Techniques d’Integration , pages multiples

3. Substitution trigonométrique

Les expressions algébriques qui suivent nécessitent la méthode de substitution trigonométrique:

$(a^{2}-u^{2})^{\frac{1}{2}}$

$(u^{2}-a^{2})^{\frac{3}{2}}$

$\frac{1}{(a^{2}+u^{2})^{2}}$

Cas trigonométriques

Integral Involving	Use Substitution	Then Identity
$a^{2}-u^{2}$	$u=a \sin \theta$	$1-\sin^{2} \theta=\cos^{2} \theta$
$a^{2}+u^{2}$	$u=a \tan \theta$	$1+\tan^{2} \theta=\sec^{2} \theta$
$u^{2}-a^{2}$	$u=a \sec \theta$	$\sec^{2} \theta-1=\tan^{2} \theta$

Ou Substitution Hyperbolique :

Integral Involving	Use Substitution	Then Identity
$a^{2}-u^{2}$	$u=a \tanh \theta$	$1-\tanh^{2} \theta=sech^{2} \theta$
$a^{2}+u^{2}$	$u=a \sinh \theta$	$1+\sinh^{2} \theta=\cosh^{2} \theta$
$u^{2}-a^{2}$	$u=a \cosh \theta$	$\cosh^{2} \theta-1=\sinh^{2} \theta$

Problème 19
Evaluer: $\int \frac{dx}{(4-x^{2})^{3}}$

Nous retrouvons $a^{2}-u^{2}$

Dans ce cas, $a=2$

Soit $x=2 \sin \theta$

$dx=2 \cos \theta d\theta$

$4-x^{2}=4-4 \sin^{2}\theta=4(1-\sin^{2} \theta)=4\cos^{2}\theta$

De retour à l’équation:

(1) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{(4-x^{2})^{3}} &={\displaystyle \int \frac{2 \cos \theta \,d\theta}{(4\cos^{2}\theta)^{3}}}\\ &={\displaystyle \frac{2}{64}\int \frac{\cos \theta \,d\theta}{\cos^{6}\theta}}\\ &={\displaystyle\frac{1}{32}\int sec^{5} \theta \, d\theta} \end{split} \end{equation*}$

Maintenant calculons ce qui suit:

$\int sec^{5} \theta \, d\theta$

Combinant avec les méthodes connues:

(2) $\begin{equation*} \begin{split} \displaystyle \int sec^{5} \theta \, d\theta &=\displaystyle \frac{1}{4} \sec^{3} \theta \tan \theta+\frac{3}{4}\int \sec^{3}\theta \, d\theta \\ &= {\displaystyle \frac{1}{4} \sec^{3} \theta \tan \theta+\frac{3}{4}\left (\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2} \int sec \theta \, d\theta \right )}\\ &={\displaystyle \frac{1}{4} \sec^{3} \theta \tan \theta+\frac{3}{4}\left (\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2} \ln | \sec \theta +\tan \theta| \right ) +C} \\ &={\displaystyle \frac{1}{4} \sec^{3} \theta \tan \theta+\frac{3}{8}\sec \theta \tan \theta +\frac{3}{8} \ln | \sec \theta +\tan \theta| +C} \end{split} \end{equation*}$

De retour sur $x$ on applique ce qui suit:

$x=2 \sin \theta \Rightarrow \sin \theta =\frac{x}{2}$

$\sin \theta =\frac{x}{2} \Rightarrow \cos \theta=\sqrt{1-(\frac{x}{2})^{2}}=\sqrt{1-\frac{x^{2}}{4}}=\sqrt{\frac{4-x^{2}}{4}}=\frac{1}{2}\sqrt{4-x^{2}}$

$\cos \theta=\frac{1}{2}\sqrt{4-x^{2}}$

$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{x}{\sqrt{4-x^{2}}}$

$\sec \theta=\frac{1}{\cos \theta}=\frac{2}{\sqrt{4-x^{2}}}$

En appliquant au résultat:

$\int \frac{dx}{(4-x^{2})^{3}} }={\displaystyle\frac{1}{32}\int sec^{5} \theta \, d\theta$

