I wanted to return to the example we were considering in class today, before too much time passes and we forget where we were.

Recall, we were trying to evaluate the integral $\int \frac{\sqrt{x^2-9}}{x}dx$. To do so, we tried making a substitution where $x = 3 \sec(\theta)$. In order to make this substitution, we used $x = 3 \sec(\theta)$ to create a dictionary that would take us from an integral in terms of $x$ to an integral in terms of $\theta$.

We saw that if $x = 3 \sec(\theta)$, then (after a little work) $\sqrt{x^2-9} = 3 \tan(\theta)$ and $dx = 3 \sec(\theta) \tan(\theta)d \theta.$

This means that $\int \frac{\sqrt{x^2-9}}{x}dx = \int \frac{3 \tan(\theta)}{3\sec(\theta)}3\sec(\theta)\tan(\theta)d\theta.$

After simplifying, this is equal to $3\int \tan^2(\theta)d\theta$.

This is how far we got in class. The point is that this is an integral we know how to evaluate.

To continue, $3\int \tan^2(\theta)\d\theta = 3 \int(\sec^2(\theta) – 1)d\theta = 3(\tan(\theta) – \theta) + C.$

To finish, we have to go back to our dictionary to write our answer in terms of $x$ instead of $\theta$. Remember, $x = 3 \sec(\theta)$, which means that $\frac{x}{3} = \sec(\theta)$. We have a $\theta$ to replace AND a $\tan(\theta)$ to replace.

To replace the $\tan(\theta)$ we don’t have to get $\theta$ by itself. We use $\frac{x}{3} = \sec(\theta)$ to label the sides of a right triangle. Draw a right triangle and label one of the acute angles by $\theta$. Since $\sec(\theta)$ is defined as the length of the hypotenuse over the length of the adjacent side, we can label the hypotenuse by $x$ and the side adjacent to $\theta$ by $3$. We can use the Pythagorean theorem to find the length of the side opposite to $\theta$: it’s $\sqrt{x^2-3}$. Then the $\tan(\theta)$ term in our answer is opposite over adjacent, which is $\frac{\sqrt{x^2-3}}{3}$.

To replace $\theta$ we do have to get $\theta$ by itself. We apply $\sec^{-1}$ to both sides of the equation $\frac{x}{3} = \sec(\theta)$ to get $\sec^{-1}(\frac{x}{3}) = \theta$.

This means our final answer is:

$\int \frac{\sqrt{x^2-9}}{x}dx = 3(\tan(\theta) – \theta) + C = 3(\frac{\sqrt{x^2-3}}{3} –  \sec^{-1}(\frac{x}{3})) + C

= \sqrt{x^2-3} – 3\sec^{-1}(\frac{x}{3}) + C$.

 

A few people asked during class, “Where does $x = 3 \sec(\theta)$ come from?” That’s a tricky question to answer, because it doesn’t really *come from* anywhere. But we see from this example, that by making this substitution, we’re able to evaluate the integral. In some sense it *comes from* the Pythagorean identity $\tan^2(\theta) = \sec^2(\theta) – 1$.

In general, to make a trigonometric substitution:

  • if you see $x^2 – a^2$ let $x=a\sec(\theta)$,
  • if you see $a^2 – x^2$ let $x=a\sin(\theta)$,
  • if you see $x^2 + a^2$ let $x=a\tan(\theta)$.

This should be enough for you to get a solid start on the Trigonometric Substitution Webwork set, which is due on March 1.