# How do you find the antiderivative of int x^2/sqrt(4-x^2)dx?

Feb 22, 2017

$2 a r c \sin \left(\frac{x}{2}\right) - \frac{x}{2} \sqrt{4 - {x}^{2}} + C .$

#### Explanation:

We know that,

$\left(1\right) : \int \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = \frac{x}{2} \sqrt{{a}^{2} - {x}^{2}} + {a}^{2} / 2 a r c \sin \left(\frac{x}{a}\right) + {c}_{1.}$

$\left(2\right) : \int \frac{1}{\sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} = a r c \sin \left(\frac{x}{a}\right) + {c}_{2.}$

Hence, $I = \int {x}^{2} / \sqrt{4 - {x}^{2}} \mathrm{dx}$

$= - \int \frac{- {x}^{2}}{\sqrt{4 - {x}^{2}}} \mathrm{dx} = - \int \frac{\left(4 - {x}^{2}\right) - 4}{\sqrt{4 - {x}^{2}}} \mathrm{dx} ,$

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

$= - \int \sqrt{4 - {x}^{2}} \mathrm{dx} + 4 a r c \sin \left(\frac{x}{2}\right) , \ldots \ldots \ldots \ldots \left[\because , \left(2\right)\right]$

$= - \left\{\frac{x}{2} \sqrt{4 - {x}^{2}} + \frac{4}{2} a r c \sin \left(\frac{x}{2}\right)\right\} + 4 a r c \sin \left(\frac{x}{2}\right) ,$

$= 2 a r c \sin \left(\frac{x}{2}\right) - \frac{x}{2} \sqrt{4 - {x}^{2}} + C .$

Enjoy Maths.!