(3) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{(4-x^{2})^{3}} }&={\displaystyle\frac{1}{32}\left ( \frac{1}{4} \sec^{3} \theta \tan \theta+\frac{3}{8}\sec \theta \tan \theta +\frac{3}{8} \ln | \sec \theta +\tan \theta| \right ) +C}\\ &={\displaystyle \frac{1}{32} \cdot \frac{1}{4}(\frac{2}{\sqrt{4-x^{2}}})^{3} \cdot \frac{x}{\sqrt{4-x^{2}}}+\frac{1}{32} \cdot \frac{3}{8}\cdot \frac{2}{\sqrt{4-x^{2}}} \cdot \frac{x}{\sqrt{4-x^{2}}}+ \frac{1}{32} \cdot \frac{3}{8} \cdot \ln |\frac{2}{\sqrt{4-x^{2}}}+ \frac{x}{\sqrt{4-x^{2}}}| +C}\\ &={\displaystyle \frac{1}{128}\cdot \frac{8}{(\sqrt{4-x^{2}})^{3}}\cdot \frac{x}{\sqrt{4-x^{2}}}+\frac{3}{256} \cdot \frac{2}{\sqrt{4-x^{2}}} \cdot \frac{x}{\sqrt{4-x^{2}}}+ \frac{3}{256} \cdot \ln |\frac{2}{\sqrt{4-x^{2}}}+ \frac{x}{\sqrt{4-x^{2}}}| +C} \\ &= {\displaystyle \frac{1}{256}\cdot \frac {16x}{(4-x^{2})^{2}}+\frac{3}{256}\cdot \frac{2x}{4-x^{2}}+ \frac{3}{256}\cdot \ln |\frac{2+x}{\sqrt{4-x^{2}}} | +C}\\ &= {\displaystyle \frac{1}{256}\cdot \frac{16x+6x(4-x^{2})}{(4-x^{2})^{2}}+ \frac{3}{256}\cdot \ln |\frac{\sqrt{(2+x)(2+x)}}{\sqrt{(2+x)(2-x)}} | +C}\\ &= {\displaystyle \frac{1}{256}\cdot \frac{40x-6x^{3}}{(4-x^{2})^{2}}+\frac{3}{256}\cdot \frac{1}{2} \cdot \ln |\frac{2+x}{2-x} | +C }\\ &= {\displaystyle \frac{20x-3x^{3}}{(4-x^{2})^{2}}+\frac{3}{512}\ln |2+x|- \frac{3}{512}\ln |2-x| +C} \end{split} \end{equation*}$

Finalement:

$\int \frac{dx}{(4-x^{2})^{3}} }={\displaystyle \frac{20x-3x^{3}}{(4-x^{2})^{2}}+\frac{3}{512}\ln |2+x|- \frac{3}{512}\ln |2-x| +C$

Problème 20
Evaluer: $\int \frac{dx}{\sqrt{9+4x^{2}}}$

Ce cas contient $a^{2}+u^{2}$

Dans ce cas, $a=3$

Ici $u=2x$

Soit $u=3 \tan \theta \Rightarrow 2x=3 \tan \theta$

$2dx=3 \sec^{2} \theta d\theta \Rightarrow dx=\frac{3}{2}\sec^{2} \theta \, d\theta$

$9+u^{2}=9+9\tan^{2}\theta=9(1+\tan^{2} \theta)=9\sec^{2}\theta$

De retour sur l’équation

(4) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{\sqrt{9+4x^{2}}} }&={\displaystyle \frac{3}{2} \int \frac{\sec^{2} \theta \, d\theta}{\sqrt{9\sec^{2}\theta} } }\\ &={\displaystyle \frac{3}{2} \int \frac{\sec^{2} \theta \, d\theta}{3\sec \theta} }\\ &={\displaystyle \frac{1}{2}\int \sec \theta \, d\theta }\\ &= {\displaystyle \frac{1}{2} \ln | \sec \theta + \tan \theta |+C }\end{align*}$ \end{split} \end{equation*}$

De retour sur $x$

$\tan \theta=\frac{2x}{3}$

$\cos \theta=\frac{3}{\sqrt{9+4x^{2}}}$

$\sec \theta=\frac{\sqrt{9+4x^{2}}}{3}$

(5) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{\sqrt{9+4x^{2}}} }&= {\displaystyle \frac{1}{2} \ln \left |\frac{\sqrt{9+4x^{2}}}{3} + \frac{2x}{3}\right |+C }\\ &={\displaystyle \frac{1}{2} \ln \left |\frac{\sqrt{9+4x^{2}}}{3} + \frac{2x}{3}\right |+C } \end{split} \end{equation*}$

