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

${[-x\cos x]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\cos ={[\sin x-x\cos x]}_0^{\frac{\pi }{2}}=1$

 

Exercice 13-2

${\int }_0^{\frac{\pi }{4}}{t}^2\cos t\mathrm{d}t$

Solution

${[t^2\sin t]}_0^{\frac{\pi }{4}}-2{\int }_0^{\frac{\pi }{4}}t\sin t\mathrm{d}t={[t^2\sin t+2t\cos t]}_0^{\frac{\pi }{4}}-2{\int }_0^{\frac{\pi }{4}}\cos ={[t^2\sin t+2t\cos t-2\sin t]}_0^{\frac{\pi }{4}}=\frac{1}{\sqrt{2}}(\frac{{\pi }^2}{16}+\frac{\pi }{2}-2)$.

 

Exercice 13-3

${\int }_0^{\frac{\pi }{4}}t\sin 3t\mathrm{d}t$

Solution

${[-\frac{t\cos 3t}{3}]}_0^{\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}\frac{\cos 3t}{3}\mathrm{d}t={[-\frac{t\cos 3t}{3}+\frac{\sin 3t}{9}]}_0^{\frac{\pi }{4}}=\frac{1}{3\sqrt{2}}(\frac{\pi }{4}+\frac{1}{3})$.

 

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

${[x\tan x]}_0^{\frac{\pi }{3}}-{\int }_0^{\frac{\pi }{3}}\tan ={[x\tan x+\ln \cos x]}_0^{\pi /3}=\frac{\pi \sqrt{3}}{3}-\ln 2$.

 

Exercice 13-6

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

Solution

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

 

Exercice 13-7

${\int }_0^{\frac{\pi }{2}}(x{\cos }^{2}x+{\sin }^{4}x)\mathrm{d}x$

Solution

${\int }_0^{\frac{\pi }{2}}(x\frac{1+\cos 2x}{2}+\frac{3}{8}-\frac{\cos 2x}{2}+\frac{\cos 4x}{8})\mathrm{d}x=$

${\left[x(\frac{x}{2}+\frac{\sin 2x}{4})\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(\frac{x}{2}+\frac{\sin 2x}{4})\mathrm{d}x+{[\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}]}_0^{\frac{\pi }{2}}=$

${[\frac{x^2}{4}+\frac{x\sin 2x}{4}+\frac{\cos 2x}{8}+\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{16}-\frac{1}{4}+\frac{3\pi }{16}$.

 

Exercice 13-8

${\int }_0^{\frac{\pi }{2}}(2{\cos }^{2}t\sin t+4{\sin }^{2}t)\mathrm{d}t$

Solution

${\int }_1^0-2x^2\mathrm{d}x+2{\int }_0^{\frac{\pi }{2}}(1-\cos 2t)\mathrm{d}t=\frac{2}{3}+\pi$.

 

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

${\int }_1^{\frac{1}{\sqrt{2}}}\frac{-\mathrm{d}y}{\sqrt{2}{y}^2+2y(1-y^2)}={\int }_{\frac{1}{\sqrt{2}}}^1(\frac{1}{2y}-\frac{1}{6(y-\sqrt{2})}-\frac{1}{3(y+1/\sqrt{2})})\mathrm{d}y=\frac{1}{6}{[3\ln y-\ln (\sqrt{2}-y)-2\ln (y+1/\sqrt{2})]}_{\frac{1}{\sqrt{2}}}^1=\frac{\ln 2}{2}-\frac{\ln (1+\sqrt{2})}{6}$.