5.9 Intégration par parties
Énoncé indispensable
5.9.1
Soient \( f \) et \( g \) des fonctions dérivables. Alors,
\[ \int _a^b f'(t) g(t) \, dt = \left[ f(t) g(t) \right]_a^b - \int _a^b f(t) g'(t) \, dt \]
Preuve: \( (f g)' = f' g + f g' \) Donc,
\[ \int _a^b \left( f'(t) g(t) + f(t) g'(t) \right) \, dt = \int _a^b f'(t) g(t) \, dt + \int _a^b f(t) g'(t) \, dt \]
Ainsi,
\[ \int _a^b (f(t) g(t))' \, dt = \left[ f(t) g(t) \right]_a^b \]
Exemple
5.9.2
\[ \int _0^{\frac{\pi }{2}} \underbrace{\cos x}_{f'} \underbrace{\sin x}_{g} \, dx = \left[ f \cdot g \right]_0^{\frac{\pi }{2}} - \int _0^{\frac{\pi }{2}} f \cdot g' \]
\[ = \left[ \sin x \cdot \sin x \right]_{\ 0}^{\frac{\pi }{2}} - \int _0^{\frac{\pi }{2}} \sin x \cdot \cos x \, dx \]
\[ f = \sin x, \quad g = \sin x \quad \Rightarrow \quad g' = \cos x \]
\[ \Rightarrow \int _0^{\frac{\pi }{2}} \cos x \sin x \, dx = 1 - \int _0^{\frac{\pi }{2}} \sin x \cdot \cos x \, dx \]
\[ \Rightarrow \int _0^{\frac{\pi }{2}} \cos x \sin x \, dx + \int _0^{\frac{\pi }{2}} \sin x \cdot \cos x \, dx = 1 \]
\[ \Rightarrow 2 \int _0^{\frac{\pi }{2}} \cos x \sin x \, dx = 1 \]
\[ \Rightarrow \int _0^{\frac{\pi }{2}} \cos x \sin x \, dx = \frac{1}{2} \]
Exercice
5.9.3
Calculer \(\int _0^1 x. e^x dx\).
Exercice
5.9.4
Calculer \(\int _0^1 x^2. e^x dx\).
Exercice
5.9.5
Calculer \(\int _0^{2\pi } \sin ^2(x).dx\).
Exercice
5.9.6
Calculate (with an error of not more than 10\(\% \) of the answer)
\[ \int _0^{2\pi } \sin ^{100}(x) \, dx \]