Section 3.3 Changement de variable
¶Exercice 3.3.1.
Calculez les intégrales indéfinies suivantes.
\(\displaystyle\int(x^3+1)x^2\;dx\)
\(\displaystyle\int(x^3+1)^{150}x^2\;dx\)
\(\displaystyle\int\frac{5x+15}{(x^2+6x)^3}\;dx\)
\(\displaystyle\int\sqrt[3]{x^5-2x^3}\;dx\)
\(\displaystyle\int\frac{e^\frac{1}{x}}{x^2}\;dx\)
\(\displaystyle\int\frac{1}{(2x-3)\ln(2x-3)}\;dx\)
\(\displaystyle\int\frac{8x^6}{\sqrt{1-x^{14}}}\;dx\)
\(\displaystyle\int A\sin(\omega t+\phi)\;dt\)
\(\displaystyle\int\frac{1}{\sqrt[n]{ax+b}}\;dx\)
\(\displaystyle\frac{x^6}{6}+\frac{x^3}{3}+C\)
\(\displaystyle\frac{(x^3+1)^{151}}{453}+C\)
\(\displaystyle-\frac{5}{4(x^2+6x)^2}+C\)
\(\displaystyle\frac{3(x^2-2)^\frac{4}{3}}{8}+C\)
\(\displaystyle-e^\frac{1}{x}+C\)
\(\displaystyle\frac{1}{2}\ln\left|\ln(2x-3)\right|+C\)
\(\displaystyle\frac{8}{7}\arcsin(x^7)+C\)
\(\displaystyle-\frac{A}{\omega}\cos(\omega t+\phi)+C\)
\(\displaystyle\frac{n(ax+b)^\frac{n-1}{n}}{a(n-1)}+C\)
Exercice 3.3.2.
Utilisez un changement de variable pour calculer les intégrales indéfinies suivantes.
\(\displaystyle\int(4-3x)^8\;dx\)
\(\displaystyle\int\sin(7t+2)\;dt\)
\(\displaystyle\int x^3\cos\left(x^4+2\right)\;dx\)
\(\displaystyle\int e^x\cos(e^x)\;dx\)
\(\displaystyle\int e^{5x}\;dx\)
\(\displaystyle\int\frac{e^u}{(1-e^u)^2}\;du\)
\(\displaystyle\int\frac{x}{\sqrt{1-4x^2}}\;dx\)
\(\displaystyle\int\frac{\sin\sqrt{x}}{\sqrt{x}}\;dx\)
\(\displaystyle\int\frac{\left(\ln x\right)^2}{x}\;dx\)
\(\displaystyle\int\cos^3\theta\sin\theta\;d\theta\)
\(\displaystyle\int\cot s\;ds\)
\(\displaystyle\frac{1}{27}(3x-4)^9+C\)
\(\displaystyle-\frac{1}{7}\cos(7t+2)+C\)
\(\displaystyle\frac{1}{4}\sin(x^4+2)+C\)
\(\displaystyle\sin(e^x)+C\)
\(\displaystyle\frac{e^{5x}}{5}+C\)
\(\displaystyle\frac{1}{1-e^u}+C\)
\(\displaystyle-\frac{1}{4}\sqrt{1-4x^2}+C\)
\(\displaystyle-2\cos\left(\sqrt{x}\right)+C\)
\(\displaystyle\frac{1}{3}\left(\ln x\right)^3+C\)
\(\displaystyle-\frac{1}{4}\cos^4\theta+C\)
\(\displaystyle\ln|\sin s|+C\)
Exercice 3.3.3.
Calculez les intégrales indéfinies suivantes.
\(\displaystyle\int\frac{15x^2+x+5}{5x^2+1}\;dx\)
\(\displaystyle\int(x+7)\sqrt[3]{3-2x}\;dx\)
\(\displaystyle\int\cos^3\theta\;d\theta\)
\(\displaystyle\int\sqrt{1+\sqrt{x}}\;dx\)
\(\displaystyle 3x+\frac{2\sqrt{5}}{5}\arctan(\sqrt{5}x)+\frac{1}{10}\ln(5x^2+1)+C\)
\(\displaystyle\frac{3}{28}(3-2x)^{\frac{7}{3}}-\frac{51}{16}(3-2x)^{\frac{4}{3}}+C\)
\(\displaystyle\sin\theta-\frac{1}{3}\sin^3\theta+C\)
\(\displaystyle\frac{4}{5}(1+\sqrt{x})^{\frac{5}{2}}-\frac{4}{3}(1+\sqrt{x})^{\frac{3}{2}}+C\)
Exercice 3.3.4.
Évaluez les intégrales définies suivantes à l'aide d'un changement de variable.
\(\displaystyle\int_1^2x(x^2-4)^3\;dx\)
\(\displaystyle\int_{(\ln 5)^2}^{(\ln 15)^2}\frac{e^\sqrt{x}}{5\sqrt{x}}\;dx\)
\(\displaystyle\int_e^{e^e}\frac{-1}{x\ln\sqrt{x}}\;dx\)
\(\displaystyle\int_0^{\frac{\pi}{8}}\sin^3(4\theta)\cos(4\theta)\;d\theta\)
\(\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\cos\theta}\;d\theta\)
\(\displaystyle-\frac{81}{8}\)
\(\displaystyle 4\)
\(\displaystyle-2\)
\(\displaystyle\frac{1}{16}\)
\(\displaystyle 4\frac{\sqrt{3}}{3}-2\)
Exercice 3.3.5.
Calculez \(\int\sec(x)\,dx\text{.}\)
Utilisez \(\displaystyle\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}=1\) puis le changement de variable \(u=\sec(x)+\tan(x)\text{.}\)
\(\displaystyle\ln|\sec(x)+\tan(x)|+C\)