L'intégrale simple

Posons \(t = \tan x \Leftrightarrow dt = 1 + \tan^2x dx = 1 + t^2 dx\)

or \(\sin^2x=\cos^2x\tan^2x=\frac{\tan^2x}{1+\tan^2x}\Leftrightarrow\color{blue}\sin2x=\frac{t^2}{1+t^2}~~\color{red}\text{ (1 pt)}\)

d'où

\(I_2=\int\frac{dt}{(1+t^2)(1+\frac{t^2}{1+t^2})}=\int\frac{dt}{1+2t^2}~~\color{red}\text{ (2 pts)}\)

Pour \(u=\sqrt2t\Leftrightarrow du=\sqrt2dt\)

\(\color{blue}I_2\color{black}=\frac1{\sqrt2}\int\frac{du}{1+u^2}\)

\(=\frac{1}{\sqrt2}\arctan u + C\)

\(=\color{blue}\frac1{\sqrt2}\arctan(\sqrt2\tan x) + C~~\color{red}\text{ (2 pts)}\)