# How do I evaluate int\sqrt{20x^{2}-5}dx?

Jun 28, 2018

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} x \sqrt{4 {x}^{2} - 1} - \frac{\sqrt{5}}{4} \ln \left(2 \left\mid x \right\mid + \sqrt{4 {x}^{2} - 1}\right) + C$

#### Explanation:

Substitute $x = \frac{1}{2} \sec t$ with $t \in \left(0 , \frac{\pi}{2}\right)$, $\mathrm{dx} = \frac{1}{2} \sec t \tan t \mathrm{dt}$:

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{1}{2} \int \sqrt{20 {\left(\sec \frac{t}{2}\right)}^{2} - 5} \sec t \tan t \mathrm{dt}$

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{1}{2} \int \sqrt{5 {\sec}^{2} t - 5} \sec t \tan t \mathrm{dt}$

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} \int \sqrt{{\sec}^{2} t - 1} \sec t \tan t \mathrm{dt}$

Use now the trigonometric identity:

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

and as $\tan x > 0$ for $t \in \left(0 , \frac{\pi}{2}\right)$:

$\sqrt{{\sec}^{2} t - 1} = \tan t$

Then:

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} \int \sec t {\tan}^{2} t \mathrm{dt}$

and using the same identity:

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} \int \sec t \left({\sec}^{2} t - 1\right) \mathrm{dt}$

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} \int {\sec}^{3} t \mathrm{dt} - \frac{\sqrt{5}}{2} \int \sec t \mathrm{dt}$

Solve now the integrals:

$\int \sec t = \ln \left\mid \sec t + \tan t \right\mid + C$

and:

$\int {\sec}^{3} t \mathrm{dt} = \int \sec t \frac{d}{\mathrm{dt}} \left(\tan t\right) \mathrm{dt}$

Integrate by parts:

$\int {\sec}^{3} t \mathrm{dt} = \sec t \tan t - \int \tan t \frac{d}{\mathrm{dt}} \left(\sec t\right) \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \sec t \tan t - \int \sec t {\tan}^{2} t \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \sec t \tan t - \int \sec t \left({\sec}^{2} t - 1\right) \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \sec t \tan t - \int {\sec}^{3} t \mathrm{dt} + \int \sec t \mathrm{dt}$

Solve now for the original integral:

$2 \int {\sec}^{3} t \mathrm{dt} = \sec t \tan t + \int \sec t \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \frac{\sec t \tan t}{2} + \frac{1}{2} \int \sec t \mathrm{dt}$

$\int {\sec}^{3} t \mathrm{dt} = \frac{\sec t \tan t}{2} + \frac{1}{2} \ln \left\mid \sec t + \tan t \right\mid + C$

Then:

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{4} \sec t \tan t - \frac{\sqrt{5}}{4} \ln \left\mid \sec t + \tan t \right\mid + C$

Undo the substitution:

$\tan t = \sqrt{{\sec}^{2} t - 1} = \sqrt{4 {x}^{2} - 1}$

$\sec t = 2 x$

so:

$\int \sqrt{20 {x}^{2} - 5} \mathrm{dx} = \frac{\sqrt{5}}{2} x \sqrt{4 {x}^{2} - 1} - \frac{\sqrt{5}}{4} \ln \left(2 \left\mid x \right\mid + \sqrt{4 {x}^{2} - 1}\right) + C$