L'intégrale simple

Posons \(x = (4/5) \sin t \Rightarrow t = \arcsin (5x/4)\) et \(dx = (4/5) \cos t dt\)

d'où

\(\begin{array}{l l}x_1=0&t_1=\arcsin 0 = 0\\x_2=\frac25&t_2=\arcsin\frac{5\times2}{4\times 5}=\arcsin \frac12=\frac{\pi}6\end{array}~~\color{red}\text{ (1 pt)}\)

\(I_3=\int_0^{\pi/6}\sqrt{16-\frac{25\times16}{16}\sin^2t}.\frac45\cos t dt\)

\(=\frac{16}5\int_0^{\pi/6}\sqrt{1-\sin^2t}\cos tdt\)

or \(\sqrt{1-\sin^2t}=|\cos t|= \cos t\)pour \(t \in [ 0 ; \pi/6]\) d'où

\(I_3=\frac{16}5\int_0^{\pi/6}\cos^2tdt~~\color{red}\text{ (2 pts)}\)

Sachant que \(\cos^2t = (1 + \cos2t) / 2\)

\(I_3=\frac{16}{5}[\frac12\int_0^{\pi/6}dt+\frac12\int_0^{\pi/6}\cos2tdt]\)

\(=\frac85[t+\frac{\sin2t}2]_0^{\pi/6}=\frac85[\frac{\pi}6+\frac12\sin\frac{\pi}3]\)

\(=\frac85[\frac{\pi}6+\frac{\sqrt3}4]\)

\(\color{blue}I_3=\frac{4\pi}{15}+\frac{2\sqrt3}{5}~~\color{red}\text{ (2 pts)}\)