Posons :

${\displaystyle x=6y\Rightarrow \mathrm {d} x=6\mathrm {d} y\qquad \qquad 0\to x\to \pi \Rightarrow 0\to y\to {\frac {\pi }{6}}}$.

On obtient alors :

${\displaystyle \int _{0}^{\pi }{\frac {1-\cos {\frac {x}{3}}}{\sin {\frac {x}{2}}}}\,\mathrm {d} x=6\int _{0}^{\frac {\pi }{6}}{\frac {1-\cos(2y)}{\sin(3y)}}\,\mathrm {d} y=6\int _{0}^{\frac {\pi }{6}}{\frac {2(1-\cos ^{2}y)}{(4\cos ^{2}y-1)\sin y}}\,\mathrm {d} y=3\int _{0}^{\frac {\pi }{6}}{\frac {4\sin y}{4\cos ^{2}y-1}}\,\mathrm {d} y}$.

Posons alors :

${\displaystyle z=\cos y\Rightarrow \mathrm {d} z=-\sin y\,\mathrm {d} y\qquad \qquad 0\to x\to {\frac {\pi }{6}}\Rightarrow 1\to y\to {\frac {\sqrt {3}}{2}}}$.

Nous obtenons :

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

Nous pouvons conclure que :

Nous sommes dans le « cas hybride » des règles de Bioche, où les trois changements de variable y = cos x, y = sin x et t = tan x sont fructueux mais où un changement plus intéressant est u = cos(2x).

Pour mettre en forme ce changement de variable, remarquons que

${\displaystyle {\frac {\tan x}{1+\sin ^{2}x}}={\frac {\sin x\cos x}{\cos ^{2}x\left(1+\sin ^{2}x\right)}}={\frac {{\frac {1}{2}}\sin(2x)}{{\frac {1}{2}}\left(1+\cos(2x)\right){\frac {1}{2}}\left(3-\cos(2x)\right)}}}$.

En posant u = cos(2x), on obtient donc :

${\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{4}}{\frac {\tan x}{1+\sin ^{2}x}}\,\mathrm {d} x&=\int _{1}^{0}{\frac {-\mathrm {d} u}{(1+u)(3-u)}}={\frac {1}{4}}\int _{0}^{1}\left({\frac {1}{1+u}}+{\frac {1}{3-u}}\right)\,\mathrm {d} u\\&={\frac {1}{4}}\left[\ln {\frac {1+u}{3-u}}\right]_{0}^{1}={\frac {\ln 3}{4}}.\end{aligned}}}$

a) En posant ${\displaystyle y={\sqrt {x^{3}+1}}}$, nous avons ${\displaystyle y^{2}=x^{3}+1}$ donc ${\displaystyle 2y\,\mathrm {d} y=3x^{2}\,\mathrm {d} x}$ et

${\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {\sqrt {x^{3}+1}}{x}}\,\mathrm {d} x&=\int _{1}^{2}{\frac {\sqrt {x^{3}+1}}{x^{3}}}\,x^{2}\mathrm {d} x={\frac {1}{3}}\int _{\sqrt {2}}^{3}{\frac {2y^{2}}{y^{2}-1}}\,\mathrm {d} y\\&={\frac {1}{3}}\int _{\sqrt {2}}^{3}\left(2-{\frac {1}{y+1}}+{\frac {1}{y-1}}\right)\,\mathrm {d} y\\&={\frac {1}{3}}\left[2y-\ln {(y+1)}+\ln {(y-1)}\right]_{\sqrt {2}}^{3}\\&={\frac {1}{3}}\left[2y+\ln {\frac {y-1}{y+1}}\right]_{\sqrt {2}}^{3}\\&={\frac {6-2{\sqrt {2}}-\ln {(6-4{\sqrt {2}})}}{3}}.\end{aligned}}}$

b) ${\displaystyle 4x^{2}+x+1=\left(2x+{\frac {1}{4}}\right)^{2}+{\frac {15}{16}}={\frac {15}{16}}\left(y^{2}+1\right)}$ avec ${\displaystyle y={\frac {4}{\sqrt {15}}}\left(2x+{\frac {1}{4}}\right)={\frac {8x+1}{\sqrt {15}}}}$ donc posons

${\displaystyle \tan \theta ={\frac {8x+1}{\sqrt {15}}},\quad {\frac {\mathrm {d} x}{\sqrt {4x^{2}+x+1}}}={\frac {\mathrm {d} \theta }{2\cos \theta }}}$.

