Evaluate: integral Square root of {x^2 - 64} dx

Question:

Evaluate: {eq}\displaystyle \int \sqrt {x^2 - 64} \ dx {/eq}

Indefinite integrals:


Integration is the process of finding the anti - derivative.

In this integration, we are going to use an identity as follows,

{eq}\displaystyle \int \sqrt{x^2 - a^2}\ dx = \frac {x}{2}\sqrt{x^2 - a^2} - \frac {a^2}{2}\ln |x + \sqrt{x^2 - a^2}| + C {/eq}

Here {eq}C {/eq} is the constant of integration.

Answer and Explanation:


We have,

$$\displaystyle \begin{align*} \displaystyle \int \sqrt {x^2 - 64} \ dx &= \int \sqrt {x^2 - 8^2}\ dx \end{align*} $$


Use the identity, {eq}a = 8 {/eq},

$$\displaystyle \begin{align*} \int \sqrt {x^2 - 64} \ dx &= \frac {x}{2}\sqrt{x^2 - 64} - \frac {64}{2}\ln |x + \sqrt{x^2 - 64}| + C \\ \int \sqrt {x^2 - 64} \ dx &= \frac {x}{2}\sqrt{x^2 - 64} - 32 \ln |x + \sqrt{x^2 - 64}| + C \end{align*} $$

Here {eq}C {/eq} is the constant of integration.


Learn more about this topic:

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Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11
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