On a:
\(=\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x\)
=\(a \int_{0}^{a} \sqrt{1-\left(\frac{x}{a}\right)^{2}} d x\)
Posons :
\(\forall t \in [0, \frac{\pi}{2}]\,\forall x \in[0, a]\)
\(\frac{x}{a}=\sin t\) donc \(d x=a \cos t d t\)
On a:
\(x=0 \Leftrightarrow t=0\) et \(x=a \Leftrightarrow t=\frac{\pi}{2}\)
Donc:
\(\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x\)
=\(a \int_{0}^{a} \int_{1-\left(\frac{x}{a}\right)^{2}} d x\)
\(=a\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin ^{2} t} a \cos t d t\)
=\(a^{2} \int_{0}^{\frac{\pi}{2}} \cos ^{2} t d t\)
=\(a^{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos (2 t)+1}{2}\)
=\(\frac{a^{2}}{2}\left[\frac{1}{2} \sin (2 t)+t\right]_{0}^{\frac{\pi}{2}}\)
=\(\frac{a^{2} \pi}{4}\)
Et par suite la bonne réponse est D.