91影视

The integral is as follows.

${鈭�}_3^5鈥�\sqrt{x^2鈭�9}dx$

The objective is to solve the integral.

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

Now, derivative of above equation is solved below.

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

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

So,

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

The integral is solved below.

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

The definite integral is solved below.

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

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