# How do you integrate sqrt(1-x^2)?

Mar 29, 2018

The answer is $= \frac{1}{2} \arcsin x + \frac{1}{2} x \sqrt{1 - {x}^{2}} + C$

#### Explanation:

Let $x = \sin \theta$, $\implies$, $\mathrm{dx} = \cos \theta d \theta$

$\cos \theta = \sqrt{1 - {x}^{2}}$

$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 x \sqrt{1 - {x}^{2}}$

Therefore, the integral is

$I = \int \sqrt{1 - {x}^{2}} \mathrm{dx} = \int \cos \theta \cdot \cos \theta d \theta$

$= \int {\cos}^{2} \theta d \theta$

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1$

${\cos}^{2} \theta = \frac{1 + \cos 2 \theta}{2}$

Therefore,

$I = \frac{1}{2} \int \left(1 + \cos 2 \theta\right) d \theta$

$= \frac{1}{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)$

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