Question 2
Durée : 5 mn
Note maximale : 5
Question
Calculer la primitive \(I_2=\int\sinh^5xdx\)
Solution
La puissance de \(\sinh x\) étant impaire, posons \(u = \cosh x \Leftrightarrow du = \sinh x dx\)
alors
\(I_2=\int\sinh^4x\sinh xdx=\int(\cosh^2x-1)^2\sinh xdx\)
\(I_2=\int(u^2-1)^2du=\int(u^4-2u^2+1)du\)
\(=\frac{u^5}5-2\frac{u^3}3+u+C~~\color{red}\text{ (2 pts)}\)
\(\color{blue}I_2=\frac{\cosh^5x}5-2\frac{\cosh^3x}3+\cosh x+C~~\color{red}\text{ (3 pts)}\)