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

May 25, 2018

$\textcolor{g r e e n}{\int \sqrt{{x}^{2} - 25} \cdot \mathrm{dx} = - \frac{25 \cdot \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right) - x \cdot \sqrt{{x}^{2} - 25}}{2}}$

#### Explanation:

The integral is $\int \sqrt{{x}^{2} - 25} \cdot \mathrm{dx}$

Let suppose:

$x = 5 \sec \theta$

$\mathrm{dx} = 5 \cdot \sec \theta \cdot \tan \theta \cdot d \left(\theta\right)$

$\sqrt{{x}^{2} - 25} = \sqrt{25 \left({\sec}^{2} \theta - 1\right)} = 5 \tan \theta$

the integral become after suppose:

$\int \sqrt{{x}^{2} - 25} \cdot \mathrm{dx} = \int 5 \tan \theta \cdot 5 \sec \theta \cdot \tan \theta \cdot d \left(\theta\right)$

$25 \int \sec \theta \cdot {\tan}^{2} \theta \cdot d \left(\theta\right) = 25 \int \sec \theta \cdot \left({\sec}^{2} \theta - 1\right) \cdot d \left(\theta\right)$

$25 \int {\sec}^{3} \theta - \sec \theta \cdot d \left(\theta\right) = 25 \int {\sec}^{3} \theta - 25 \int \sec \theta \cdot d \left(\theta\right)$

Firstly let solve color(red)[intsec^3theta*d(theta) by using integral y parts:

color(red)(I=intsec^3(theta)d(theta)...to(1)

$\text{Using "color(blue)"Integration by Parts}$

$\int \left(u \cdot v\right) \mathrm{dx} = u \int v \mathrm{dx} - \int \left(u ' \int v \mathrm{dx}\right) d \left(\theta\right)$

Let $u = \sec \left(\theta\right) \mathmr{and} v = {\sec}^{2} \left(\theta\right)$

$\implies u ' = \sec \left(\theta\right) \tan \left(\theta\right) \mathmr{and} \int v d \left(\theta\right) = \tan \left(\theta\right)$

$I = \sec \left(\theta\right) \times \tan \left(\theta\right) - \int \sec \left(\theta\right) \tan \theta \times \tan \left(\theta\right) d \left(\theta\right)$

$= \sec \theta \tan \left(\theta\right) - \int \sec \left(\theta\right) {\tan}^{2} \left(\theta\right) \mathrm{dx} + c$

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

$I = \sec \left(\theta\right) \tan \left(\theta\right) - \textcolor{red}{\int {\sec}^{3} \left(\theta\right) \mathrm{dx}} + \int \sec \left(\theta\right) d \left(\theta\right) + c$

$I = \sec \left(\theta\right) \tan \left(\theta\right) - \textcolor{red}{I} + \int \sec \left(\theta\right) d \left(\theta\right) + c \ldots \to$from1

$I + I = \sec \left(\theta\right) \tan \left(\theta\right) + \int \sec \left(\theta\right) d \left(\theta\right) + c$

$2 I = \sec \left(\theta\right) \tan \left(\theta\right) + \ln | \sec \left(\theta\right) + \tan \left(\theta\right) | + C$

$I = \frac{1}{2} \sec \left(\theta\right) \tan \left(\theta\right) + \frac{1}{2} \ln | \sec \left(\theta\right) + \tan \left(\theta\right) | + C$

secondly let find color(blue)[intsectheta*d(theta)

$\int \sec \theta \cdot d \left(\theta\right) = \int \sec \theta \cdot \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} \cdot d \left(\theta\right)$

$\int \frac{{\sec}^{2} \theta + \sec \theta \cdot \tan \theta}{\sec \theta + \tan \theta} \cdot d \left(\theta\right) = \ln | \sec \theta + \tan \theta | + c$

$\textcolor{g r e e n}{25 \int {\sec}^{3} \theta - 25 \int \sec \theta \cdot d \left(\theta\right)} =$

$25 \left[\frac{1}{2} \sec \left(\theta\right) \tan \left(\theta\right) + \frac{1}{2} \ln | \sec \left(\theta\right) + \tan \left(\theta\right) |\right] - 25 \left[\ln | \sec \theta + \tan \theta |\right] + c$

$= \frac{x \cdot \sqrt{{x}^{2} - 25}}{2} - \frac{25 \cdot \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right)}{2}$

After simplified it we will get:

$= - \frac{25 \cdot \ln \left(\left\mid \sqrt{{x}^{2} - 25} + x \right\mid\right) - x \cdot \sqrt{{x}^{2} - 25}}{2}$