Exercice 90
\(\begin{align*} \int_0^{\pi/2} \frac{sin^3x}{cos^3x+sin^3x}dx\end{align*}\) |
Connaissances:
- Méthode intégrales de Wallis
\(\begin{align*} I & = \int_0^{\pi/2} \frac{sin^3x}{sin^3x+ cos^3x}dx \end{align*}\)
Posons: \(x=\pi/2-t\):
alors:\(\begin{cases} dx= -dt \\ sinx= sin(\pi/2 - t)= cos t \\ cosx= cos(\pi/2 - t)= sin t \end{cases}\)
et \(\begin{cases} x = 0 \Rightarrow t=\pi/2 \\ x = \pi/2 \Rightarrow t=0 \\ dx=-dt \end{cases}\)
\(\begin{align*} I & = \int_0^{\pi/2} \frac{sin^3x}{sin^3x+ cos^3x}dx \\
& = \int_{\pi/2}^0 \frac{cos^3t}{sin^3t+cos^3t}.(-dt) \\
& = \int_0^{\pi/2} \frac{cos^3t}{sin^3t+cos^3t}.dt \\\end{align*}\)
\(\begin{align*}2I & = I + I \\
& = \int_0^{\pi/2} \frac{sin^3x}{sin^3x+ cos^3x}dx + \int_0^{\pi/2} \frac{cos^3t}{sin^3t+cos^3t}.dt \\
& = \int_{\pi/2}^0 dt = \pi/2 \\ \\
\Rightarrow I & = \pi/4 \end{align*}\)
\[\boxed {\begin{align*} I &=\frac{\pi}{4} \end{align*}}\]
Exercice 89
\(\begin{align*} \int \frac{\sqrt{x+4}}{x}dx\end{align*}\) |
Connaissances:
- Changement de variable
- Division polynômiale
- Décomposition en éléments simples
Posons le changement de variable: \(u = \sqrt{x+4} \Rightarrow x = u²-4 \Rightarrow dx = 2u.du\)
\(\begin{align*} I & = \int \frac{\sqrt{x+4}}{x}dx = \int \frac{u}{u²-4}2u.du \\
& = \int \frac{2u²}{u²-4}du\\ \end{align*}\)
Procédons à une division polynômiale:
\(\begin{align*} & 2u² & +0u & +0 & \lvert & \underline{u²-4 } \\
& 2u² & +0u & -8 & \lvert & 2 \\
& 0u² & +0u & +8 & \lvert \end{align*}\)
\(\begin{align*}2u² & = 2(u²-4)+8 \\
\frac{2u²}{u²-4} & = 2+\frac{8}{u²-4}\end{align*}\)
\(\begin{align*} I & = \int \frac{2u²}{u²-4}du = \int(2+\frac{8}{u²-4})du \\
& =2\int du+8 \int \frac{1}{u²-4}du \\
& =2\int du+8 \int \frac{1}{(u-2)(u+2)}du \\\end{align*}\)
Procédons à une décomposition en éléments simples:
\(\begin{align*} & \frac{1}{(u-2)(u+2)} = \frac{a}{u-2} + \frac{b}{u+2} \\ \\
& \begin{cases} \times(u-2) \text{ et }u=2 \Rightarrow & a = 1/4 \\ \times(u+2) \text{ et }u=- 2 \Rightarrow & b=-1/4 \end{cases} \\ \\
I & = 2\int du+8 \int \frac{1}{(u-2)(u+2)}du \\
& = 2\int du+8 \bigg[ \frac{1}{4} \int \frac{1}{u-2}du - \frac{1}{4} \int \frac{1}{u+2}du \bigg] \\
& = 2\int du+ 2 \int \frac{1}{u-2}du - 2 \int \frac{1}{u+2}du \\
& = 2u+ 2 ln \lvert u-2 \rvert - 2 ln \lvert u+2 \rvert +C \\
& = 2u+ 2 ln \lvert \frac{u-2}{u+2} \rvert +C \\
& = 2\sqrt{x+4}+ 2 ln \lvert \frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} \rvert +C \\
& = 2\sqrt{x+4}+ 2 ln \bigg( \frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} \bigg) +C \end{align*}\)
\[\boxed {\begin{align*} I &= 2\sqrt{x+4}+ 2 ln \bigg( \frac{\sqrt{x+4}-2}{\sqrt{x+4}+2} \bigg) +C(\in \mathbb R)\end{align*}}\]
Exercice 88
\(\begin{align*} \int \frac{\sqrt{x²+4}}{x²} dx\end{align*}\) |
Connaissances:
- Changement de variable trigonométrique
- Trigonométrie
- Trigonométrie notation anglosaxonne
Posons le changement de variable: \(x=2.tant \Rightarrow dx = 2.sec²t.dt \)
\(\begin{align*} I & = \int \frac{\sqrt{x²+4}}{x²}dx \\
& = \int \frac{\sqrt{4.tan²t+4}}{4tan²t}2.sec²t.dt \\
& = \int \frac{2\sqrt{tan²t+1}}{4.tan²t}.2.sec²t.dt \\
& = \int \frac{\sqrt{sec²t}}{tan²t}.sec²t.dt \\
& = \int \frac{sect}{tan²t}.sec²t.dt \\
