Calculer :

${\displaystyle \int _{0}^{1}{\frac {\mathrm {d} x}{{\sqrt {x+1}}+{\sqrt {3-x}}-2}}}$.

On pose :

${\displaystyle x={\frac {3-1}{2}}-{\frac {3+1}{2}}\cos(2\theta )=1-2\cos(2\theta )\qquad \mathrm {d} x=4\sin(2\theta )\,\mathrm {d} \theta }$

On a donc :

${\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {\mathrm {d} x}{{\sqrt {x+1}}+{\sqrt {3-x}}-2}}&=\int _{\frac {\pi }{6}}^{\frac {\pi }{4}}{\frac {4\sin(2\theta )\,\mathrm {d} \theta }{{\sqrt {2-2\cos(2\theta )}}+{\sqrt {2+2\cos(2\theta )}}-2}}\\&=\int _{\frac {\pi }{6}}^{\frac {\pi }{4}}{\frac {4\sin(2\theta )\,\mathrm {d} \theta }{{\sqrt {4\sin ^{2}\theta }}+{\sqrt {4\cos ^{2}\theta }}-2}}\\&=\int _{\frac {\pi }{6}}^{\frac {\pi }{4}}{\frac {2\sin(2\theta )\,\mathrm {d} \theta }{\sin \theta +\cos \theta -1}}=2\int _{\frac {\pi }{6}}^{\frac {\pi }{4}}{\frac {\sin(2\theta )\,\mathrm {d} \theta }{{\sqrt {2}}\sin \left(\theta +{\frac {\pi }{4}}\right)-1}}.\end{aligned}}}$

Posons :

${\displaystyle y=\theta +{\frac {\pi }{4}}}$.

On obtient alors :

${\displaystyle \sin(2\theta )=\sin \left(2y-{\frac {\pi }{2}}\right)=-\cos(2y)=2\sin ^{2}y-1=\left({\sqrt {2}}\sin y-1\right)\left({\sqrt {2}}\sin y+1\right)}$

donc

${\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {\mathrm {d} x}{{\sqrt {x+1}}+{\sqrt {3-x}}-2}}&=2\int _{\frac {5\pi }{12}}^{\frac {\pi }{2}}\left(1+{\sqrt {2}}\sin y\right)\,\mathrm {d} y=2\left[y-{\sqrt {2}}\cos y\right]_{\frac {5\pi }{12}}^{\frac {\pi }{2}}\\&=2\left({\frac {\pi }{2}}-{\frac {5\pi }{12}}-{\sqrt {2}}\cos {\frac {\pi }{2}}+{\sqrt {2}}\cos {\frac {5\pi }{12}}\right)\\&=\pi -{\frac {5\pi }{6}}-0+2{\sqrt {2}}\,{\frac {{\sqrt {3}}-1}{2{\sqrt {2}}}}\\&={\frac {\pi }{6}}+{\sqrt {3}}-1.\end{aligned}}}$