Finalement:

$\int \frac{dx}{\sqrt{9+4x^{2}}} }={\displaystyle \frac{1}{2} \ln \left |\frac{\sqrt{9+4x^{2}}}{3} + \frac{2x}{3}\right |+C}={\displaystyle \frac{1}{2} \ln \left |\sqrt{9+4x^{2}} + 2x\right |+C_{1}$

Ou bien:

${\displaystyle \int \frac{dx}{\sqrt{9+4x^{2}}} }=\frac{1}{2}\sinh^{-1} (\frac{2x}{3})+C$

En utilisant simplement la substitution hyperbolique:

Ici $u=2x$

Soit $u=3\sinh \theta$

Nous avonst $2x=3\sinh \theta \Rightarrow 2dx=3 \cosh \theta \, d\theta$

$\sqrt{9+4x^{2}}=\sqrt{9+9\sinh^{2} \theta}=\sqrt{9(1+\sinh^{2} \theta)}=3\sqrt{1+\sinh^{2} \theta}=3\cosh \theta$

On remplace:

(6) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{\sqrt{9+4x^{2}}} }&= {\displaystyle \frac{3}{2} \int \frac{cosh \theta \, d\theta}{3\cosh \theta}}\\ &={\displaystyle \frac{1}{2} \int d\theta }\\ &=\theta +C \end{split} \end{equation*}$

Mais nous avons vu que:

$2x=3\sinh \theta \Rightarrow \theta=\sinh^{-1} (\frac{2x}{3})$

Alors:

${\displaystyle \int \frac{dx}{\sqrt{9+4x^{2}}} }=\frac{1}{2}\sinh^{-1} (\frac{2x}{3})+C$

Problème 21
Evaluate: $\int \sqrt{9+16x^{2}} \, dx$

Ce cas contient $a^{2}+u^{2}$

Ici, $a=3$

Alors $u=4x$

Soit $u=3 \sinh \theta \Rightarrow 4x=3 \sinh \theta$

$4dx=3 \cosh \theta d\theta \Rightarrow dx=\frac{3}{4} \cosh \theta d\theta$

$\sqrt{9+16x^{2}}=\sqrt{9+9\sinh^{2} \theta}=\sqrt{9(1+\sinh^{2} \theta)}=3\sqrt{1+\sinh^{2} \theta}=3\cosh \theta$

De retour sur l’équation

(7) $\begin{equation*} \begin{split} \displaystyle \int \sqrt{9+16x^{2}} \, dx }&= {\displaystyle \frac{3}{4}\int 3\cosh \theta \cosh \theta \, d\theta }\\ &= {\displaystyle \frac{9}{4}\int \cosh^{2} \theta \, d\theta } \end{split} \end{equation*}$

Mais nous savons que:

$\cosh 2\theta=2 \cosh \theta -1$

$\cosh^{2} \theta =\frac{\cosh 2\theta + 1}{2}$

(8) $\begin{equation*} \begin{split} \displaystyle \int \sqrt{9+16x^{2}} \, dx }&={\displaystyle \frac{9}{4}\int \cosh^{2} \theta \, d\theta } \\ &={\displaystyle \frac{9}{8} \int \cosh 2\theta \, d\theta +\frac{9}{8} \int d\theta }\\ &={\displaystyle \frac{9}{16} \sinh 2\theta +\frac{9}{8} \theta +C}\\ &= \frac{9}{16} \cdot 2\sinh \theta \cosh \theta +\frac{9}{8}\theta +C \end{split} \end{equation*}$

De retour sur $x$

$\sinh \theta =\frac{4x}{3}$

$\cosh \theta= \sqrt{1+(\frac{4x}{3})^{2}}=\frac{1}{3}\sqrt{9+16x^{2}}$

Sur l’équation d’origine:

(9) $\begin{equation*} \begin{split} \displaystyle \int \sqrt{9+16x^{2}} \, dx }&={\displaystyle \frac{9}{16}\cdot 2 \cdot \frac{4x}{3}\cdot \frac{1}{3}\sqrt{9+16x^{2}} + \frac{9}{8}\sinh^{-1} \left ( \frac{4x}{3}\right)+C} \end{split} \end{equation*}$

