# Changement de variable en calcul intégral/Exercices/Changement de variable difficile

Les changements de variable présentés dans cette page demandent de la réflexion. Plusieurs changement de variables successifs peuvent être nécessaires pour mener les calculs à leur terme.

## Exercice 4-1

Calculer :
$a)\quad \int _{\frac {1}{4}}^{1}{\sqrt {\frac {1-{\sqrt {x}}}{\sqrt {x}}}}\,\mathrm {d} x$
a) Posons :
$x=y^{2}\Rightarrow \mathrm {d} x=2ydy\qquad \qquad {\frac {1}{4}}\rightarrow x\rightarrow 1\Rightarrow {\frac {1}{2}}\rightarrow y\rightarrow 1$
On obtient :
$\int _{\frac {1}{4}}^{1}{\sqrt {\frac {1-{\sqrt {x}}}{\sqrt {x}}}}\,\mathrm {d} x=\int _{\frac {1}{2}}^{1}2y{\sqrt {\frac {1-y}{y}}}\,\mathrm {d} y=\int _{\frac {1}{2}}^{1}{\sqrt {4y^{2}{\frac {1-y}{y}}}}\,\mathrm {d} y=\int _{\frac {1}{2}}^{1}{\sqrt {4y-4y^{2}}}\,\mathrm {d} y=\int _{\frac {1}{2}}^{1}{\sqrt {1-(2y-1)^{2}}}\,\mathrm {d} y$
Posons ensuite:
$2y-1=\sin {z}\Rightarrow \mathrm {d} y={\frac {\cos {z}}{2}}dz\qquad \qquad {\frac {1}{2}}\rightarrow y\rightarrow 1\Rightarrow 0\rightarrow z\rightarrow {\frac {\pi }{2}}$
On obtient :
${\begin{aligned}\int _{\frac {1}{4}}^{1}{\sqrt {\frac {1-{\sqrt {x}}}{\sqrt {x}}}}\,\mathrm {d} x&=\int _{\frac {1}{2}}^{1}{\sqrt {1-(2y-1)^{2}}}\,\mathrm {d} y=\int _{0}^{\frac {\pi }{2}}{\sqrt {1-\sin ^{2}(z)}}{\frac {\cos {z}}{2}}\,\mathrm {d} z={\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\cos ^{2}(z)\,\mathrm {d} z={\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}\cos(2z)+1\,\mathrm {d} z\\&={\frac {1}{4}}\left[{\frac {\sin(2z)}{2}}+z\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{8}}\end{aligned}}$
Nous pouvons conclure que :

## Exercice 4-2

Calculer :
    $\int _{0}^{\frac {\pi }{4}}{\frac {\sin x}{\cos x{\sqrt {\cos(2x)}}}}\,\mathrm {d} x$.
On a :
    ${\frac {\sin x}{\cos x{\sqrt {\cos 2x}}}}={\frac {\sin x\cos x}{\cos ^{2}x{\sqrt {\cos(2x)}}}}={\frac {{\frac {1}{2}}\sin(2x)}{{\frac {1}{2}}\left(1+\cos(2x)\right){\sqrt {\cos(2x)}}}}={\frac {\sin(2x)}{\left(1+\cos(2x)\right){\sqrt {\cos(2x)}}}}$.
Posons :
    $y=\cos(2x)\Rightarrow \mathrm {d} y=-2\sin(2x)\,\mathrm {d} x\qquad \qquad 0\to x\to {\frac {\pi }{4}}\Rightarrow 1\to y\to 0$.
On obtient :
    $\int _{0}^{\frac {\pi }{4}}{\frac {\sin(2x)\,\mathrm {d} x}{\left(1+\cos(2x)\right){\sqrt {\cos(2x)}}}}=\int _{1}^{0}{\frac {-{\frac {1}{2}}\,\mathrm {d} y}{(1+y){\sqrt {y}}}}=\int _{0}^{1}{\frac {\mathrm {d} y}{2(1+y){\sqrt {y}}}}$.
Posons ensuite :
    $y=z^{2}\Rightarrow \mathrm {d} y=2z\,\mathrm {d} z\qquad \qquad 0\to y\to 1\Rightarrow 0\to z\to 1$.
On obtient :
    $\int _{0}^{1}{\frac {\mathrm {d} y}{2(1+y){\sqrt {y}}}}=\int _{0}^{1}{\frac {2z\,\mathrm {d} z}{2(1+z^{2})z}}=\int _{0}^{1}{\frac {\mathrm {d} z}{1+z^{2}}}=\left[\arctan \right]_{0}^{1}={\frac {\pi }{4}}$.
Nous pouvons conclure que :