${\displaystyle \int _{1}^{+\infty }{\frac {\mathrm {d} x}{x{\sqrt {4x^{2}+x+1}}}}=4\int _{\arctan \left(3{\sqrt {3/5}}\right)}^{\frac {\pi }{2}}{\frac {\mathrm {d} \theta }{{\sqrt {15}}\sin \theta -\cos \theta }}}$.

Posons enfin ${\displaystyle t=\tan {\frac {\theta }{2}},\quad \cos \theta ={\frac {1-t^{2}}{1+t^{2}}},\quad \sin \theta ={\frac {2t}{1+t^{2}}},\quad \mathrm {d} \theta ={\frac {2\mathrm {d} t}{1+t^{2}}},\quad T=\tan {\frac {\arctan \left(3{\sqrt {3/5}}\right)}{2}}={\frac {4{\sqrt {2}}-{\sqrt {5}}}{3{\sqrt {3}}}}}$.

${\displaystyle 4\int _{\arctan \left(3{\sqrt {3/5}}\right)}^{\frac {\pi }{2}}{\frac {\mathrm {d} \theta }{{\sqrt {15}}\sin \theta -\cos \theta }}=8\int _{T}^{1}{\frac {\mathrm {d} t}{t^{2}+2{\sqrt {15}}t-1}}=\int _{T}^{1}\left({\frac {1}{t+{\sqrt {15}}-4}}-{\frac {1}{t+{\sqrt {15}}+4}}\right)\,\mathrm {d} t=\ln {\frac {\left({\sqrt {15}}-3\right)\left(T+{\sqrt {15}}+4\right)}{\left({\sqrt {15}}+5\right)\left(T+{\sqrt {15}}-4\right)}}}$.

Finalement,

${\displaystyle \int _{1}^{+\infty }{\frac {\mathrm {d} x}{x{\sqrt {4x^{2}+x+1}}}}=\ln {\frac {\left({\sqrt {15}}-3\right)\left({\sqrt {2}}+2{\sqrt {5}}+3{\sqrt {3}}\right)}{\left({\sqrt {15}}+5\right)\left({\sqrt {2}}+2{\sqrt {5}}-3{\sqrt {3}}\right)}}\approx 0{,}457}$.

a) Pour ${\displaystyle 0\leq x\leq {\frac {\pi }{2}}}$, ${\displaystyle \cos x\geq 0}$ donc ${\displaystyle {\sqrt {1+\cos 2x}}={\sqrt {1+2\cos ^{2}x-1}}={\sqrt {2\cos ^{2}x}}={\sqrt {2}}\cos x}$.

Par conséquent,

${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos x}{\sqrt {1+\cos 2x}}}\,\mathrm {d} x=\int _{0}^{\frac {\pi }{2}}{\frac {1}{\sqrt {2}}}\,\mathrm {d} x={\frac {\pi }{2{\sqrt {2}}}}}$.

b) L'intégration du a) n'ayant pas posé de problème, nous allons faire un changement de variable pour avoir un dénominateur identique au a). Pour cela, nous posons :

${\displaystyle x=y+{\frac {\pi }{4}}\Rightarrow \mathrm {d} x=\mathrm {d} y\qquad \qquad 0\to x\to {\frac {\pi }{2}}\Rightarrow -{\frac {\pi }{4}}\to y\to {\frac {\pi }{4}}}$.

