# How do you find the indefinite integral of ∫ dx/ (√5-x^2) ^3 ?

Apr 15, 2018

$= \frac{1}{5} \cdot \frac{x}{\sqrt{5 - {x}^{2}}} + C$

#### Explanation:

$I = \int \frac{1}{\sqrt{5 - {x}^{2}}} ^ 3 \mathrm{dx}$

$x = \sqrt{5} \sin \theta$

$\mathrm{dx} = \sqrt{5} \cos \theta \cdot d \left(\theta\right)$

Substitute

$\int \frac{1}{\sqrt{5 - {x}^{2}}} ^ 3 \mathrm{dx}$=$\int \frac{1}{\sqrt{5 - 5 {\sin}^{2} \theta}} ^ 3 \sqrt{5} \cos \theta d \left(\theta\right)$

color(green) (5-5sin^2theta=5cos^2theta

=$\int \frac{1}{{\left(\sqrt{5 {\cos}^{2} \theta}\right)}^{3}} \sqrt{5} \cos \theta d \left(\theta\right)$

Simplify

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

=$\frac{1}{5} \tan \theta + C$

$= \frac{1}{5} \sin \frac{\theta}{\cos} \theta + C$

$= \frac{1}{5} \sin \frac{\theta}{\sqrt{\left(1 - {\sin}^{2} \theta\right)}} + C \ldots \to w h e r e , \sin \theta = \frac{x}{\sqrt{5}}$

Reverse the substitution

$= \frac{1}{5} \frac{\frac{x}{\sqrt{5}}}{\sqrt{1 - {\left(\frac{x}{\sqrt{5}}\right)}^{2}}} + C$

$= \frac{1}{5} \frac{\frac{x}{\sqrt{5}}}{\sqrt{1 - {x}^{2} / 5}} + C$

$= \frac{1}{5} \cdot \frac{x}{\sqrt{5 - {x}^{2}}} + C$