Textbook Problem

Finding or Evaluating an Integral In Exercises 1–8, use the basic integration rules to find or evaluate the integral.

${\displaystyle \int x\sqrt{x^2-36}dx}$

To determine

To calculate: The value of the indefinite integral ${\displaystyle \int x\sqrt{x^2-36}dx}$.

Answer to Problem 1RE

Solution:

The value of the indefinite integral ${\displaystyle \int x\sqrt{x^2-36}dx}$ is $\frac{1}{3}{\left(x^2-36\right)}^{3/2}+C$.

Explanation of Solution

Given:

The integral is ${\displaystyle \int x\sqrt{x^2-36}dx}$.

Formula used:

The power formula,

${\displaystyle \int u^{n}du=\frac{u^{n+1}}{n+1}+C}$

Calculation:

Consider the integral $y={\displaystyle \int x\sqrt{x^2-36}dx}$.

Rewrite the equation as,

$y={\displaystyle \int \sqrt{x^2-36}\left(xdx\right)}$

Let $u=x^2-36$.

Differentiate both sides with respect to x.

$\begin{array}{c}\frac{du}{dx}=\frac{d}{dx}\left(x^2-36\right)\\ \frac{du}{dx}=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(36\right)\\ \frac{du}{dx}=2x+0\\ du=2xdx\end{array}$

Substitute $u\text{for}{x}^2-36$ and $\frac{du}{2}\text{for}xdx$.

$\begin{array}{c}y={\displaystyle \int \sqrt{u}\left(\frac{du}{2}\right)}\\ =\frac{1}{2}{\displaystyle \int \sqrt{u}du}\\ =\frac{1}{2}{\displaystyle \int u^{1/2}du}\end{array}$

Solve the above integral as below,

$\begin{array}{c}y=\frac{1}{2}\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+C\\ =\frac{1}{2}\left[\frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]+C\\ =\frac{1}{2}\left[\frac{2}{3}{u}^{\frac{3}{2}}\right]+C\\ =\frac{1}{3}{u}^{\frac{3}{2}}+C\end{array}$

Resubstitute $x^2-36\text{for}u$.

$y=\frac{1}{3}{\left(x^2-36\right)}^{3/2}+C$

Hence, the value of the indefinite integral ${\displaystyle \int x\sqrt{x^2-36}dx}$ is $\frac{1}{3}{\left(x^2-36\right)}^{3/2}+C$.