Nous obtenons :

• ${\displaystyle {\sqrt {1+\sin 2x}}={\sqrt {1+\sin \left(2y+{\frac {\pi }{2}}\right)}}={\sqrt {1+\cos(2y)}}={\sqrt {2}}\cos y}$ ;
• ${\displaystyle \sin x=\sin \left(y+{\frac {\pi }{4}}\right)={\frac {\cos y+\sin y}{\sqrt {2}}}}$

donc (par imparité de ${\displaystyle \sin }$)

${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\sin x}{\sqrt {1+\sin 2x}}}\,\mathrm {d} x=\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\cos y+\sin y}{{\sqrt {2}}{\sqrt {2}}\cos y}}\,\mathrm {d} y=\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\cos y}{2\cos y}}\,\mathrm {d} y={\frac {\pi }{4}}}$.

a) Posons :

${\displaystyle y={\sqrt {\frac {x-1}{x+1}}}\Leftrightarrow x={\frac {1+y^{2}}{1-y^{2}}}\Rightarrow \mathrm {d} x={\frac {4y}{(y^{2}-1)^{2}}}\,\mathrm {d} y\qquad \qquad 1\to x\to 2\Rightarrow 0\to y\to {\frac {1}{\sqrt {3}}}}$.

On a donc :

${\displaystyle {\begin{aligned}\int _{1}^{2}x{\sqrt {\frac {x-1}{x+1}}}\,\mathrm {d} x&=\int _{0}^{\frac {1}{\sqrt {3}}}y{\frac {1+y^{2}}{1-y^{2}}}{\frac {4y}{(y^{2}-1)^{2}}}\,\mathrm {d} y=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {4y^{2}(1+y^{2})}{(1-y^{2})^{3}}}\,\mathrm {d} y\\&=\int _{0}^{\frac {1}{\sqrt {3}}}\left({\frac {1}{2(y+1)}}-{\frac {3}{2(y+1)^{2}}}+{\frac {1}{(y+1)^{3}}}-{\frac {1}{2(y-1)}}-{\frac {3}{2(y-1)^{2}}}-{\frac {1}{(y-1)^{3}}}\right)\,\mathrm {d} y\\&=\left[{\frac {1}{2}}\ln {\frac {1+y}{1-y}}+{\frac {3y+2}{2(y+1)^{2}}}+{\frac {3y-2}{2(y-1)^{2}}}\right]_{0}^{\frac {1}{\sqrt {3}}}={\frac {\ln(2+{\sqrt {3}})}{2}}.\end{aligned}}}$

b) Posons :

${\displaystyle y={\sqrt {\frac {x+1}{x+3}}}\Leftrightarrow x={\frac {1-3y^{2}}{y^{2}-1}}\Rightarrow \mathrm {d} x={\frac {4y}{(y^{2}-1)^{2}}}\mathrm {d} y\qquad \qquad -1\to x\to 0\Rightarrow 0\to y\to {\frac {1}{\sqrt {3}}}}$.

On a donc :

${\displaystyle {\begin{aligned}\int _{-1}^{0}\arctan {\sqrt {\frac {x+1}{x+3}}}\,\mathrm {d} x&=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {4y}{(y^{2}-1)^{2}}}\arctan y\,\mathrm {d} y\\&=\left[{\frac {-2}{y^{2}-1}}\arctan y\right]_{0}^{\frac {1}{\sqrt {3}}}+2\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{(y^{2}-1)(y^{2}+1)}}\,\mathrm {d} y\\&=\left[{\frac {2}{1-y^{2}}}\arctan y\right]_{0}^{\frac {1}{\sqrt {3}}}+\int _{0}^{\frac {1}{\sqrt {3}}}\left({\frac {1}{2(y-1)}}-{\frac {1}{2(y+1)}}-{\frac {1}{y^{2}+1}}\right)\,\mathrm {d} y\\&=\left[{\frac {1+y^{2}}{1-y^{2}}}\arctan y+{\frac {1}{2}}\ln {\frac {1-y}{1+y}}\right]_{0}^{\frac {1}{\sqrt {3}}}\\&={\frac {4/3}{2/3}}\,{\frac {\pi }{6}}+{\frac {1}{2}}\ln {\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}\\&={\frac {\pi }{3}}+{\frac {\ln(2-{\sqrt {3}})}{2}}.\end{aligned}}}$