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[ID: 1810] [Date de publication: 12 mai 2021 12:14] [Catégorie(s): Changement de variable ] [ Nombre commentaires: 1] [nombre d'éditeurs: 1 ] [Editeur(s): Emmanuel Vieillard-Baron ] [nombre d'auteurs: 3 ] [Auteur(s): Emmanuel Vieillard-Baron Alain Soyeur François Capaces ]

Solution(s)

Solution(s)

Exercice 525
Par Emmanuel Vieillard-Baron Alain Soyeur François Capaces le 12 mai 2021 12:14

1. On pose \(u=\ln x\). \(u\) est bien \(\mathcal{C}^{1}\) et \(\int_{1}^{e} {\scriptstyle\ln x\over\scriptstyle x}\,\textrm{d}x = \int_{0}^{1} u\,\textrm{d}u=\left[{\scriptstyle u^2\over\scriptstyle 2}\right]_{0}^{1}=\boxed{{\scriptstyle 1\over\scriptstyle 2}}\).
   Point n’est besoin de changement de variable puisque
   \(\displaystyle\int\dfrac{\ln x}{x}\,\textrm dx = \dfrac12 \ln^2x + C\).

2. On pose \(\begin{cases}u=\sqrt{x+1} \\du=\dfrac{dx}{2\sqrt{x+1}}\end{cases}\) et \(x=u^2-1\). On a : \(\int_{0}^{1} {\scriptstyle x^3\over\scriptstyle\sqrt{x+1}}\,\textrm{d}x=\int_{1}^{\sqrt{2}} \dfrac{\left(u^2-1\right)^3} {2}\,\textrm{d}u=\boxed{\dfrac{16}{35}-\dfrac{9}{35}\sqrt{2}}\)

3. On pose \(\begin{cases}u=\cos x \\ du=-\sin x dx \end{cases}\), on obtient : \(\int_{0}^{{\scriptstyle\pi\over\scriptstyle 2}} \sin x \cos^2 x\,\textrm{d}x=\int_{0}^{1} u^2\,\textrm{d}u=\boxed{\dfrac{1}{3}}\).

4. On pose \(\begin{cases}u=\tan x \\ du=\left(1+\tan^2 x\right)dx=\left(1+u^2\right)dx \end{cases}\), on obtient : \(\int_{0}^{{\scriptstyle\pi\over\scriptstyle 4}} \tan^4 x\,\textrm{d}x=\int_{0}^{1} \dfrac{u^4}{1+u^2}\,\textrm{d}u=\int_{0}^{1} \left(\dfrac{u^4-1}{1+u^2} +\dfrac{1}{1+u^2}\right)\,\textrm{d}u=\int_{0}^{1} \left(u^2-1\right)\,\textrm{d}u+\int_{0}^{1} \dfrac{1}{1+u^2}\,\textrm{d}u=\Bigl[ \dfrac{u^3}{3}-u+\operatorname{arctan} u \Bigr]_{0}^{1}=\boxed{-\dfrac{2}{3}+\dfrac{\pi}{4}}\) .

5. On pose \(\begin{cases}x=\cos u\\ dx=-\sin u du \end{cases}\), on obtient : \(\int_{0}^{1} \sqrt{1-x^2}\,\textrm{d}x=-\int_{{\scriptstyle\pi\over\scriptstyle 2}}^{0} \sqrt{1-\cos^2 u}\sin u\,\textrm{d}u=\int_{0}^{{\scriptstyle\pi\over\scriptstyle 2}} \sin^2 u\,\textrm{d}u=\int_{0}^{{\scriptstyle\pi\over\scriptstyle 2}} \dfrac{1-\cos 2u}{2}\,\textrm{d}u=\boxed{\dfrac{\pi}{4}}\).

6. On pose \(\begin{cases}u=\sqrt{x}\\du=\dfrac{dx}{2\sqrt x} \end{cases}\), on obtient : \(\int_{1}^{e} {\scriptstyle\ln x\over\scriptstyle\sqrt x}\,\textrm{d}x=\int_{1}^{\sqrt e} 2\ln u^2\,\textrm{d}u=\int_{1}^{\sqrt e} 4\ln u\,\textrm{d}u=4\Bigl[ u\ln u -u \Bigr]_{1}^{\sqrt e}=\boxed{4-2\sqrt{e}}\).