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

Jan 12, 2017

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

#### Explanation:

As the integrand function is defined for $x \in \left[- 1 , 1\right]$, you can substitute:

$x = \sin t$ with $t \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

$\mathrm{dx} = \cos t \mathrm{dt}$

so the integral becomes:

$\int \sqrt{1 - {x}^{2}} \mathrm{dx} = \int \sqrt{1 - {\sin}^{2} t} \cos t \mathrm{dt} = \int \sqrt{{\cos}^{2} t} \cos t \mathrm{dt}$

In the given interval $\cos t$ is positive, so $\sqrt{{\cos}^{2} t} = \cos t$:

$\int \sqrt{1 - {x}^{2}} \mathrm{dx} = \int {\cos}^{2} t \mathrm{dt}$

Now we can use the identity:

${\cos}^{2} t = \frac{1 + \cos \left(2 t\right)}{2}$

$\int \sqrt{1 - {x}^{2}} \mathrm{dx} = \int \frac{1 + \cos \left(2 t\right)}{2} \mathrm{dt} = \int \frac{\mathrm{dt}}{2} + \frac{1}{4} \int \cos \left(2 t\right) d \left(2 t\right) = \frac{1}{2} t + \frac{1}{4} \sin 2 t = \frac{1}{2} \left(t + \sin t \cos t\right)$

To substitute back $x$ we note that:

$x = \sin t$ for $t \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \implies t = \arcsin x$

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

Finally:

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