Correction exercice

a) Soit à calculer al primitives des fonctions suivantes :
Posons : \(u = x\) et \(dv = \sin xdx\) \( \Rightarrow v = - \cos x\). D’après la formule de d’intégration par parties \(\int {udv = uv - \int {vdu} } \), nous avons :
\(I = \int {x\sin xdx} = \) \( - x\cos x + \) \(\int {\cos xdx} \)
\(I = - x\cos x\) \( + \sin x + cte\)

b) \(I = \int {\arctan xdx} \)
Posons \(u = \arctan x\) \( \Rightarrow du = \) \(\frac{{dx}}{{1 + {x^2}}}\) et \(dv = dx\) \( \Rightarrow v = x\)
Ainsi \(I = \) \(\int {\arctan xdx} \) \( = x\arctan x - \) \(\int {\frac{x}{{1 + {x^2}}}dx} \)
\(I = x\arctan x\) \( - \frac{1}{2}Log\left| {1 + {x^2}} \right|\) \( + cte\)

c) \(I = \int {{x^2}{e^x}dx} \)
Posons \(\left\{ \begin{array}{l}u = {x^2}\\dv = {e^x}dx\end{array} \right.\) \( \Rightarrow \) \(\left\{ \begin{array}{l}du = 2xdx\\v = {e^x}\end{array} \right.\)
Ainsi \(I = \int {{x^2}{e^x}dx} \) \( = {x^2}{e^x} - \) \(2\int {x{e^x}dx} \)
Avec \(\int {x{e^x}dx} = \) \(x{e^x} - \int {{e^x}dx} \) \( = x{e^x} - {e^x}\) en procèdent également par une intégration par parties
\(I = {x^2}{e^x} - \) \(2x{e^x} - 2{e^x}\) \( + cte\)

d) \(I = \int {\left( {{x^2} + 7x - 5} \right)} \) \(\cos 2xdx\)
Posons \(\left\{ \begin{array}{l}u = {x^2} + 7x - 5\\dv = \cos 2xdx\end{array} \right.\) \( \Rightarrow \) \(\left\{ \begin{array}{l}du = \left( {2x + 7} \right)dx\\v = \frac{{\sin 2x}}{2}\end{array} \right.\)
Ainsi \(I = \) \(\left( {{x^2} + 7x - 5} \right)\) \(\frac{{\sin 2x}}{2} - \) \(\int {\left( {2x + 7} \right)\frac{{\sin 2x}}{2}dx} \)
\(\int {\left( {2x + 7} \right)\frac{{\sin 2x}}{2}dx} \) \( = \left( {2x + 7} \right)\frac{{\cos 2x}}{4}\) \( - \frac{{\sin 2x}}{4}\) en utilisant également l’intégration par parties
\(I = \) \(\left( {{x^2} + 7x - 5} \right)\) \(\frac{{\sin 2x}}{2} = \) \(\left( {2x + 7} \right)\) \(\frac{{\cos 2x}}{4} - \) \(\frac{{\sin 2x}}{4} + cte\)

e) \(I = \int {{x^n}Logx} dx\)
Posons \(\left\{ \begin{array}{l}du = {x^n}dx\\v = Logx\end{array} \right.\) \( \Rightarrow \) \(\left\{ \begin{array}{l}u = {x^{n + 1}}\\dv = \frac{{dx}}{x}\end{array} \right.\)
\(I = \int {{x^n}Logx} dx\) \( = \frac{{{x^{n + 1}}}}{{n + 1}}Logx - \) \(\int {\frac{{{x^n}}}{{n + 1}}} dx\)
\(I = \int {{x^n}Logx} dx\) \( = \frac{{{x^{n + 1}}}}{{n + 1}}\) \(\left( {Logx - \frac{1}{{n + 1}}} \right)\) \( + cte\)