Evaluate the following integral. integral frac squareroot(x^2 - 25)/x dx, x > 5 What substitution...

Question:

Evaluate the following integral. {eq}\int \frac {\sqrt{x^2 - 25}}{x} dx, x > 5 {/eq}

What substitution will be the most helpful for evaluating this integral?

A. {eq}x = 5 \sec \theta {/eq}

B. {eq}x = 5 \tan \theta{/eq}

C. {eq}x = 5 \sin \theta{/eq}

Find dx.

{eq}dx = ({/eq}_____{eq})d\theta{/eq}

Rewrite the given integral using this substitution.

{eq}\int \frac{\sqrt{x^2 - 25}}{x} dx = \int ( {/eq} _____{eq})d \theta{/eq}

Evaluate the indefinite integral.

{eq}\int \frac{\sqrt{x^2 - 25}}{x} dx = {/eq} _____

(Use C as the arbitrary constant.)

Integrals by trigonometric substitution

Integrals by trigonometric substitution seem to be difficult but with the correct substitution they can be clear enough to be able to master this type of integrals

There are several in the integration by trigonometric substitution and in each of these applying the correct substitution is essential

When in an integral we have the form {eq}\sqrt (x^2-a^2) {/eq} the replacement of {eq}x=a \sec \theta {/eq}.Remembering that (a) will always have a value greater than zero.

Evaluate the following integral. {eq}\int \frac {\sqrt{x^2 - 25}}{x} dx, x > 5 {/eq}

What substitution will be the most helpful for evaluating this integral?

The most useful substitution to evaluate this integral is A. {eq}x = 5 \sec \theta {/eq}

Find dx.

{eq}dx = 5\sec \theta \tan \theta \,\, d\theta {/eq}

Rewrite the given integral using this substitution.

{eq}I= \int \frac{\sqrt{x^2 - 25}}{x} dx = \int \frac{\sqrt{25\sec^2 \theta - 25}}{5\sec \theta} 5\sec \theta \tan \theta \,\, d\theta\\ I= \int \frac{\sqrt{25(\sec^2 \theta - 1)}}{5\sec \theta} 5\sec \theta \tan \theta \,\, d\theta\\ I= \int \frac{5\tan \theta}{5\sec \theta} 5\sec \theta \tan \theta \,\, d\theta\\ I= \int 5\tan^2 \theta \,\, d\theta\\ {/eq}

Evaluate the indefinite integral.

{eq}I= \int 5\tan^2 \theta \,\, d\theta\\ \textrm {Apply trigonometric identity} \,\,\,\, \sec^2 \theta -\tan^2 \theta=1 \,\,\,\, \rightarrow \,\,\,\,\, \tan^2 \theta= \sec^2 \theta -1\\ I= 5\int \sec^2 \theta -1 \,\, d\theta\\ I=5\tan (\theta) -5\theta +C \textrm {Replace the variable}\\ x = 5 \sec \theta \,\,\,\, \rightarrow \,\,\,\,\, \theta = \sec^{-1} (\frac{x}{5})\\ I=5\tan (\sec^{-1} (\frac{x}{5})) -5\sec^{-1} (\frac{x}{5}) +C\\ {/eq} 