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Section 7.3 Trig Substitution

Sometime we have to integrate equations that contain one of the following:

\begin{equation*} \sqrt{a^2 - x^2} \mbox{ , } \sqrt{x^2 - a^2} \mbox{ ,or } \sqrt{x^2 + a^2} \end{equation*}

where a is a positive constant.

To evaluate these, we will use a method called trig substitution.

Note 7.3.1.
To get rid of the square root in \(\sqrt{a^2 - x^2}\text{,}\) we can make the substitution \(x = a \sin(\theta)\text{.}\)

\begin{align*} \sqrt{a^2 - x^2} \amp = \sqrt{a^2 - (a \sin(\theta))^2} \\ \amp = \sqrt{a^2 - a^2\sin^2(\theta)}\\ \amp = \sqrt{a^2 ( 1 - \sin^2(\theta))}\\ \amp = \sqrt{a^2} \cdot \sqrt{\cos^2(\theta)}\\ \amp = a \cdot \cos(\theta) \end{align*}

Similartly, we make the substitution \(x = a \sec(\theta)\) in \(\sqrt{x^2 - a^2}\) to get \(a\tan(\theta)\text{,}\) and \(x = a \tan(\theta)\) in \(\sqrt{x^2 + a^2}\) to get \(a \sec(\theta)\text{.}\)

So to recap:

\begin{align*} \sqrt{a^2 - x^2} \amp = a \cos(\theta) \amp \sqrt{x^2 - a^2} \amp = a \tan(\theta) \amp \sqrt{x^2 + a^2} \amp = a \sec(\theta) \end{align*}

Now let's try and use it in an integral. Let's use \(\displaystyle \int \dfrac{1}{\sqrt{x^2 + 9}}dx\text{.}\)

\begin{align*} \int \dfrac{1}{\sqrt{x^2 + 9}}dx \amp = \int \dfrac{1}{\sqrt{\sec(\theta))^2 - 9}} 3 \sec(\theta)\tan(\theta)d\theta \\ \amp = \int \dfrac{1}{3\tan(\theta)} 3 \sec(\theta)\tan(\theta)d\theta \\ \amp = \int \sec(\theta) d \theta\\ \amp = \ln | \sec(\theta)\tan(\theta) | + c \\ \amp = \ln | \dfrac{x}{3} + \dfrac{\sqrt{x^2 - 9}}{3} | + c \end{align*}

Table 7.3.2.
\(x = 3 \sec(\theta)\)
\(dx = 3 \sec(\theta) \tan(\theta) d\theta\)

Note 7.3.3.

Notice that trig substitution functions similarly to u-substitution but backwards.