\(\begin{align*} \int \frac{\sqrt{x²+4}}{x²} dx\end{align*}\) |
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:
\(tant = x/2 \Rightarrow \begin{cases} t = angle \\ opposé = x \\ adjascent = 2 \end{cases} \Rightarrow \begin{cases} hypothénuse = \sqrt{x²+4} \\ tant = x/2 \\ sint = \frac{x}{\sqrt{x²+4}} \\ cost = \frac{2}{\sqrt{x²+4}}\end{cases} \)
\(\begin{align*} I & = ln \lvert sec t + tant \rvert- \frac{1}{sint}+C \\
& = ln \lvert \frac{\sqrt{x²+4}}{2} + \frac{x}{2} \rvert- \frac{\sqrt{x²+4}}{x}+C \\
& = ln \bigg( \frac{\sqrt{x²+4}}{2} + \frac{x}{2} \bigg)- \frac{\sqrt{x²+4}}{x}+C \\
& = ln\frac{1}{2} + ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C \\
& = ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C_2 \end{align*}\)
\[\boxed {\begin{align*} I &= ln \bigg( \sqrt{x²+4} + x \bigg)- \frac{\sqrt{x²+4}}{x}+C(\in \mathbb R)\end{align*}}\]