# How do you integrate int sqrt(x^2-25)/x dx using trigonometric substitution?

Jun 8, 2018

$\sqrt{{x}^{2} - 25} - 5 a r c \sec \left(\frac{x}{5}\right) + C .$

#### Explanation:

Suppose that, $I = \int \frac{\textcolor{red}{\sqrt{{x}^{2} - 25}}}{\textcolor{g r e e n}{x}} \mathrm{dx}$, and, let, $\textcolor{g r e e n}{x = 5 \sec u}$.

$\therefore \textcolor{b l u e}{\mathrm{dx} = 5 \sec u \tan u \mathrm{du}}$.

$\therefore I = \int \frac{\textcolor{red}{\sqrt{25 {\sec}^{2} u - 25}}}{\textcolor{g r e e n}{5 \sec u}} \cdot \textcolor{b l u e}{5 \sec u \tan u \mathrm{du}}$,

$= \int \textcolor{red}{5 \tan u} \cdot \textcolor{b l u e}{\tan u \mathrm{du}}$,

$= 5 \int {\tan}^{2} u \mathrm{du}$,

$= \int \left({\sec}^{2} u - 1\right) \mathrm{du}$,

$= 5 \left(\tan u - u\right)$.

Here, color(red)(5tanu=sqrt(x^2-25),) and, color(green)(x=5secu rArr u=arcsec(x/5).

$\Rightarrow I = \sqrt{{x}^{2} - 25} - 5 a r c \sec \left(\frac{x}{5}\right) + C ,$ as desired!

Enjoy Maths.!

Jun 8, 2018

Here is a Second Method to find the Integral.

#### Explanation:

Let, $I = \int \frac{\sqrt{25 - {x}^{2}}}{x} \mathrm{dx} = \int \sqrt{\frac{{x}^{2} - 25}{x} ^ 2} \mathrm{dx}$.

$\therefore I = \int \sqrt{1 - {\left(\frac{5}{x}\right)}^{2}} \mathrm{dx}$.

We subst. $\frac{5}{x} = \sin t . \therefore \sqrt{1 - {\left(\frac{5}{x}\right)}^{2}} = \cos t$.

$\text{Also, } \frac{5}{x} = \sin t . \therefore x = 5 \csc t . \therefore \mathrm{dx} = - 5 \csc t \cot t \mathrm{dt}$.

$\therefore I = \int \left(\cos t\right) \left(- 5 \csc t \cot t\right) \mathrm{dt}$,

$= - 5 \int \left\{\cos t \cdot \frac{1}{\sin} t \cdot \cos \frac{t}{\sin} t\right\} \mathrm{dt}$,

$= - 5 \int {\cot}^{2} t \mathrm{dt}$,

$= - 5 \int \left({\csc}^{2} t - 1\right) \mathrm{dt}$,

$= - 5 \left(- \cot t - t\right)$,

$= 5 \left\{\left(\cos \frac{t}{\sin} t\right) + t\right\}$,

$= 5 \left\{\frac{\sqrt{1 - {\left(\frac{5}{x}\right)}^{2}}}{\frac{5}{x}} + \arcsin \left(\frac{5}{x}\right)\right\}$.

$\Rightarrow I = \sqrt{{x}^{2} - 25} + 5 \arcsin \left(\frac{5}{x}\right) + C$.

I leave it to the Questioner to show that both the Answers are

equivalent!

Enjoy Maths.!