# Evaluate using trigonometric substitution. Sketch and label the associated right triangle. \int...

## Question:

Evaluate using trigonometric substitution. Sketch and label the associated right triangle.

{eq}\int \frac {x}{\sqrt {x^2 - 7}} dx {/eq}

## Trigonometric Substitutions

To evaluate an integral like {eq}\displaystyle \int \sqrt{x^2-a^2}dx, {/eq} we can use the trigonometric substitution

{eq}\displaystyle x=a\sec \theta, dx=a\sec \theta \tan \theta d\theta {/eq} and the integral becomes a trigonometric integral.

We may also need to use the following trigonometric formulae

{eq}\displaystyle \sec^2 \theta =1+\tan^2 \theta\\ \displaystyle \frac{d}{d\theta}(\sec \theta) =\sec \theta \tan \theta\\ \displaystyle \frac{d}{d\theta}(\tan \theta) =\sec^2 \theta. {/eq}

To evaluate the integral {eq}\displaystyle \int\frac{x}{\sqrt{x^2-7}}\ dx, {/eq} we will use the following trigonometric substitution

{eq}\displaystyle x=\sqrt{7}\ \sec \theta, dx=\sqrt{7}\ \sec \theta \tan \theta d\theta\\ \displaystyle \sqrt{x^2-7}=\sqrt{7} \sqrt{\sec^2 \theta -1} =\sqrt{7}\tan \theta. {/eq}

Therefore,

{eq}\displaystyle \begin{align}\int\frac{x}{\sqrt{x^2-7}}\ dx&= \int \frac{7\sec^2 \theta \tan \theta }{\sqrt{7}\ \tan \theta }\ d\theta\\ &= \sqrt{7}\int \sec^2 \theta \ d\theta\\ &= \sqrt{7} \tan \theta +C, C- \text{ constant of integration }\\ &\text{ substituting back }x=\sqrt{7}\ \sec \theta \iff \tan \theta =\sqrt{\sec^2 \theta -1}=\frac{1}{\sqrt{7}}\sqrt{ x^2-7}\\ &=\boxed{\sqrt{ x^2-7} +C} \end{align} {/eq}