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Explicit more about intégration par changement de variables

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flyingscorpio@pinebookpro 2021-09-16 10:25:32 +02:00
parent 6ec9eead21
commit 3cdfc6787f
1 changed files with 10 additions and 6 deletions
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analyse/main.tex
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@ -258,15 +258,19 @@
\paragraph{Exemple} \paragraph{Exemple}
\begin{align*} \begin{align*}
\int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{,\quad on pose }u=e^x \int_0^{\ln{\sqrt{3}}} \frac{1}{e^x + e^{-x}}\,\mathrm{d}x\text{,\quad on pose }u&=e^x \\
u' &= e^x = \frac{\mathrm{d}u}{\mathrm{d}x} \\
\mathrm{d}x &= \frac{\mathrm{d}u}{e^x} = \frac{\mathrm{d}u}{u}
\end{align*} \end{align*}
Cela nous donne~:
\begin{align*} \begin{align*}
&= \int_{e^0}^{e^{\ln{\sqrt{3}}}} \frac{1}{u + \frac{1}{u}}\,\frac{\mathrm{d}u}{u} \\\\ \int_{e^0}^{e^{\ln{\sqrt{3}}}} \frac{1}{u + \frac{1}{u}}\,\frac{\mathrm{d}u}{u} \\
&= \int_1^{\sqrt{3}} \frac{1}{u^2 + 1}\,\mathrm{d}u &= \int_1^{\sqrt{3}} \frac{1}{u^2 + 1}\,\mathrm{d}u \\
= [\arctan{u}]_1^{\sqrt{3}} &= [\arctan{u}]_1^{\sqrt{3}} \\
= \arctan{\sqrt{3}} - \arctan{1} &= \arctan{\sqrt{3}} - \arctan{1} \\
= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} &= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \\
\end{align*} \end{align*}
\clearpage \clearpage