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

##### 1 Answer
Jul 26, 2016

${\left({x}^{2} - {a}^{2}\right)}^{\frac{3}{2}} / \left(3 {a}^{2} {x}^{3}\right) + C$.

#### Explanation:

We take subst. $x = a \sec t , s o , \mathrm{dx} = a \sec t \cdot \tan t \mathrm{dt}$

Also, $\frac{\sqrt{{x}^{2} - {a}^{2}}}{x} ^ 4$

$= \frac{\sqrt{{a}^{2} {\sec}^{2} t - {a}^{2}}}{{a}^{4} {\sec}^{4} t}$

$= \frac{a \tan t}{{a}^{4} {\sec}^{4} t} = \frac{1}{a} ^ 3 \cdot \sin \frac{t}{\cos} t \cdot {\cos}^{4} t = \frac{\sin t {\cos}^{3} t}{a} ^ 3$

$\therefore I = \int \frac{\sqrt{{x}^{2} - {a}^{2}}}{x} ^ 4 \mathrm{dx} = \int \frac{\sin t {\cos}^{3} t}{a} ^ 3 \cdot a \sec t \cdot \tan t \mathrm{dt}$

$= \frac{1}{a} ^ 2 \int \sin t {\cos}^{3} t \cdot \frac{1}{\cos} t \cdot \sin \frac{t}{\cos} t \mathrm{dt}$

$= \frac{1}{a} ^ 2 \int {\sin}^{2} t \cos t \mathrm{dt} \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$

$= \frac{1}{a} ^ 2 \int {\left(\sin t\right)}^{2} d \left(\sin t\right)$

$= \frac{1}{a} ^ 2 \cdot {\left(\sin t\right)}^{3} / 3$

Now, $\sec t = \frac{x}{a} \Rightarrow \cos t = \frac{a}{x} \Rightarrow \sin t = \sqrt{1 - {a}^{2} / {x}^{2}} = \frac{\sqrt{{x}^{2} - {a}^{2}}}{x}$

$\therefore I = \frac{1}{3 {a}^{2}} \cdot {\left(\frac{\sqrt{{x}^{2} - {a}^{2}}}{x}\right)}^{3} = {\left({x}^{2} - {a}^{2}\right)}^{\frac{3}{2}} / \left(3 {a}^{2} {x}^{3}\right) + C$.

To proceed further from $\left(\star\right)$, we can use another subst. $\sin t = y$,

so, $\cos t \mathrm{dt} = \mathrm{dy}$, thus giving,

$I = \frac{1}{a} ^ 2 \int {y}^{2} \mathrm{dy} = {y}^{3} / \left(3 {a}^{2}\right) = {\left(\sin t\right)}^{3} / \left(3 {a}^{2}\right)$