\[ \definecolor{blue}{RGB}{0,0,255} \int\frac{\sqrt{{\color{blue}x}^2-4}}{\color{blue}x}\, d{\color{blue}x} = \int\frac{\sqrt{ ({\color{blue}2\sec\theta})^2 - 4 }}{{\color{blue}2\sec\theta}} \, d \,{\color{blue}2\sec\theta}. \]
\[ = \int\frac{\sqrt{4\tan^2\theta}}{2\sec\theta}\, 2\sec\theta\tan\theta \, d\theta \]
\[ = \int\frac{2\tan\theta}{2\sec\theta}\, 2\sec\theta\tan\theta\, d\theta \]
\[ = \int 2 \tan^2\theta \, d\theta. \]
We now have to find the integral \[\int 2\tan^2\theta\, d\theta.\]
But we can write \[ \tan^2\theta = \sec^2\theta - 1\]
So \[ \int 2\tan^2\theta \, d\theta = 2\int\tan^2\theta\, d\theta \]
\[ = 2\int(\sec^2\theta - 1)\, d\theta \]
\[ = 2(\tan\theta - \theta) + C. \]
So using \( x = 2\sec\theta \), we have found \[ \int\frac{\sqrt{x^2-4}}x\, dx = \int 2\tan^2\theta\, d\theta = 2(\tan\theta - \theta) + C \]
So we have found the integral, but we must change variables back to \(x\).
We began with \( x = 2\sec\theta \), but what is \(\tan\theta\) in terms of \(x\)?
We can solve this problem by thinking of a right triangle:
This should work since \[ \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{x}{2}. \]
But we need \(\tan\theta\).
Use Pythagorus to find the missing side of the triangle.
So \(2^2 + y^2 = x^2\) which gives \(y=\sqrt{x^2-4}.\)
So we have:
So \[\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\sqrt{x^2-4}}{2}\]
We also need to convert \(\theta\) back to \(x\).
Since \[ \sec\theta = \frac x2\]
we might write \[ \theta = \sec^{-1}(x/2)\]
But this function might not be convenient (it's not on typical calculator keyboards).
So instead write \[ \cos\theta = \frac 2x\]
and then \[ \theta = \cos^{-1}(2/x)\]
So we finally can write
\[ \int\frac{\sqrt{x^2-4}}x\, dx = \int 2\tan^2\theta\, d\theta = 2(\tan\theta - \theta) + C \]
\[ = 2\left(\frac{\sqrt{x^2-4}}{2} - \cos^{-1}(2/x)\right) + C\]
\[ = \sqrt{x^2-4} - 2\cos^{-1}(2/x) + C\]
So we are done.