# How do I integrate intsqrt(x²-49)dx?

Jul 9, 2015

$\sqrt{{x}^{2} - 49} \propto \sqrt{{x}^{2} - {a}^{2}}$

Let:
$x = a \sec \theta$ with $a = 7$:
$\mathrm{dx} = 7 \sec \theta \tan \theta d \theta$
$\sqrt{{x}^{2} - 49} = \sqrt{{7}^{2} {\sec}^{2} \theta - {7}^{2}} = 7 \sqrt{{\sec}^{2} \theta - 1}$

$= 7 \tan \theta$

Thus:

$= \int 7 \tan \theta \cdot 7 \sec \theta \tan \theta d \theta$

$= 49 \int \sec \theta {\tan}^{2} \theta d \theta$

$= 49 \int \sec \theta \left({\sec}^{2} \theta - 1\right) d \theta$

$= \textcolor{h i g h l i g h t}{49} \int {\sec}^{3} \theta - \sec \theta d \theta$

Solving these individually:

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

$= \int \frac{{\sec}^{2} \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} d \theta$

u-substitution. Let:
$u = \sec \theta + \tan \theta$
$\mathrm{du} = \sec \theta \tan \theta + {\sec}^{2} \theta d \theta$

$\implies \int \frac{1}{u} \mathrm{du}$

$= \ln | u | = \textcolor{g r e e n}{\ln | \sec \theta + \tan \theta |}$

The other one:

$\int {\sec}^{3} \theta d \theta$

Integration by Parts. Let:
$u = \sec \theta$
$\mathrm{du} = \sec \theta \tan \theta$
$\mathrm{dv} = {\sec}^{2} \theta d \theta$
$v = \tan \theta$

$\implies u v - \int v \mathrm{du}$

$= \sec \theta \tan \theta - \int \sec \theta {\tan}^{2} \theta d \theta$

$= \sec \theta \tan \theta - \int {\sec}^{3} \theta - \sec \theta d \theta$

$\int {\sec}^{3} \theta d \theta = \sec \theta \tan \theta - \int {\sec}^{3} \theta d \theta + \int \sec \theta d \theta$

Add the ${\sec}^{3} \theta$ over. No need to evaluate it. We also know $\int \sec \theta d \theta$ already.

$2 \int {\sec}^{3} \theta d \theta = \sec \theta \tan \theta + \int \sec \theta d \theta$

$2 \int {\sec}^{3} \theta d \theta = \sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |$

$\int {\sec}^{3} \theta d \theta = \textcolor{g r e e n}{\frac{1}{2} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]}$

Overall:

$= {\overbrace{\frac{1}{2} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]}}^{\int {\sec}^{3} \theta d \theta} - {\overbrace{\ln | \sec \theta + \tan \theta |}}^{\int \sec \theta d \theta}$

$= \frac{1}{2} \left[\sec \theta \tan \theta - \ln | \sec \theta + \tan \theta |\right]$

Recall that $\sec \theta = \frac{x}{7}$ and $\tan \theta = \frac{\sqrt{{x}^{2} - 49}}{7}$, and don't forget the $\textcolor{h i g h l i g h t}{49}$:

$= 49 \left\{\frac{1}{2} \left[\frac{x}{7} \cdot \frac{\sqrt{{x}^{2} - 49}}{7} - \ln | \frac{x}{7} + \frac{\sqrt{{x}^{2} - 49}}{7} |\right]\right\}$

$= \frac{1}{2} \left[\left(x \sqrt{{x}^{2} - 49}\right) - 49 \ln | \frac{x}{7} + \frac{\sqrt{{x}^{2} - 49}}{7} |\right]$

$= \frac{1}{2} \left[\left(x \sqrt{{x}^{2} - 49}\right) - 49 \ln | \left(\frac{1}{7}\right) \left(x + \sqrt{{x}^{2} - 49}\right) |\right]$

$= \frac{1}{2} \left[\left(x \sqrt{{x}^{2} - 49}\right) - 49 \left(\ln | x + \sqrt{{x}^{2} - 49} | + \ln \left(\frac{1}{7}\right)\right)\right] + C$

The remaining $- \frac{49}{2} \ln \left(\frac{1}{7}\right)$ gets embedded into $C$:

$= \textcolor{b l u e}{\frac{1}{2} \left[x \sqrt{{x}^{2} - 49} - 49 \ln | x + \sqrt{{x}^{2} - 49} |\right] + C}$