Question 2

Durée : 10 mn

Note maximale : 5

Question

Calculer la primitive \(I_2=\int\cos^3\theta\sin^2\theta d \theta\)

Solution

Sachant que \(\cos^3\theta = \cos^2\theta \cos\theta = (1 - \sin^2\theta ) \cos\theta\)

nous avons \(I_2=\int(1-\sin^2\theta )\sin^2\theta \cos\theta d\theta~~\color{red}\text{ (2 pts)}\)

d'où

\(I_2=\int\sin^2\theta\cos\theta d\theta-\int\sin^4\theta\cos\theta d \theta\)

\(\color{blue}I_2=\frac13\sin^3\theta-\frac15\sin^5\theta+C~~\color{red}\text{ (3 pts)}\)