Integration by Trigonometric Substitution Examples 1

We will now look at further examples of Integration by Trigonometric Substitution. For more examples, see the Integration by Trigonometric Substitution Examples 2 page. Also, 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 following integral $\int \frac{\sqrt{9 - x^2}}{x^2} \: dx$.

We have a function containing the form $a^2 - x^2$, hence we can make the substitution $x = 3 \sin \theta$. It thus follows that $dx = 3 \cos \theta \: d \theta$. Hence we get that:

(1)

\begin{align} \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{\sqrt{9 - 9\sin ^2 \theta}}{9\sin ^2 \theta} 3 \cos \theta \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{\sqrt{9(1 - \sin ^2 \theta)}}{9\sin ^2 \theta} 3 \cos \theta \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{\sqrt{9\cos ^2 \theta}}{9\sin ^2 \theta} 3 \cos \theta \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{3\cos \theta}{9\sin ^2 \theta} 3 \cos \theta \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{9\cos ^2 \theta}{9\sin ^2 \theta} \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \frac{\cos ^2 \theta}{\sin ^2 \theta} \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \cot ^2 \theta \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = \int \csc ^2 \theta - 1 \: d \theta \\ \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = -\cot \theta - \theta + C \end{align}

We will now construct a triangle knowing that $3\sin \theta = x$, or more appropriately $\sin \theta = \frac{x}{3}$:

We can infer from our triangle that $\cot \theta = \frac{\sqrt{9 - x^2}}{x^2}$. We also know that $\theta = \sin ^{-1} (\frac{x}{3})$. Hence it follows that:

(2)

\begin{align} \int \frac{\sqrt{9 - x^2}}{x^2} \: dx = -\frac{\sqrt{9 - x^2}}{x^2} - \sin ^{-1} (\frac{x}{3}) + C \end{align}

Example 2

Evaluate the following integral $\int \frac{x^2}{\sqrt{25 - x^2}} \: dx$.

Notice that in our integrand we have the form $a^2 - x^2$, so we will attempt to integrate this function using trigonometric substitution. First let $a^2 = 25$ so that $a = 5$. We thus construct our substitution for $x$ to be $x = a \sin \theta$ or rather, $x = 5 \sin \theta$. We will also be applying the trigonometric identity $\cos ^2 \theta = 1 - \sin ^2 \theta$.

We will also note that $\frac{x}{5} = \sin \theta$, $\theta = \sin ^{-1} \left ( \frac{x}{5} \right )$ and $dx = 5 \cos \theta d \theta$. Now we will begin using our substitution that $x = 5 \sin \theta$ in our integral and simplify to get:

(3)

\begin{align} \int \frac{(5 \sin \theta)^2}{\sqrt{25 - (5 \sin \theta)^2}} \cdot 5 \cos \theta \: d \theta \\ = \int \frac{25 \sin ^2 \theta}{\sqrt{25 - 25 \sin ^2 \theta}} \cdot 5 \cos \theta \: d \theta \\ = \int \frac{25 \sin ^2 \theta}{\sqrt{25(1 - \sin ^2 \theta)}} \cdot 5 \cos \theta \: d \theta \\ = \int \frac{25 \sin ^2 \theta}{\sqrt{25\cos ^2 \theta}} \cdot 5 \cos \theta \: d \theta \quad [\mathrm{By\:trigonometric\:identity}]\\ = \int \frac{25 \sin ^2 \theta}{5 \cos \theta} \cdot 5 \cos \theta \: d \theta \quad = \int 25 \sin ^2 \theta \: d \theta = 25 \int \sin ^2 \theta \: d \theta \end{align}

We will now use an integral property that we have already learned. Recall that $\int \sin ^2 \theta \: d \theta = \frac{\theta - \sin \theta \cos \theta}{2}$. Applying this we get that:

(4)

\begin{align} = 25 \int \frac{\theta - \sin \theta \cos \theta}{2} \: d \theta \\ = \frac{25}{2} \int \theta - \sin \theta \cos \theta \: d \theta \\ \end{align}

Now recall that $\theta = \sin ^{-1} \left ( \frac{x}{5} \right )$, and $\sin \theta = \frac{x}{5}$. We need to construct a triangle to figure out what $\cos \theta$ equals in terms of $x$. The triangle below illustrates our case:

From this we can get that $\cos \theta = \frac{\sqrt{25 - x^2}}{5}$. Putting these into our integral we get:

(5)

\begin{align} \quad \int \frac{x^2}{\sqrt{25 - x^2}} \: dx = \frac{25}{2} \left ( \sin ^{-1} \left( \frac{x}{5} \right ) - \frac{x}{5} \cdot \frac{\sqrt{25 - x^2}}{5} \right )\\ \quad \int \frac{x^2}{\sqrt{25 - x^2}} \: dx = \frac{25}{2} \sin ^{-1} \left( \frac{x}{5} \right ) - \frac{1}{2} x \sqrt{25 - x^2} + C \\ \end{align}