# Using trigonometric substitution what is the integral of intdx/(x^2sqrt(x^2-1) ?

## intdx/(x^2sqrt(x^2-1)

Feb 8, 2018

$I = \sqrt{1 - \frac{1}{x} ^ 2} + C$

#### Explanation:

We want to solve

$I = \int \frac{1}{{x}^{2} \cdot \sqrt{{x}^{2} - 1}} \mathrm{dx}$

Let $x = \sec \left(u\right) \implies \frac{\mathrm{dx}}{\mathrm{du}} = \sec \left(u\right) \tan \left(u\right)$

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

Use the identity ${\sec}^{2} \left(a\right) - 1 = {\tan}^{2} \left(a\right)$

$I = \int \frac{1}{{\sec}^{2} \left(u\right) \cdot \tan \left(u\right)} \sec \left(u\right) \tan \left(u\right) \mathrm{du}$

$= \int \frac{1}{\sec} \left(u\right) \mathrm{du}$

$= \int \cos \left(u\right) \mathrm{du}$

$= \sin \left(u\right) + C$

Substitute $u = {\sec}^{-} 1 \left(x\right)$

$I = \sin \left({\sec}^{-} 1 \left(x\right)\right) + C$

Use $\sin \left({\sec}^{-} 1 \left(x\right)\right) = \sqrt{1 - \frac{1}{x} ^ 2}$

$I = \sqrt{1 - \frac{1}{x} ^ 2} + C$

Feb 8, 2018

$\int \frac{\mathrm{dx}}{{x}^{2} \cdot \sqrt{{x}^{2} - 1}} = \frac{\sqrt{{x}^{2} - 1}}{x} + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{{x}^{2} \cdot \sqrt{{x}^{2} - 1}}$

After using $x = \cosh u$ and $\mathrm{dx} = \sinh u \cdot \mathrm{du}$ transforms, this integral became

$\int \frac{\sinh u \cdot \mathrm{du}}{{\left(\cosh u\right)}^{2} \cdot \sqrt{{\left(\cosh u\right)}^{2} - 1}}$

=$\int \frac{\sinh u \cdot \mathrm{du}}{{\left(\cosh u\right)}^{2} \cdot \sqrt{{\left(\sinh u\right)}^{2}}}$

=$\int \frac{\sinh u \cdot \mathrm{du}}{{\left(\cosh u\right)}^{2} \cdot \sinh u}$

=$\int {\left(\sech u\right)}^{2} \cdot \mathrm{du}$

=$\tanh u + C$

=$\sinh \frac{u}{\cosh} u + C$

=$\frac{\sqrt{{\left(\cosh u\right)}^{2} - 1}}{\cosh} u + C$

=$\frac{\sqrt{{x}^{2} - 1}}{x} + C$