Calcular $I\ =\ \displaystyle\int_0^a \sqrt{a^2 - x^2}\ dx$, $a > 0$.

$I = a\displaystyle\int_0^a \sqrt{1 - \left(\dfrac{x}{a}\right)^2}\ dx$

Seja $y = \dfrac{x}{a}$. $dy = \dfrac{dx}{a}$

$I = a^2\displaystyle\int_0^1 \sqrt{1 - y^2}\ dy$

Seja $y = \sin \theta$, $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$. $dy = \cos \theta\ d\theta$.

$I = a^2\displaystyle\int_0^{\pi / 2} \cos^2 \theta\ d\theta\ =\ a^2\displaystyle\int_0^{\pi / 2} \dfrac{(\cos 2\theta) + 1}{2} d\theta\ =\ a^2\displaystyle\int_0^{\pi / 2} \cos 2\theta\ d\theta + \dfrac{a^2\pi}{4}$

Seja $\varphi = 2\theta$. $d\varphi = 2 d\theta$.

$I = \cancelto{0}{\dfrac{a^2}{2}\displaystyle\int_0^{\pi} \cos \varphi\ d\varphi} + \dfrac{a^2\pi}{4}\ =\ \fbox{$\dfrac{a^2\pi}{4}$}$