L'intégrale simple

Posons \(\omega(x)=\frac{\cos^3x}{\sin x+3}dx\) l'élément différentiel

et calculons \(\omega(\pi-x)=\frac{\cos^3(\pi-x)}{\sin(\pi-x)+3}d(\pi-x)=\frac{(-\cos^3x)(-dx)}{\sin x + 3}=\omega(x)\)

nous poserons donc : \(\color{blue}t = \sin x ~~\color{red}\text{ (1 pt)}\color{black}\Leftrightarrow dt = \cos x dx\)

d'où les bornes d'intégration :

\(\color{blue}t_1\color{black} = \sin 0 = \color{blue}0\)

\(\color{blue}t_2\color{black} = \sin (\pi/2) = \color{blue}1 ~~\color{red}\text{ (1 pt)}\)

alors

\(I_3=\int_0^{\pi/2}\frac{\cos^2x\cos x}{\sin x + 3}dx=\int_0^1\frac{1-t^2}{t+3}dt\)

\(=\int_0^1(-t+3-\frac8{t+3})dt\)

\(=[-\frac{t^2}2+3t-8\ln(t+3)]_0^1\)

\(\color{blue}I_3=\frac52+8\ln\frac34~~\#~~0,198~~\color{red}\text{ (3 pts)}\)