Intégration en mathématiques/Exercices/Intégrales 3

Calculer les intégrales suivantes.

Exercice 11-1

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

Solution

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

 

Exercice 11-2

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\tan }^{2}x\mathrm{d}x$

Solution

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\tan }^{2}x\mathrm{d}x={[\tan x-x]}_{-\pi /4}^{\pi /4}=2-\frac{\pi }{2}$.

 

Exercice 11-3

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\tan x}{\cos x}\mathrm{d}x$

Solution

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\tan x}{\cos x}\mathrm{d}x=0$, par imparité.

 

Exercice 11-4

${\int }_0^{\frac{\pi }{3}}\cos 2t{\sin }^{3}t\mathrm{d}t$

Solution

$\begin{array}{cc}{\int }_0^{\frac{\pi }{3}}\cos 2t{\sin }^{3}t\mathrm{d}t& ={\int }_0^{\frac{\pi }{3}}(2{\cos }^{2}t-1)({\cos }^{2}t-1)(-\sin t)\mathrm{d}t\\ & ={\int }_1^{1/2}(2x^2-1)(x^2-1)\mathrm{d}x\\ & ={\int }_1^{1/2}(2x^4-3x^2+1)\mathrm{d}x\\ & ={[\frac{2x^5}{5}-x^3+x]}_1^{1/2}\\ & =\frac{-1}{80}.\end{array}$

 

Exercice 11-5

${\int }_0^{\frac{\pi }{2}}\cos t\cos 2t\cos 3t\mathrm{d}t$

Solution

$2\cos a\cos b=\cos (a+b)+\cos (a-b)$ donc $\cos t\cos 2t\cos 3t=\frac{1}{2}(\cos 3t+\cos t)\cos 3t=\frac{1}{4}(1+\cos 6t+\cos 4t+\cos 2t)$ et

$\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(1+\cos 6t+\cos 4t+\cos 2t)\mathrm{d}t=\frac{1}{4}{[t+\frac{\sin 6t}{6}+\frac{\sin 4t}{4}+\frac{\sin 2t}{2}]}_0^{\frac{\pi }{2}}={\left[\frac{t}{4}\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}$.

 

Exercice 11-6

${\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}({\cos }^{2}t+{\sin }^{4}t)\sin t\mathrm{d}t$

Solution

${\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}({\cos }^{2}t+{\sin }^{4}t)\sin t\mathrm{d}t=-{\int }_{1/\sqrt{2}}^{-1/\sqrt{2}}(x^2+(1-x^2{)}^2)\mathrm{d}x={[\frac{x^5}{5}-\frac{x^3}{3}+x]}_{-1/\sqrt{2}}^{1/\sqrt{2}}=\sqrt{2}(\frac{1}{20}-\frac{1}{6}+1)=\frac{53}{60}$.

 

Exercice 11-7

${\int }_0^{\frac{\pi }{2}}\frac{\cos t}{5-3\sin t}\mathrm{d}t$

Solution

${\int }_0^{\frac{\pi }{2}}\frac{\cos t}{5-3\sin t}\mathrm{d}t={\int }_0^1\frac{\mathrm{d}x}{5-3x}=\frac{-1}{3}{[\ln (5-3x)]}_0^1=\frac{\ln (5/2)}{3}$.

 

Exercice 11-8

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{2\sin t}{1+{\cos }^2\frac{t}{2}}\mathrm{d}t$

Solution

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{2\sin t}{1+{\cos }^2\frac{t}{2}}\mathrm{d}t=-4{\int }_{7/4}^{3/4}\frac{\mathrm{d}x}{x}=4\ln (7/3)$.