# How do you integrate int 1/sqrt(x^2-4)^3 by trigonometric substitution?

Apr 21, 2018

$\int \setminus \frac{1}{{\sqrt{{x}^{2} - 4}}^{3}} \setminus \mathrm{dx} = - \setminus \frac{x}{4 \sqrt{{x}^{2} - 4}} + C$

#### Explanation:

Whenever we encounter a square root of the form $\sqrt{{x}^{2} - a}$, then this calls for substituting $\sec t$, and subsequently making use of the trigonometric identity ${\sec}^{2} t - 1 = {\tan}^{2} t$. Taking this line, we substitute $x \setminus \mapsto g \left(t\right)$ with

$g \left(t\right) = 2 \sec t$,

$g ' \left(t\right) = 2 \sec t \tan t$, and

${g}^{- 1} \left(x\right) = {\sec}^{- 1} \left(\frac{x}{2}\right)$:

$\int \setminus \frac{1}{{\sqrt{{x}^{2} - 4}}^{3}} \setminus \mathrm{dx} =$

$= {\left[\setminus \frac{1}{4} \int \setminus \frac{\sec t \tan t}{{\sqrt{{\sec}^{2} t - 1}}^{3}} \setminus \mathrm{dt}\right]}_{t = {\sec}^{- 1} \left(\frac{x}{2}\right)} =$

$= {\left[\setminus \frac{1}{4} \int \setminus \frac{\sec t}{{\tan}^{2} t} \setminus \mathrm{dt}\right]}_{t = {\sec}^{- 1} \left(\frac{x}{2}\right)} =$

= [ \frac{1}{4} int \frac{cos t}{sin^2 t} \ dt ]_{t = sec^{-1} ( x / 2 ).

At this point, we may interpret $\sin t$ as an inner function (with derivative $\cos t$), and can thus write the integral as

${\left[\setminus \frac{1}{4} \int \setminus \frac{1}{{u}^{2}} \setminus \mathrm{du}\right]}_{u = \sin t = \sin {\sec}^{- 1} \left(\frac{x}{2}\right)}$.

Remembering that

$\sin x = \frac{\sqrt{{\sec}^{2} x - 1}}{\sec} x$,

we simplify the resubstitution, and get

${\left[\setminus \frac{1}{4} \int \setminus \frac{1}{{u}^{2}} \setminus \mathrm{du}\right]}_{u = \frac{\sqrt{{x}^{2} / 4 - 1}}{\frac{x}{2}} = \frac{\sqrt{{x}^{2} - 4}}{x}} =$

${\left[- \setminus \frac{1}{4 u} + C\right]}_{u = \frac{\sqrt{{x}^{2} - 4}}{x}} =$

$- \setminus \frac{x}{4 \sqrt{{x}^{2} - 4}} + C$.