Integration by Trigonometric Substitution Examples 2
We will now look at some more examples of integration by trigonometric substitution. Please look back at the Integration by Trigonometric Substitution Examples 1 for more examples. Additionally, recall the following table:
Within the Integrand | Appropriate Substitution | Appropriate Trigonometric Identity |
---|---|---|
$a^2 + x^2$ | $x = a \tan \theta$, $-\frac{\pi}{2} ≤ \theta ≤ \frac{\pi}{2}$ | $\tan ^2 \theta + 1 = \sec ^2 \theta$ |
$a^2 - x^2$ | $x = a \sin \theta$, $-\frac{\pi}{2} ≤ \theta ≤ \frac{\pi}{2}$ | $1 - \sin ^2 \theta = \cos ^2 \theta$ |
$x^2 - a^2$ | $x = a \sec \theta$, $0 ≤ \theta ≤ \frac{\pi}{2}$ | $\sec ^2 \theta - 1 = \tan ^2 \theta$ |
Example 1
Evaluate the integral $\int \frac{\sqrt{25x^2 - 4}}{x} \: dx$.
We first note that we almost have the form $x^2 - a^2$, however, the $25$ in front of the $x^2$ is problematic. For that reason, we will factor out theta $25$ from the radical:
(1)We now have the form $x^2 - a^2$ like we wanted. Let $a^2 = \left (\frac{2}{5} \right)^2$ so that $a = \frac{2}{5}$. We make the substitution that $x = a \sec \theta$, or rather, $x = \frac{2}{5} \sec \theta$. The corresponding trigonometric identity that we will apply is $\tan ^2 \theta = \sec ^2 \theta - 1$.
We also note that $x = \frac{2}{5} \sec \theta$ which means that $\cos \theta = \frac{2}{5x}$. Furthermore, $\theta = \cos ^{-1} \left ( \frac{2}{5x} \right )$. Lastly, $dx = \frac{2}{5} \sec \theta \tan \theta \: d \theta$. We are now ready to apply our substitution:
(2)Now recall that $\int \tan ^2 \theta \: d \theta = \tan \theta - \theta + C$, and therefore:
(3)We know that $\theta = \cos ^{-1} \left ( \frac{2}{5x} \right )$, but we must compute $\tan \theta$. The following triangle allows us to do that:
Therefore, $\tan \theta = \frac{\sqrt{25x^2 - 4}}{2}$. Thus:
(4)Example 2
Evaluate the integral $\int \frac{x}{\sqrt{3 - 2x - x^2}} \: dx$.
Our first step is to covert the polynomial under the radical into the "complete-the-square form" as follows:
(5)Therefore, $\int \frac{x}{\sqrt{3 - 2x - x^2}} \: dx = \int \frac{x}{\sqrt{4 - (x + 1)^2}} \: dx$. We now have a function containing a part with the form $a^2 - x^2$. Let $a^2 = 4$ so that $a = 2$. We will use the trigonometric substitution $x = a \sin \theta$, that is $x + 1 = 2 \sin \theta$.
We note that $\frac{x + 1}{2} = \sin \theta$, $\theta = \sin ^{-1} \left ( \frac{x + 1}{2} \right )$, and that $dx = 2\cos \theta$. Therefore:
(6)We will now construct a triangle from the before mentioned $\sin \theta = \frac{x + 1}{2}$. By the Pythagorean theorem we get that $(x + 1)^2 + y^2 = 2^2$ or rather $x^2 +2x + 1 + y^2 = 4$, so then $y = \sqrt{3 - 2x - x^2}$:
Therefore $\cos \theta = \frac{\sqrt{3 - 2x - x^2}}{2}$, and putting this into our integral we get:
(7)