## Exercice 4-3

Calculer :
$\int _{-k}^{k}\ln \left({\sqrt {k-x}}+{\sqrt {k+x}}\right)\,\mathrm {d} x$
Nous avons :
$\int _{-k}^{k}\ln \left({\sqrt {k-x}}+{\sqrt {k+x}}\right)\,\mathrm {d} x={\frac {1}{2}}\int _{-k}^{k}\ln \left(\left({\sqrt {k-x}}+{\sqrt {k+x}}\right)^{2}\right)\,\mathrm {d} x={\frac {1}{2}}\int _{-k}^{k}\ln \left(2k+2{\sqrt {k^{2}-x^{2}}}\right)\,\mathrm {d} x$
Posons :
$x=k\sin(y)\Rightarrow \mathrm {d} x=k\cos(y)dy\qquad \qquad -k\rightarrow x\rightarrow k\Rightarrow -{\frac {\pi }{2}}\rightarrow y\rightarrow {\frac {\pi }{2}}$
On obtient :
${\begin{aligned}\int _{-k}^{k}\ln \left({\sqrt {k-x}}+{\sqrt {k+x}}\right)\,\mathrm {d} x&={\frac {1}{2}}\int _{-k}^{k}\ln \left(2k+2{\sqrt {k^{2}-x^{2}}}\right)\,\mathrm {d} x={\frac {1}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\ln \left(2k+2{\sqrt {k^{2}-k^{2}\sin ^{2}(y)}}\right)k\cos(y)\,\mathrm {d} y\\&={\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\ln \left(2k+2k\cos(y)\right)\cos(y)\,\mathrm {d} y={\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\ln \left(2k(1+\cos(y))\right)\cos(y)\,\mathrm {d} y\\&={\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\ln \left(2k\right)\cos(y)\,\mathrm {d} y+{\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\ln \left(1+\cos(y)\right)\cos(y)\,\mathrm {d} y\\&={\frac {k}{2}}\left[\ln \left(2k\right)\sin(y)\right]_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}+{\frac {k}{2}}\left[\ln \left(1+\cos(y)\right)\sin(y)\right]_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}+{\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {\sin ^{2}(y)}{1+\cos(y)}}\,\mathrm {d} y\\&=k\ln(2k)+0+{\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {1-\cos ^{2}(y)}{1+\cos(y)}}\,\mathrm {d} y=k\ln(2k)+{\frac {k}{2}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}1-\cos(y)\,\mathrm {d} y\\&=k\ln(2k)+{\frac {k}{2}}\left[y-\sin(y)\right]_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}=k\ln(2k)+{\frac {k}{2}}\left(\pi -2\right)\end{aligned}}$
Nous pouvons conclure que :

