# How do you use substitution to integrate sqrt(4-x^2) dx?

Jun 23, 2015

This can be done from trig substitution. Notice how

$\sqrt{{a}^{2} - {x}^{2}} \propto \sqrt{{a}^{2} - {a}^{2} {\sin}^{2} \theta} \propto \sqrt{4 - {x}^{2}}$
where $a = 2$

so let:
$x = 2 \sin \theta$
$\mathrm{dx} = 2 \cos \theta d \theta$
$\sqrt{4 - {x}^{2}} = 2 \cos \theta$

$\implies \int 2 \cos \theta \cdot 2 \cos \theta d \theta$

$= 4 \int {\cos}^{2} \theta d \theta$

Now you can use the identity:
${\cos}^{2} \theta = \frac{1 + \cos \left(2 \theta\right)}{2}$

Thus:
$= 2 \int d \theta + 2 \int \cos \left(2 \theta\right) d \theta$

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

$= 2 \theta + \sin \left(2 \theta\right) + C$

Since $x = 2 \sin \theta$, $\theta = \arcsin \left(\frac{x}{2}\right)$.
Since $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$:

$\sin \left(2 \theta\right) = \frac{x \sqrt{4 - {x}^{2}}}{2}$

$\implies \textcolor{b l u e}{2 \arcsin \left(\frac{x}{2}\right) + \frac{x \sqrt{4 - {x}^{2}}}{2} + C}$