Using trig substitution, integrate: {eq}\displaystyle \int \frac {2x}{\sqrt {x^2 - 9}} \ dx {/eq}.

Question:

Using trig substitution, integrate:

{eq}\displaystyle \int \frac {2x}{\sqrt {x^2 - 9}} \ dx {/eq}.

Trigonometric Substitution

The process of trigonometric substitution is a method of expressing a function in terms of a trigonometric function so the resulting form is much simpler than the initial form of the function. This method is applied in integrals since the resulting form is a common integral or can be solved using basic integration rules.

To integrate the given equation, let's first set the following values.

\begin{align} x &= 3\sec(u) \\ u &= \textrm{arcsec} \left (\frac{x}{3} \right ) \\ \\ \frac{dx}{du} &= \frac{d}{du}[3\sec(u)] \\ \frac{dx}{du} &= 3\frac{d}{du}[\sec(u)] \\ \frac{dx}{du} &= 3[\sec(u)\tan(u)] \\ dx &= 3\sec(u)\tan(u)\,du \end{align}

Substitute these values into the integral.

\begin{align} \int \frac {2x}{\sqrt {x^2 - 9}} \ dx &= \int \frac{2[3\sec(u)]}{\sqrt{\left[3\sec(u) \right ]^2 - 9}} \left [ 3\sec(u)\tan(u)\,du \right ] \\ &= \int \frac{18\sec^2 (u)\tan(u)}{\sqrt{9\sec^2(u) - 9}}\,du \\ &= \int \frac{18\sec^2 (u)\tan(u)}{\sqrt{9[\sec^2(u) - 1]}}\,du \\ &= \int \frac{18\sec^2 (u)\tan(u)}{\sqrt{9}\sqrt{ [\sec^2(u) - 1] }}\,du \\ &= \int \frac{18\sec^2 (u)\tan(u)}{3\sqrt{\sec^2(u) - 1}}\,du \\ &= \int \frac{6\sec^2 (u)\tan(u)}{\sqrt{\sec^2(u) - 1}}\,du \\ &= 6 \int \frac{\sec^2 (u)\tan(u)}{\sqrt{\sec^2(u) - 1}}\,du \end{align}

Use the trigonometric identity, {eq}\tan^2(\theta) + 1= \sec^2(\theta) {/eq}, to simplify the integrand.

\begin{align} \int \frac {2x}{\sqrt {x^2 - 9}} \ dx &= 6 \int \frac{\sec^2 (u)\tan(u)}{\sqrt{\sec^2(u) - 1}}\,du \\ &= 6 \int \frac{\sec^2 (u)\tan(u)}{\sqrt{\tan^2(u)}}\,du \\ &= 6 \int \frac{\sec^2 (u)\tan(u)}{\tan(u)}\,du \\ &= 6 \int \sec^2(u)\,du \end{align}

The resulting integral is a common integral. After finding its solution, substitute back {eq}\displaystyle u = \textrm{arcsec} \left (\frac{x}{3} \right ) {/eq} to the solution.

\begin{align} \int \frac {2x}{\sqrt {x^2 - 9}} \ dx &= 6 \int \sec^2(u)\,du \\ &= 6[\tan(u)] + C \tag{where C is a constant of integration} \\ &= 6\tan(u) + C \\ &= 6\tan \left[\textrm{arcsec}\left (\frac{x}{3} \right ) \right ] + C \end{align}

Finally, apply the identity: {eq}\displaystyle \tan \left(\textrm{arcsec} \left(\theta\right)\right)=\sqrt{\theta^2-1} {/eq}.

\begin{align} \int \frac {2x}{\sqrt {x^2 - 9}} \ dx &= 6\tan \left[\textrm{arcsec}\left (\frac{x}{3} \right ) \right ] + C \\ &= 6\left [ \sqrt{\left (\frac{x}{3} \right )^2 -1} \right ] + C \\ &= 6 \sqrt{\frac{x^2}{9} - 1} + C \\ &= 6\sqrt{\frac{x^2}{9} - \frac{9}{9}} + C \\ &= 6\sqrt{\frac{x^2 - 9}{9}} + C \\ &= 6 \frac{\sqrt{x^2 - 9}}{\sqrt{9}} + C \\ &= \frac{6}{3} \sqrt{x^2 - 9} + C \\ \bf{\int \frac {2x}{\sqrt {x^2 - 9}} \ dx} &= \boxed{\bf{2\sqrt{x^2 - 9} + C}} \end{align}