## Exercice 4-4

Calculer :
$a)\quad \int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2x+1}}}{1+{\sqrt {1-2x}}}}\,\mathrm {d} x\qquad \qquad \qquad b)\quad \int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x$
a) Nous écrirons tout d’abord :
$\int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2x+1}}}{1+{\sqrt {1-2x}}}}\,\mathrm {d} x=\int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2}}{\sqrt {x+{\frac {1}{2}}}}}{1+{\sqrt {2}}{\sqrt {{\frac {1}{2}}-x}}}}\,\mathrm {d} x$
Posons alors :
$x=-{\frac {1}{2}}\cos(2\theta )\Rightarrow \mathrm {d} x=\sin(2\theta )\mathrm {d} \theta \qquad \qquad 0\rightarrow x\rightarrow {\frac {1}{2}}\Rightarrow {\frac {\pi }{4}}\rightarrow \theta \rightarrow {\frac {\pi }{2}}$
Nous obtenons :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2x+1}}}{1+{\sqrt {1-2x}}}}\,\mathrm {d} x&=\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {1+{\sqrt {2}}{\sqrt {\sin ^{2}(\theta )}}}{1+{\sqrt {2}}{\sqrt {\cos ^{2}(\theta )}}}}\sin(2\theta )\mathrm {d} \theta =\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {1+{\sqrt {2}}\sin(\theta )}{1+{\sqrt {2}}\cos(\theta )}}\sin(2\theta )\mathrm {d} \theta \\&=\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {{\frac {1}{\sqrt {2}}}+\sin(\theta )}{{\frac {1}{\sqrt {2}}}+\cos(\theta )}}\sin(2\theta )\mathrm {d} \theta =\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\sin \left({\frac {\pi }{4}}\right)+\sin(\theta )}{\cos \left({\frac {\pi }{4}}\right)+\cos(\theta )}}\sin(2\theta )\mathrm {d} \theta \\&=\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {2\sin \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)\cos \left({\frac {\theta }{2}}-{\frac {\pi }{8}}\right)}{2\cos \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)\cos \left({\frac {\theta }{2}}-{\frac {\pi }{8}}\right)}}\sin(2\theta )\mathrm {d} \theta =\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\sin \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)}{\cos \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)}}\sin(2\theta )\mathrm {d} \theta \end{aligned}}$
Posons alors :
$y={\frac {\theta }{2}}+{\frac {\pi }{8}}\Rightarrow \theta =2y-{\frac {\pi }{4}}\Rightarrow \mathrm {d} \theta =2\mathrm {d} y\qquad \qquad {\frac {\pi }{4}}\rightarrow \theta \rightarrow {\frac {\pi }{2}}\Rightarrow {\frac {\pi }{4}}\rightarrow y\rightarrow {\frac {3\pi }{8}}$
Et donc :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2x+1}}}{1+{\sqrt {1-2x}}}}\,\mathrm {d} x&=\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\sin \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)}{\cos \left({\frac {\theta }{2}}+{\frac {\pi }{8}}\right)}}\sin(2\theta )\mathrm {d} \theta =\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}{\frac {\sin(y)}{\cos(y)}}\sin \left(4y-{\frac {\pi }{2}}\right)\mathrm {d} y\\&=-2\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}{\frac {\sin(y)}{\cos(y)}}\cos(4y)\mathrm {d} y=-2\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}{\frac {\sin(y)}{\cos(y)}}\left(1-8\cos ^{2}(y)\sin ^{2}(y)\right)\mathrm {d} y\\&=16\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}\sin ^{3}(y)\cos(y)\mathrm {d} y-2\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}{\frac {\sin(y)}{\cos(y)}}\mathrm {d} y\end{aligned}}$
Dans la première intégrale, on pose :
$z=\sin(y)\Rightarrow \mathrm {d} z=\cos(y)\mathrm {d} y\qquad \qquad {\frac {\pi }{4}}\rightarrow y\rightarrow {\frac {3\pi }{8}}\Rightarrow {\frac {1}{\sqrt {2}}}\rightarrow z\rightarrow {\frac {\sqrt {2+{\sqrt {2}}}}{2}}$