& = \int \frac{sec^3t}{tan²t}dt = \int \frac{1}{cos^3t} \times \frac{cos²t}{sin²t}dt \\
& = \int \frac{1}{cost . sin²t}dt = \int \frac{cos²t + sin²t}{cost . sin²t}dt \\
& = \int \frac{cos²t }{cost . sin²t}dt + \int \frac{ sin²t}{cost . sin²t}dt \\
& = \int \frac{cost}{sin²t}dt + \int \frac{1}{cost}dt \\
& = \int sect.dt + \int \frac{1}{u²}du \\
& = ln \lvert sec t + tant \rvert-\frac{1}{u} +C \\
& = ln \lvert sec t + tant \rvert- \frac{1}{sint}+C \end{align*}\)
Dessinons un triangle rectange:
\(tant = x/2 \Rightarrow \begin{cases} t = angle \\ opposé = x \\ adjascent = 2 \end{cases} \Rightarrow \begin{cases} hypothénuse = \sqrt{x²+4} \\ tant = x/2 \\ sint = \frac{x}{\sqrt{x²+4}} \\ cost = \frac{2}{\sqrt{x²+4}}\end{cases} \)
\(\begin{align*} I & = ln \lvert sec t + tant \rvert- \frac{1}{sint}+C \\
& = ln \lvert \frac{\sqrt{x²+4}}{2} + \frac{x}{2} \rvert- \frac{\sqrt{x²+4}}{x}+C \\
& = ln \bigg( \frac{\sqrt{x²+4}}{2} + \frac{x}{2} \bigg)- \frac{\sqrt{x²+4}}{x}+C \\
& = ln\frac{1}{2} + ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C \\
& = ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C_2 \end{align*}\)
\[\boxed {\begin{align*} I &= ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C(\in \mathbb R)\end{align*}}\]
Exercice 87
\(\begin{align*} \int (ln(x))²dx \end{align*}\) |
Connaissances:
- Changement de variable
- suivi d'une IPP multiple
- Intégration par parties directement
Changement de variable:
Posonx \(u=lnx \Rightarrow \begin{cases}x=e^u \\ du=1/x.dx \Rightarrow dx = x.du=e^udu \end{cases}\)
\(\begin{align*} I & = \int (ln(x))²dx = \int u².e^u.du\end{align*}\)
| \(D\) | \(I\) | ||
| \(+\) | \(u²\) | \(e^u\) | |
| \(\searrow\) | |||
| \(-\) | \(2u\) | \(e^u\) | |
| \(\searrow\) | |||
| \(+\) | \(2\) | \(e^u\) | |
| \(\searrow\) | |||
| \(-\) | \(0\) | \(\rightarrow\) | \(e^u\) |
\(\begin{align*} I & = \int (ln(x))²dx = \int u².e^u.du \\
& = e^u(u²-2u+2)+C \\
& = e^{lnx}((lnx)²-2lnx+2)+C \\
& = x(ln²x-2lnx+2) +C \end{align*}\)
Intégration par parties directe:
| \(D\) | \(I\) | ||
| \(+\) | \(ln²x\) | \(1\) | |
| \(\searrow\) | |||
| \(-\) | \(2\frac{lnx}{x}\) | \(\rightarrow\) | \(x\) |
\(\begin{align*} I & = \int (ln(x))²dx \\
& = x.ln²x-\int 2\frac{lnx}{x} \times x.dx \\
& = x.ln²x-2\int lnx .dx\\
\end{align*}\)
Faisons une 2ème intégration par partie:
| \(D\) | \(I\) | ||
| \(+\) | \(lnx\) | \(1\) | |
| \(\searrow\) | |||
| \(-\) | \(\frac{1}{x}\) | \(\rightarrow\) | \(x\) |
\(\begin{align*} I & = \int (ln(x))²dx \\
& = x.ln²x-2(xlnx-\int 1 \times dx ) \\
& = x.ln²x-2(xlnx-x) +C \\
& = x. ln²x-2xlnx+2x) +C \\
& = x(ln²x-2lnx+2) +C
\end{align*}\)
\[\boxed {\begin{align*} I &= x(ln²x-2lnx+2) +C(\in \mathbb R)\end{align*}}\]
Exercice 86
\(\begin{align*} \int \frac{arctan(x)}{1+x²}dx \end{align*}\) |
Connaissances:
- remarquer une forme \(u'\) et \(u\)
On remarque pour ce calcul que \((arctan(x))' = \frac{1}{1+x²}\)
Et donc l'intégrande est de la forme \(u'.u\).
Amenons la a la forme \(2u'.u\) dont on connait la primitive: \(u²\)
\(\begin{align*} I & = \int \frac{arctan(x)}{1+x²}dx \\
& = \frac{1}{2} \int 2 \times \frac{1}{1+x²} \times arctan(x)dx \\\
& = \frac{1}{2}arctan²(x) +C \end{align*}\)
\[\boxed {\begin{align*} I & = \frac{1}{2}arctan²(x) +C(\in \mathbb R)\end{align*}}\]