L'intégrale simple

Sachant que \(\sin2x = 2 \sin x \cos x\) en posant : \(t = \sin x \Leftrightarrow dt = \cos x dx,\) donc

\(\begin{array}{l l}x_1=\pi&t_1=\sin\pi=0\\x_2=\frac{3\pi}4&t_2=\sin\frac{3\pi}4=\frac{\sqrt2}2\end{array}~~\color{red}\text{ (1 pt)}\)

d'où

\(I_4=\int_0^{\sqrt2/2}\frac{2t}{1+t}dt~~\color{red}\text{ (1 pt)}\)

\(I_4=2\int_0^{\sqrt2/2}\frac{(t+1)-1}{1+t}dt\)

\(=2\int_0^{\sqrt2/2}(1-\frac1{1+t})dt\)

\(=2[t-\ln|1+t|]_0^{\sqrt2/2}\)

\(=2[\frac{\sqrt2}2-\ln(1+\frac{\sqrt{2}}2)]\)

\(\color{blue}I_4=\sqrt2-2\ln(1+\frac{\sqrt2}2)~~\color{red}\text{ (3 pts)}\)