Dans la deuxième intégrale, on pose :
$z=\cos(y)\Rightarrow \mathrm {d} z=-\sin(y)\mathrm {d} y\qquad \qquad {\frac {\pi }{4}}\rightarrow y\rightarrow {\frac {3\pi }{8}}\Rightarrow {\frac {1}{\sqrt {2}}}\rightarrow z\rightarrow {\frac {\sqrt {2-{\sqrt {2}}}}{2}}$
On obtient :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {1+{\sqrt {2x+1}}}{1+{\sqrt {1-2x}}}}\,\mathrm {d} x&=16\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}\sin ^{3}(y)\cos(y)\mathrm {d} y+2\int _{\frac {\pi }{4}}^{\frac {3\pi }{8}}{\frac {-\sin(y)}{\cos(y)}}\mathrm {d} y\\&=16\int _{\frac {1}{\sqrt {2}}}^{\frac {\sqrt {2+{\sqrt {2}}}}{2}}z^{3}\mathrm {d} z+2\int _{\frac {1}{\sqrt {2}}}^{\frac {\sqrt {2-{\sqrt {2}}}}{2}}{\frac {\mathrm {d} z}{z}}=16\left[{\frac {z^{4}}{4}}\right]_{\frac {1}{\sqrt {2}}}^{\frac {\sqrt {2+{\sqrt {2}}}}{2}}+2\left[\ln(z)\right]_{\frac {1}{\sqrt {2}}}^{\frac {\sqrt {2-{\sqrt {2}}}}{2}}\\&=16\left[{\frac {\left({\frac {\sqrt {2+{\sqrt {2}}}}{2}}\right)^{4}}{4}}-{\frac {\left({\frac {1}{\sqrt {2}}}\right)^{4}}{4}}\right]+2\left[\ln \left({\frac {\sqrt {2-{\sqrt {2}}}}{2}}\right)-\ln \left({\frac {1}{\sqrt {2}}}\right)\right]\\&=16\left({\frac {(2+{\sqrt {2}})^{2}}{64}}-{\frac {1}{16}}\right)+2\left({\frac {1}{2}}\ln(2-{\sqrt {2}})-\ln(2)+{\frac {1}{2}}\ln(2)\right)\\&={\frac {1+2{\sqrt {2}}}{2}}+\ln(2-{\sqrt {2}})-\ln(2)\\&={\sqrt {2}}+{\frac {1}{2}}+\ln \left(1-{\frac {1}{\sqrt {2}}}\right)\end{aligned}}$
En conclusion, nous avons :
b) nous avons :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x&=2{\sqrt {2}}\lim _{\alpha \to 0^{+}}\int _{\alpha }^{\frac {1}{2}}{\frac {{\sqrt {2x^{2}+1}}-{\sqrt {1-2x^{2}}}}{({\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}})({\sqrt {2x^{2}+1}}-{\sqrt {1-2x^{2}}})}}\,\mathrm {d} x\\&=2{\sqrt {2}}\lim _{\alpha \to 0^{+}}\int _{\alpha }^{\frac {1}{2}}{\frac {{\sqrt {2x^{2}+1}}-{\sqrt {1-2x^{2}}}}{2x^{2}+1-1+2x^{2}}}\,\mathrm {d} x=2{\sqrt {2}}\lim _{\alpha \to 0^{+}}\int _{\alpha }^{\frac {1}{2}}{\frac {{\sqrt {2x^{2}+1}}-{\sqrt {1-2x^{2}}}}{4x^{2}}}\,\mathrm {d} x\\&={\sqrt {2}}\lim _{\alpha \to 0^{+}}\left(\int _{\alpha }^{\frac {1}{2}}{\frac {\sqrt {2x^{2}+1}}{2x^{2}}}\,\mathrm {d} x-\int _{\alpha }^{\frac {1}{2}}{\frac {\sqrt {1-2x^{2}}}{2x^{2}}}\,\mathrm {d} x\right)\end{aligned}}$
En intégrant par parties chacune des deux intégrales, nous obtenons :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x&={\sqrt {2}}\lim _{\alpha \to 0^{+}}\left(\left[-{\frac {\sqrt {2x^{2}+1}}{2x}}\right]_{\alpha }^{\frac {1}{2}}-\int _{\alpha }^{\frac {1}{2}}-{\frac {2x}{2x\times 2{\sqrt {2x^{2}+1}}}}\,\mathrm {d} x-\left[-{\frac {\sqrt {1-2x^{2}}}{2x}}\right]_{\alpha }^{\frac {1}{2}}+\int _{\alpha }^{\frac {1}{2}}-{\frac {-4x}{2x\times 2{\sqrt {1-2x^{2}}}}}\,\mathrm {d} x\right)\\&={\sqrt {2}}\lim _{\alpha \to 0^{+}}\left(-{\sqrt {\frac {3}{2}}}+{\frac {\sqrt {2\alpha ^{2}+1}}{2\alpha }}+\int _{\alpha }^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {2x^{2}+1}}}+{\sqrt {\frac {1}{2}}}-{\frac {\sqrt {1-2\alpha ^{2}}}{2\alpha }}+\int _{\alpha }^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {1-2x^{2}}}}\right)\\&={\sqrt {2}}\lim _{\alpha \to 0^{+}}\left({\frac {1-{\sqrt {3}}}{\sqrt {2}}}+{\frac {{\sqrt {2\alpha ^{2}+1}}-{\sqrt {1-2\alpha ^{2}}}}{2\alpha }}+\int _{\alpha }^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {2x^{2}+1}}}+\int _{\alpha }^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {1-2x^{2}}}}\right)\end{aligned}}$
