Integration by Trigonometric Substitution Examples 2

# 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)
\begin{align} = \int \frac{\sqrt{25(x^2 - \frac{4}{25})}}{x} \: dx \\ = \int 5 \cdot \frac{\sqrt{ x^2 - \frac{4}{25}}}{x} \: dx \\ = 5 \int \frac{\sqrt{x^2 - \left ( \frac{2}{5} \right)^2}}{x} \: dx \end{align}

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)
\begin{align} = 5 \int \frac{\sqrt{\left( \frac{2}{5} \sec \theta\right)^2 - \left ( \frac{2}{5} \right)^2}}{\left( \frac{2}{5} \right) \sec \theta} \cdot \left ( \frac{2}{5} \right ) \sec \theta \tan \theta \: d \theta \\ = 5 \int \sqrt{ \left ( \frac{2}{5} \sec \theta \right )^2 - \left ( \frac{2}{5} \right )^2} \cdot \tan \theta \: d \theta \\ = 5 \int \sqrt{ \left ( \frac{2}{5} \right ) ^2 \left ( \sec ^2 \theta - 1 \right ) } \cdot \tan \theta \: d \theta \\ = 5 \int \sqrt{ \left ( \frac{2}{5} \right ) ^2 \tan ^2 \theta} \cdot \tan \theta \: d \theta \quad [\mathrm{By\:trigonometric\:identity}] \\ = 5 \int \frac{2}{5} \tan ^2 \theta \: d \theta \\ = 2 \int \tan ^2 \theta \: d \theta \end{align}

Now recall that $\int \tan ^2 \theta \: d \theta = \tan \theta - \theta + C$, and therefore:

(3)
\begin{align} = 2 \left( \tan \theta - \theta \right) + C \end{align}

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)
\begin{align} = 2 \left ( \frac{\sqrt{25x^2 - 4}}{2} - \cos ^{-1} \left ( \frac{2}{5x} \right ) \right) + C \\ = \sqrt{25x^2 - 4} - 2 \cos ^{-1} \left ( \frac{2}{5x} \right ) + C \end{align}

## 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)
\begin{align} 3 - 2x - x^2 = -x^2 -2x - (1) + 3 + (1) \\ 3 - 2x - x^2 = -x^2 - 2x - 1 + 4 \\ 3 - 2x - x^2 = -(x + 1)^ 2 + 4 \\ 3 - 2x - x^2 = 4 - (x + 1)^2 \end{align}

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)
\begin{align} = \int \frac{2 \sin \theta - 1}{\sqrt{4 - 4 \sin ^2 \theta}} \cdot 2 \cos \theta \: d \theta\\ = \int \frac{2 \sin \theta - 1}{\sqrt{4(1 - \sin ^2 \theta)}} \cdot 2 \cos \theta \: d \theta\\ = \int \frac{2 \sin \theta - 1}{2 \cos \theta} \cdot 2 \cos \theta \: d \theta\\ = \int 2 \sin \theta - 1 \: d \theta \\ = -2 \cos \theta - \theta + C \end{align}

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)
\begin{align} = -2 \left ( \frac{\sqrt{3 - 2x - x^2}}{2} \right ) - \sin ^{-1} \left ( \frac{x + 1}{2} \right ) + C \\ = - \sqrt{3 - 2x - x^2} - \sin ^{-1} \left ( \frac{x + 1}{2} \right ) + C \end{align}