# How do you integrate (1-x^2)^.5?

Oct 12, 2016

$\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx} = \frac{1}{2} x \sqrt{1 - {x}^{2}} + \frac{1}{2} {\sin}^{- 1} x + C$

#### Explanation:

Let $x = \sin \theta$

Then:

$\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx} = \int {\left(1 - {\sin}^{2} \theta\right)}^{0.5} \left(\frac{d \sin \theta}{d \theta}\right) d \theta$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \int {\left({\cos}^{2} \theta\right)}^{0.5} \left(\cos \theta\right) d \theta$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \int {\cos}^{2} \theta d \theta$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \int \frac{1}{2} \left(\cos 2 \theta + 1\right) d \theta$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \frac{1}{4} \sin 2 \theta + \frac{\theta}{2} + C$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \frac{1}{2} \sin \theta \cos \theta + \frac{\theta}{2} + C$

$\textcolor{w h i t e}{\int {\left(1 - {x}^{2}\right)}^{0.5} \mathrm{dx}} = \frac{1}{2} x \sqrt{1 - {x}^{2}} + \frac{1}{2} {\sin}^{- 1} x + C$