Question

Solve given definite integral.

${\int }_3^5 \sqrt{x^2-9}dx$

Short answer

$-\frac{9}{2}{\cosh }^{-1}\left(\frac{5}{3}\right)+\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{5}{3}\right)\right)$

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_3^5 \sqrt{x^2-9}dx$

The objective is to solve the integral.

Step2. Assumptions

To solve the integral, let $x=3\cos h\left(u\right)$

Now, derivative of above equation is solved below.

$$\begin{gathered} x=3\cos h\left(u\right) \\ dx=3\sin h\left(u\right)du \end{gathered}$$

Step3. Solution

Now, use the identity ${\cosh }^2\left(u\right)-1={\sinh }^2\left(u\right).$

So,

$\begin{array}{r}\sqrt{x^2-9}=\sqrt{9{\cosh }^2\left(u\right)-9}\\ =3\sqrt{{\cosh }^2\left(u\right)-1}\\ =3\sqrt{{\sinh }^2\left(u\right)}\\ =3\sinh \left(u\right)\end{array}$

Step4. Solution

The integral is solved below.

$\begin{array}{r}\int 9{\sinh }^2\left(u\right)du\\ =9\int {\sinh }^2\left(u\right)du\\ =9\int \left(\frac{1}{2}\cosh \left(2u\right)-\frac{1}{2}\right)du\\ =-9\int \frac{1}{2}du+\frac{9}{2}\int \cosh \left(2u\right)du\\ =-\frac{9}{2}u+\frac{9}{4}\sinh \left(2u\right)\\ =-\frac{9}{2}\left({\cosh }^{-1}\left(\frac{x}{3}\right)\right)+\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{x}{3}\right)\right)\end{array}$

Step5. Solution

The definite integral is solved below.

$\begin{array}{r}{\int }_3^5 \sqrt{x^2-9}dx\\ ={\left[\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{x}{3}\right)\right)-\frac{9}{2}{\cosh }^{-1}\left(\frac{x}{3}\right)\right]}_3^5\\ =\left[\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{5}{3}\right)\right)-\frac{9}{2}{\cosh }^{-1}\left(\frac{5}{3}\right)\right]\\ -\left[\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{3}{3}\right)\right)-\frac{9}{2}{\cosh }^{-1}\left(\frac{3}{3}\right)\right]\\ =-\frac{9}{2}{\cosh }^{-1}\left(\frac{5}{3}\right)+\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{5}{3}\right)\right)\end{array}$

Therefore, the value is$-\frac{9}{2}{\cosh }^{-1}\left(\frac{5}{3}\right)+\frac{9}{4}\sinh \left(2{\cosh }^{-1}\left(\frac{5}{3}\right)\right)$