\(\begin{align*} \int_0^{\pi/2} \frac{sin^3x}{cos^3x+sin^3x}dx\end{align*}\) |
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 \(\begin{cases} x = 0 \Rightarrow t=\pi/2 \\ x = \pi/2 \Rightarrow t=0 \\ dx=-dt \end{cases}\)
\(\begin{align*} I & = \int_0^{\pi/2} \frac{sin^3x}{sin^3x+ cos^3x}dx \\
& = \int_{\pi/2}^0 \frac{cos^3t}{sin^3t+cos^3t}.(-dt) \\
& = \int_0^{\pi/2} \frac{cos^3t}{sin^3t+cos^3t}.dt \\\end{align*}\)
\(\begin{align*}2I & = I + I \\
& = \int_0^{\pi/2} \frac{sin^3x}{sin^3x+ cos^3x}dx + \int_0^{\pi/2} \frac{cos^3t}{sin^3t+cos^3t}.dt \\
& = \int_{\pi/2}^0 dt = \pi/2 \\ \\
\Rightarrow I & = \pi/4 \end{align*}\)
\[\boxed {\begin{align*} I &=\frac{\pi}{4} \end{align*}}\]