# How do you evaluate the integral int dx/(1-x^2)^(5/2)?

Aug 15, 2017

$\frac{x \left(3 - 2 {x}^{2}\right)}{3 {\left(1 - {x}^{2}\right)}^{\frac{3}{2}}} + C$

#### Explanation:

Let $x = \sin \theta$. This implies that $\mathrm{dx} = \cos \theta d \theta$.

$\int \frac{\mathrm{dx}}{1 - {x}^{2}} ^ \left(\frac{5}{2}\right)$

$= \int \frac{\cos \theta d \theta}{1 - {\sin}^{2} \theta} ^ \left(\frac{5}{2}\right)$

$= \int \cos \frac{\theta}{{\cos}^{2} \theta} ^ \left(\frac{5}{2}\right) d \theta$

$= \int {\sec}^{4} \theta d \theta$

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

Let $u = \tan \theta$ so $\mathrm{du} = {\sec}^{2} \theta d \theta$.

$= \int \left(1 + {u}^{2}\right) \mathrm{du}$

$= u + \frac{1}{3} {u}^{3} + C$

$= \tan \theta + \frac{1}{3} {\tan}^{3} \theta + C$

$= \frac{1}{3} \tan \theta \left(3 + {\tan}^{2} \theta\right) + C$

Where $x = \sin \theta$, we have a triangle where the opposite side is $x$ and the hypotenuse is $1$. Thus, the adjacent side is $\sqrt{1 - {x}^{2}}$ and $\tan \theta = x / \sqrt{1 - {x}^{2}}$.

$= \frac{x}{3 \sqrt{1 - {x}^{2}}} \left(3 + {x}^{2} / \left(1 - {x}^{2}\right)\right)$

$= \frac{1}{3 \sqrt{1 - {x}^{2}}} \left(\frac{3 - {x}^{2}}{1 - {x}^{2}}\right)$

$= \frac{x \left(3 - 2 {x}^{2}\right)}{3 {\left(1 - {x}^{2}\right)}^{\frac{3}{2}}} + C$