# How do you integrate int sqrt(x^2-25)/x using trig substitutions?

Oct 21, 2016

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx}$

$\textcolor{red}{x = 5 \sec u}$

color(red)((dx)/(du)=5secutanu

$I = \int \frac{\sqrt{25 {\sec}^{2} u - 25}}{5 \sec u} \times 5 \sec u \tan u \mathrm{du}$

$I = \int \frac{\sqrt{25 \left({\sec}^{2} u - 1\right)}}{5 \sec u} \times 5 \sec u \tan u \mathrm{du}$

now $\textcolor{red}{{\sec}^{2} u = 1 + {\tan}^{2} u}$

$\therefore \textcolor{red}{{\sec}^{2} u - 1 = {\tan}^{2} u}$

$\implies I = \int \left(\frac{\sqrt{25 {\tan}^{2} u}}{5 \sec u}\right) \times 5 \sec u \tan u \mathrm{du}$

$I = \int \frac{5 \tan u}{5 \sec u} \times 5 \sec u \tan u \mathrm{du}$

now simplifying the algebra.

$I = \int \frac{\cancel{5} \tan u}{\cancel{5} \cancel{\sec u}} \times 5 \cancel{\sec u} \tan u \mathrm{du}$

$I = 5 \int \left({\tan}^{2} u\right) \mathrm{du}$

using $\textcolor{red}{{\sec}^{2} u = 1 + {\tan}^{2} u}$ once more.

$I = 5 \int \left({\sec}^{2} u - 1\right) \mathrm{du}$

$I = 5 \tan u - 5 u + C$

but

$\textcolor{red}{x = 5 \sec u \implies u = {\sec}^{-} 1 \left(\frac{x}{5}\right)}$

$\therefore I = 5 \tan \left({\sec}^{-} 1 \left(\frac{x}{5}\right)\right) - 5 {\sec}^{-} 1 x + C$

If you are used to hyperbolic functions the substitution

$x = 5 \cosh u$ would make it less messy!