Calculer les intégrales suivantes en utilisant un changement de variable adéquat :

  1. \(\int_{0}^{1} \dfrac{1}{1+e^x}\,\textrm{d}x\)

  2. \(\int_{0}^{1} x^2\sqrt{1-x^2}\,\textrm{d}x\)

  3. \(\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} \sqrt{\dfrac{\operatorname{arcsin} x}{1-x^2}}\,\textrm{d}x\)

  4. \(\int_{0}^{{\scriptstyle\pi\over\scriptstyle 4}} \dfrac{1}{\cos^2 x \sqrt{\tan x}}\,\textrm{d}x\)

  5. \(\int_{1}^{4} \dfrac{1}{x+\sqrt{x}}\,\textrm{d}x\)

  6. \(\int_{0}^{1} \dfrac{1}{1+x+x^2}\,\textrm{d}x\)


Barre utilisateur

[ID: 1814] [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 243
Par emmanuel le 12 mai 2021 12:14
  1. On pose \(\begin{cases}u=e^x \\ du=e^x dx=u dx\end{cases}\), on obtient : \(\int_{0}^{1} \dfrac{1}{1+e^x}\,\textrm{d}x=\int_{1}^{e} \dfrac{1}{u\left(1+u\right)}\,\textrm{d}u=\int_{1 }^{e} \dfrac{1}{u}-\dfrac{1}{u+1}\,\textrm{d}u=\Bigl[ \ln u-\ln \left(1+u\right) \Bigr]_{1}^{e}=\boxed{\ln 2-\ln\left(e+1\right)+1}\).

  2. On pose \(\begin{cases}u=\cos x \\ du=-\sin x dx\end{cases}\), on obtient : \(\int_{0}^{1} x^2\sqrt{1-x^2}\,\textrm{d}x=\int_{0}^{{\scriptstyle\pi\over\scriptstyle 2}} \cos^2 u \sin^2 u\,\textrm{d}u=\int_{0}^{{\scriptstyle\pi\over\scriptstyle 2}} \dfrac{1-\cos 4u}{8}\,\textrm{d}u=\boxed{\dfrac{\pi}{16}}\).

  3. On pose \(\begin{cases}u=\operatorname{arcsin} x \\ du= \dfrac{dx}{\sqrt{1-x^2}}\end{cases}\), on obtient : \(\int_{0}^{{\scriptstyle 1\over\scriptstyle 2}} \sqrt{\dfrac{\operatorname{arcsin} x}{1-x^2}}\,\textrm{d}x=\int_{0}^{{\scriptstyle\pi\over\scriptstyle 6}} \sqrt{\dfrac{u}{1-\sin^2 u}} \sqrt{1-\sin^2 u}\,\textrm{d}u= \int_{0}^{{\scriptstyle\pi\over\scriptstyle 6}} \sqrt{ u}\,\textrm{d}u=\boxed{\dfrac{\pi^{{\scriptstyle 3\over\scriptstyle 2}} \sqrt 6}{54}}\).

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

  5. On pose \(\begin{cases}u=\sqrt x \\ du= \dfrac{dx}{2\sqrt{x}}\end{cases}\), on obtient : \(\int_{1}^{4} \dfrac{1}{x+\sqrt{x}}\,\textrm{d}x=\int_{1}^{2} \dfrac{2}{1+u}\,\textrm{d}u=\Bigl[ 2\ln\left(1+u\right) \Bigr]_{1}^{2}=\boxed{2\ln\left(\dfrac{3}{2}\right) }\).

  6. On a :\(\int_{0}^{1} \dfrac{1}{1+x+x^2}\,\textrm{d}x =\int_{0}^{1} \dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} }\,\textrm{d}x= \int_{0}^{1} \dfrac{4}{3}\dfrac{1}{\left(\dfrac{2}{\sqrt{3}}\left(x+\dfrac{1}{2}\right)\right)^2+1 }\,\textrm{d}x\) On pose \(\begin{cases}u= \dfrac{2}{\sqrt{3}}\left(x+\dfrac{1}{2}\right) \\ du=\dfrac{2}{\sqrt{3}} dx\end{cases}\), on obtient : \(\int_{0}^{1} \dfrac{1}{1+x+x^2}\,\textrm{d}x = \dfrac{2\sqrt{3}}{3}\int_{{\scriptstyle\sqrt{3}\over\scriptstyle 3}}^{\sqrt{3}} \dfrac{1}{1+u^2}\,\textrm{d}u ={\scriptstyle 2\sqrt 3\over\scriptstyle 3}\Bigl[ \operatorname{arctan} u \Bigr]_{{\scriptstyle\sqrt{3}\over\scriptstyle 3}}^{\sqrt{3}}=\boxed{\dfrac{\pi\sqrt{3}}{9} }\).


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