# How do you integrate int sqrt(1-x^2) by trigonometric substitution?

Nov 28, 2016

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

#### Explanation:

Let $I = \int \sqrt{1 - {x}^{2}} \mathrm{dx}$

The way to approach these types of problems is try and make a comparison with a known trig identity. In the case we will use:

${\sin}^{2} A + {\cos}^{2} A \equiv 1 \implies {\cos}^{2} A = 1 - {\sin}^{2} A$

If you compare to the integrand, we make the following substitution:

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

And $\frac{\mathrm{dx}}{d \theta} = \cos \theta$

Substituting into the integral we have:
$I = \int \cos \theta \cos \theta d \theta$ =
$\therefore I = \int {\cos}^{2} \theta d \theta$

Now $\cos 2 A \equiv {\cos}^{2} A - {\sin}^{2} A = 2 {\cos}^{2} A - 1$
$\therefore {\cos}^{2} A = \frac{1}{2} \left(1 + \cos 2 A\right)$

And so;
$I = \int \frac{1}{2} \left(1 + \cos 2 \theta\right) d \theta$
$2 I = \int \left(1 + \cos 2 \theta\right) d \theta$
$2 I = \theta + \frac{\sin 2 \theta}{2} + C '$

We now need to use the identity $\sin 2 A \equiv 2 \sin A \cos A$
$2 I = \theta + \sin \theta \cos \theta + C '$
$\therefore 2 I = {\sin}^{-} 1 x + x \sqrt{1 - {x}^{2}} + C '$
$\therefore I = \frac{{\sin}^{-} 1 x + x \sqrt{1 - {x}^{2}} + C '}{2}$
$\therefore I = \frac{{\sin}^{-} 1 x + x \sqrt{1 - {x}^{2}}}{2} + C$