Comme :
${\begin{aligned}\lim _{\alpha \to 0^{+}}\left({\frac {{\sqrt {2\alpha ^{2}+1}}-{\sqrt {1-2\alpha ^{2}}}}{2\alpha }}\right)&=\lim _{\alpha \to 0^{+}}\left({\frac {({\sqrt {2\alpha ^{2}+1}}-{\sqrt {1-2\alpha ^{2}}})({\sqrt {2\alpha ^{2}+1}}+{\sqrt {1-2\alpha ^{2}}})}{2\alpha ({\sqrt {2\alpha ^{2}+1}}+{\sqrt {1-2\alpha ^{2}}})}}\right)\\&=\lim _{\alpha \to 0^{+}}\left({\frac {2\alpha ^{2}+1-1+2\alpha ^{2}}{2\alpha ({\sqrt {2\alpha ^{2}+1}}+{\sqrt {1-2\alpha ^{2}}})}}\right)\\&=\lim _{\alpha \to 0^{+}}\left({\frac {4\alpha ^{2}}{2\alpha ({\sqrt {2\alpha ^{2}+1}}+{\sqrt {1-2\alpha ^{2}}})}}\right)\\&=\lim _{\alpha \to 0^{+}}\left({\frac {2\alpha }{{\sqrt {2\alpha ^{2}+1}}+{\sqrt {1-2\alpha ^{2}}}}}\right)\\&=0\end{aligned}}$
il nous reste :
$\int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x=1-{\sqrt {3}}+{\sqrt {2}}\int _{0}^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {2x^{2}+1}}}+{\sqrt {2}}\int _{0}^{\frac {1}{2}}{\frac {\mathrm {d} x}{\sqrt {1-2x^{2}}}}$
Dans la première intégrale du second membre, posons :
$x{\sqrt {2}}=\tan(\theta )\Rightarrow \mathrm {d} x={\frac {\mathrm {d} \theta }{{\sqrt {2}}\cos ^{2}(\theta )}}\qquad \qquad 0\rightarrow x\rightarrow {\frac {1}{2}}\Rightarrow 0\rightarrow \theta \rightarrow \arctan \left({\frac {1}{\sqrt {2}}}\right)$
Dans la deuxième intégrale du second membre, posons :
$x{\sqrt {2}}=\sin(\theta )\Rightarrow \mathrm {d} x={\frac {\cos(\theta )}{\sqrt {2}}}\mathrm {d} \theta \qquad \qquad 0\rightarrow x\rightarrow {\frac {1}{2}}\Rightarrow 0\rightarrow \theta \rightarrow {\frac {\pi }{4}}$
On obtient :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x&=1-{\sqrt {3}}+{\sqrt {2}}\int _{0}^{\arctan \left({\frac {1}{\sqrt {2}}}\right)}{\frac {\mathrm {d} \theta }{{\sqrt {2}}\cos ^{2}(\theta ){\sqrt {\tan ^{2}+1}}}}+{\sqrt {2}}\int _{0}^{\frac {\pi }{4}}{\frac {\cos(\theta )}{{\sqrt {2}}{\sqrt {1-\sin ^{2}(\theta )}}}}\mathrm {d} \theta \\&=1-{\sqrt {3}}+\int _{0}^{\arctan \left({\frac {1}{\sqrt {2}}}\right)}{\frac {\mathrm {d} \theta }{\cos(\theta )}}+\int _{0}^{\frac {\pi }{4}}\mathrm {d} \theta \\&=1-{\sqrt {3}}+\int _{0}^{\arctan \left({\frac {1}{\sqrt {2}}}\right)}{\frac {\cos(\theta )\mathrm {d} \theta }{\cos ^{2}(\theta )}}+\left[\theta \right]_{0}^{\frac {\pi }{4}}\\&=1-{\sqrt {3}}+\int _{0}^{\arctan \left({\frac {1}{\sqrt {2}}}\right)}{\frac {\cos(\theta )\mathrm {d} \theta }{1-\sin ^{2}(\theta )}}+{\frac {\pi }{4}}\end{aligned}}$
Posons :
$y=\sin(\theta )\Rightarrow \mathrm {d} y=\cos(\theta )\mathrm {d} \theta \qquad \qquad 0\rightarrow \theta \rightarrow \arctan \left({\frac {1}{\sqrt {2}}}\right)\Rightarrow 0\rightarrow \theta \rightarrow {\frac {1}{\sqrt {3}}}$
On a alors :
${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {2{\sqrt {2}}}{{\sqrt {2x^{2}+1}}+{\sqrt {1-2x^{2}}}}}\mathrm {d} x&=1-{\sqrt {3}}+\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {\mathrm {d} y}{1-y^{2}}}+{\frac {\pi }{4}}\\&=1-{\sqrt {3}}+\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {\mathrm {d} y}{(1-y)(1+y)}}+{\frac {\pi }{4}}\\&=1-{\sqrt {3}}+{\frac {1}{2}}\int _{0}^{\frac {1}{\sqrt {3}}}\left({\frac {1}{1-y}}+{\frac {1}{1+y}}\right)\mathrm {d} y+{\frac {\pi }{4}}\\&=1-{\sqrt {3}}+{\frac {1}{2}}\left[\ln \left({\frac {1+y}{1-y}}\right)\right]_{0}^{\frac {1}{\sqrt {3}}}+{\frac {\pi }{4}}\\&=1-{\sqrt {3}}+{\frac {1}{2}}\ln(2+{\sqrt {3}})+{\frac {\pi }{4}}\end{aligned}}$
En conclusion, nous avons :
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