$\int \sqrt{9+16x^{2}} \, dx }={\displaystyle \frac{x}{2}\sqrt{9+16x^{2}} +\frac{9}{8} \sinh^{-1} \left ( \frac{4x}{3}\right)+C$

Finalement:

$\int \sqrt{9+16x^{2}} \, dx }={\displaystyle \frac{x}{2}\sqrt{9+16x^{2}} +\frac{9}{8} \sinh^{-1} \left ( \frac{4x}{3}\right)+C$

OU:

$\int \sqrt{9+16x^{2}} \, dx }={\displaystyle \frac{x}{2}\sqrt{9+16x^{2}} +\frac{9}{8} \ln \left | 4x+ \sqrt{9+16x^{2}} \right |+C_{1}$

Problème 22
Evaluer: $\int \frac{dx}{x^{2}\sqrt{4x^{2}-9}}$

Avec la substitution trigonométrique

Ce cas contient $u^{2}-a^{2}$

Ici, $a=3$

Alors $u=2x$

Soit $u=3 \sec \theta \Rightarrow 2x=3 \sec \theta$

$2dx=3 \sec \theta \tan \theta d\theta \Rightarrow dx= \frac{3}{2}\sec \theta \tan \theta d\theta$

$x=\frac{3}{2} \sec \theta$

$x^{2}=\frac{9}{4} \sec^{2} \theta$

$\sqrt{4x^{2}-9}=\sqrt{9 \sec^{2} \theta-1}=3 \sqrt{\tan^{2} \theta}=3 \tan \theta$

De retour sur l’équation

(10) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{x^{2}\sqrt{4x^{2}-9}} }&= {\displaystyle \frac{3}{2} \int \frac{\sec \theta \tan \theta\, d\theta}{\frac{9}{4}\sec^{2} \theta \cdot 3 \tan \theta} }\\ &={\displaystyle \frac{2}{9} \int \cos \theta \, d\theta}\\ &={\displaystyle \frac{2}{9} \sin \theta + C} \end{split} \end{equation*}$

$2x=3 \sec \theta \Rightarrow \cos \theta=\frac{3}{2x}$

$\sin \theta=\sqrt{1-(\frac{3}{2x})^{2}}=\sqrt{1-\frac{9}{4x^{2}}}=\sqrt{\frac{4x^{2}-9}{4x^{2}}}=\frac{1}{2x}\sqrt{4x^{2}-9}$

Nous obtenons:

${\displaystyle \int \frac{dx}{x^{2}\sqrt{4x^{2}-9}} }=\frac{2}{9} \cdot \frac{1}{2x}\sqrt{4x^{2}-9}+C$

Finalement:

$\int \frac{dx}{x^{2}\sqrt{4x^{2}-9}} }={\displaystyle \frac{\sqrt{4x^{2}-9}}{9x}+C$

Problème 23
Evaluer: $\int \frac{dx}{x^{2}\sqrt{4x^{2}-9}}$

Utilisant la substitution hyperbolique

Ce cas contient $u^{2}-a^{2}$

Ici, $a=3$

Alors $u=2x$

Soit $u=3 \cosh \theta \Rightarrow 2x=3 \cosh \theta$

$x=\frac{3}{2}\cosh \theta$

$dx=\frac{3}{2}\sinh \theta \, d\theta$

$x^{2}=\frac{9}{4}\cosh^{2} \theta$

$\sqrt{4x^{2}-9}=\sqrt{9 \cosh^{2} \theta-9}=3 \sqrt{\cosh^{2} \theta-1}=3 \sqrt{\sinh^{2} \theta}=3 \sinh \theta$

De retour sur l’équation

(11) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{x^{2}\sqrt{4x^{2}-9}} }&= {\displaystyle \frac{3}{2} \int \frac{\sinh \theta \, d\theta}{\frac{9}{4}\cosh^{2} \theta \cdot 3 \sinh \theta} }\\ &={\displaystyle \frac{2}{9} \int sech^{2} \theta \, d\theta}\\ &={\displaystyle \frac{2}{9} \tanh \theta + C} \end{split} \end{equation*}$