Intégration en mathématiques/Exercices/Intégrales 5

Calculer les intégrales suivantes.

Exercice 13-1

$\int _{0}^{\frac {\pi }{2}}x\sin x\,\mathrm {d} x$

Solution

$\left[-x\cos x\right]_{0}^{\frac {\pi }{2}}+\int _{0}^{\frac {\pi }{2}}\cos =\left[\sin x-x\cos x\right]_{0}^{\frac {\pi }{2}}=1$

 

Exercice 13-2

$\int _{0}^{\frac {\pi }{4}}t^{2}\cos t\,\mathrm {d} t$

Solution

$\left[t^{2}\sin t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}t\sin t\,\mathrm {d} t=\left[t^{2}\sin t+2t\cos t\right]_{0}^{\frac {\pi }{4}}-2\int _{0}^{\frac {\pi }{4}}\cos =\left[t^{2}\sin t+2t\cos t-2\sin t\right]_{0}^{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}\left({\frac {\pi ^{2}}{16}}+{\frac {\pi }{2}}-2\right)$.

 

Exercice 13-3

$\int _{0}^{\frac {\pi }{4}}t\sin 3t\,\mathrm {d} t$

Solution

$\left[-{\frac {t\cos 3t}{3}}\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\cos 3t}{3}}\,\mathrm {d} t=\left[-{\frac {t\cos 3t}{3}}+{\frac {\sin 3t}{9}}\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3{\sqrt {2}}}}\left({\frac {\pi }{4}}+{\frac {1}{3}}\right)$.

 

Exercice 13-4

$\int _{0}^{2\pi }t^{2}\sin 2t\,\mathrm {d} t$

Solution

$\left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\int _{0}^{2\pi }t\cos 2t\,\mathrm {d} t=\left[{\frac {-t^{2}\cos 2t}{2}}\right]_{0}^{2\pi }+\left[{\frac {t\sin 2t}{2}}\right]_{0}^{2\pi }-\int _{0}^{2\pi }{\frac {\sin 2t}{2}}\,\mathrm {d} t=-2\pi ^{2}+0-0=-2\pi ^{2}$.

 

Exercice 13-5

$\int _{0}^{\frac {\pi }{3}}{\frac {x}{\cos ^{2}x}}\,\mathrm {d} x$

Solution

$\left[x\tan x\right]_{0}^{\frac {\pi }{3}}-\int _{0}^{\frac {\pi }{3}}\tan =\left[x\tan x+\ln \cos x\right]_{0}^{\pi /3}={\frac {\pi {\sqrt {3}}}{3}}-\ln 2$.

 

Exercice 13-6

$\int _{\frac {\pi }{6}}^{\frac {\pi }{3}}\left(\tan t+{\frac {1}{\tan t}}\right)\,\mathrm {d} t$

Solution

$\left[\ln \tan \right]_{\pi /6}^{\pi /3}=\ln 3$.

 

Exercice 13-7

$\int _{0}^{\frac {\pi }{2}}\left(x\cos ^{2}x+\sin ^{4}x\right)\,\mathrm {d} x$

Solution

$\int _{0}^{\frac {\pi }{2}}\left(x{\frac {1+\cos 2x}{2}}+{\frac {3}{8}}-{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}\right)\,\mathrm {d} x=$

$\left[x\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\right]_{0}^{\frac {\pi }{2}}-\int _{0}^{\frac {\pi }{2}}\left({\frac {x}{2}}+{\frac {\sin 2x}{4}}\right)\,\mathrm {d} x+\left[{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}=$

$\left[{\frac {x^{2}}{4}}+{\frac {x\sin 2x}{4}}+{\frac {\cos 2x}{8}}+{\frac {3x}{8}}-{\frac {\sin 2x}{4}}+{\frac {\sin 4x}{32}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi ^{2}}{16}}-{\frac {1}{4}}+{\frac {3\pi }{16}}$.

 

Exercice 13-8

$\int _{0}^{\frac {\pi }{2}}\left(2\cos ^{2}t\sin t+4\sin ^{2}t\right)\,\mathrm {d} t$

Solution

$\int _{1}^{0}-2x^{2}\,\mathrm {d} x+2\int _{0}^{\frac {\pi }{2}}\left(1-\cos 2t\right)\,\mathrm {d} t={\frac {2}{3}}+\pi$.

 

Exercice 13-9

$\int _{0}^{\frac {\pi }{2}}{\frac {\mathrm {d} x}{2+\sin x}}$

Solution

$\int _{0}^{1}{\frac {\frac {2\,\mathrm {d} t}{1+t^{2}}}{2+{\frac {2t}{1+t^{2}}}}}=\int _{0}^{1}{\frac {\mathrm {d} t}{1+t+t^{2}}}=\int _{1/2}^{3/2}{\frac {\mathrm {d} s}{s^{2}+3/4}}={\frac {2}{\sqrt {3}}}\left[\arctan \right]_{1/{\sqrt {3}}}^{\sqrt {3}}={\frac {\pi }{3{\sqrt {3}}}}$.

 

Exercice 13-10

$\int _{0}^{\frac {\pi }{4}}{\frac {\tan x}{{\sqrt {2}}\cos x+2\sin ^{2}x}}\,\mathrm {d} x$

Solution

En posant (selon les règles de Bioche) $y=\cos x$, l'intégrale devient

$\int _{1}^{\frac {1}{\sqrt {2}}}{\frac {-\mathrm {d} y}{{\sqrt {2}}y^{2}+2y(1-y^{2})}}=\int _{\frac {1}{\sqrt {2}}}^{1}\left({\frac {1}{2y}}-{\frac {1}{6(y-{\sqrt {2}})}}-{\frac {1}{3(y+1/{\sqrt {2}})}}\right)\,\mathrm {d} y={\frac {1}{6}}\left[3\ln y-\ln({\sqrt {2}}-y)-2\ln(y+1/{\sqrt {2}})\right]_{\frac {1}{\sqrt {2}}}^{1}={\frac {\ln 2}{2}}-{\frac {\ln(1+{\sqrt {2}})}{6}}$.