# How do you integrate int (25+x^2)/sqrt(x^2-4)dx using trigonometric substitution?

Mar 5, 2018

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 27 \ln \left\mid x + \sqrt{4 - {x}^{2}} \right\mid + \frac{x + \sqrt{4 - {x}^{2}}}{2} + C$

#### Explanation:

Substitute $x = 2 \sec t$ $\mathrm{dx} = 2 \sec t \tan t \mathrm{dt}$, and consider first $x \in \left(2 , + \infty\right)$ so that $t \in \left(0 , \frac{\pi}{2}\right)$

int (25+x^2)/sqrt(x^2-4)dx = 2int ((25+4sec^2t)sect tant)/(sqrt(4sec^2t-4) dt

Use now the trigonometric identity:

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

and note that for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive so: $\sqrt{{\sec}^{2} - 1} = \tan t$

and:

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = \frac{1}{2} \int \frac{\left(25 + 4 {\sec}^{2} t\right) \sec t \tan t}{\sqrt{{\sec}^{2} t - 1}} \mathrm{dt} = \int \left(25 + 4 {\sec}^{2} t\right) \sec t \mathrm{dt}$

using the linearity of the integral:

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 25 \int \sec t \mathrm{dt} + 4 \int {\sec}^{3} t \mathrm{dt}$

This are fairly known integrals, but we can go through the solution:

$\int \sec t \mathrm{dt} = \int \sec t \frac{\sec t + \tan t}{\sec t + \tan t} \mathrm{dt}$

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

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

$\int \sec t \mathrm{dt} = \ln \left\mid \sec t + \tan t \right\mid + C$

and:

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

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

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

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

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

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

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

Putting it together:

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 25 \ln \left\mid \sec t + \tan t \right\mid + 2 \sec t \tan t + 2 \ln \left\mid \sec t + \tan t \right\mid + C$

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 27 \ln \left\mid \sec t + \tan t \right\mid + 2 \sec t \tan t + C$

and undoing the substitution, considering that:

$\sec t = \frac{x}{2}$

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

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 27 \ln \left\mid \frac{x}{2} + \sqrt{1 - {x}^{2} / 4} \right\mid + 2 \left(\frac{x}{2} \sqrt{1 - {x}^{2} / 4}\right) + C$

$\int \frac{25 + {x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx} = 27 \ln \left\mid x + \sqrt{4 - {x}^{2}} \right\mid + \frac{x + \sqrt{4 - {x}^{2}}}{2} + C$