# How do you evaluate the integral: sqrt(x^2-a^2)/x dx?

Apr 18, 2018

$\int \frac{\sqrt{{x}^{2} - {a}^{2}}}{x} \mathrm{dx} = \sqrt{{x}^{2} - {a}^{2}} - a {\sec}^{-} 1 \left(\frac{x}{a}\right) + C$

#### Explanation:

For the integral involving the root $\sqrt{{x}^{2} - {a}^{2}} ,$ we use the substitution:

$x = a \sec \theta$

$\mathrm{dx} = a \sec \theta \tan \theta d \theta$

So, we get

int(cancelacancelsecthetatanthetasqrt(a^2(sec^2theta-1)))/(cancelacancelsectheta))d theta

Recalling the identity ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$, we get

$\int a \tan \theta \sqrt{{\tan}^{2} \theta} = \int a {\tan}^{2} \theta d \theta$

To integrate this, we'll use the identity ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$ again.

$\int a \left({\sec}^{2} \theta - 1\right) d \theta = \int a {\sec}^{2} \theta d \theta - \int a d \theta = a \tan \theta - a \theta + C$

We need to get things in terms of $x .$ Recalling that $x = a \sec \theta , \sec \theta = \frac{x}{a} , \theta = {\sec}^{-} 1 \left(\frac{x}{a}\right)$

To find the tangent, we'll use the identity

${\tan}^{2} \theta = {\sec}^{2} \theta - 1$

${\tan}^{2} \theta = {x}^{2} / {a}^{2} - 1$

$\tan \theta = \frac{\sqrt{{x}^{2} - {a}^{2}}}{a}$

Thus,

$\int \frac{\sqrt{{x}^{2} - {a}^{2}}}{x} \mathrm{dx} = \frac{a \sqrt{{x}^{2} - {a}^{2}}}{a} - a {\sec}^{-} 1 \left(\frac{x}{a}\right) + C$

$\int \frac{\sqrt{{x}^{2} - {a}^{2}}}{x} \mathrm{dx} = \sqrt{{x}^{2} - {a}^{2}} - a {\sec}^{-} 1 \left(\frac{x}{a}\right) + C$

Apr 18, 2018

$I = \sqrt{{x}^{2} - {a}^{2}} - a {\sec}^{-} 1 \left(\frac{x}{a}\right) + c$

#### Explanation:

Here,

$I = \int \frac{\sqrt{{x}^{2} - {a}^{2}}}{x} \mathrm{dx}$

Let ,$x = a \sec u \implies \mathrm{dx} = a \sec u \tan u \mathrm{du}$

$\mathmr{and} \sec u = \frac{x}{a} \implies u = {\sec}^{-} 1 \left(\frac{x}{a}\right)$

$\therefore I = \int \frac{\sqrt{{a}^{2} {\sec}^{2} u - {a}^{2}}}{\cancel{\left(a \sec u\right)}} \left(\cancel{a \sec u} \tan u\right) \mathrm{du}$

$= \int \sqrt{{a}^{2} {\tan}^{2} u} \cdot \tan u \mathrm{du}$

$= \int \left(a \tan u\right) \cdot \tan u \mathrm{du}$

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

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

$= a \left(\tan u - u\right) + c$

$= a \left(\sqrt{{\sec}^{2} u - 1} - u\right) + c$, where, $\sec u = \frac{x}{a} , u = {\sec}^{-} 1 \left(\frac{x}{a}\right)$

$= a \left[\sqrt{{\left(\frac{x}{a}\right)}^{2} - 1} - {\sec}^{-} 1 \left(\frac{x}{a}\right)\right] + c$

$= a \left[\frac{\sqrt{{x}^{2} - {a}^{2}}}{a} - {\sec}^{-} 1 \left(\frac{x}{a}\right)\right] + c$

$= \sqrt{{x}^{2} - {a}^{2}} - a {\sec}^{-} 1 \left(\frac{x}{a}\right) + c$