bonjour comment faire la partie 1 .merci d avance

Bonjour ;

1)

[tex]\alpha = 2\pi ft \Rightarrow d\alpha = 2\pi f dt .[/tex]

[tex]\textit{Pour } t = 0 \textit{ on a : } \alpha = 0 \textit{ et pour } t = T = \dfrac{1}{f} \textit{ on a : } \alpha = 2\pi ; [/tex]

[tex]\textit{donc : } V_{eff} = \sqrt{\dfrac{1}{T}\int_{t=0}^{t=T}u^2(t)dt} = \sqrt{f\int_{\alpha=0}^{\alpha=2\pi}u^2(\alpha)\dfrac{d\alpha}{2\pi f}} \\\\\\ = \sqrt{\dfrac{1}{2\pi}\int_{\alpha=0}^{\alpha=2\pi}u^2(\alpha)d\alpha}}= \sqrt{\dfrac{1}{2\pi}\int_{\alpha=0}^{\alpha=2\pi}u_0^2cos^2(\alpha)d\alpha}}\\\\\\ = \sqrt{\dfrac{u_0^2}{2\pi}\int_{\alpha=0}^{\alpha=2\pi}cos^2(\alpha)d\alpha}} .[/tex]

2)

[tex]\textit{On a : } cos^2(\alpha) = \dfrac{1 + cos(\alpha)}{2} \\\\ \Rightarrow \int_{\alpha=0}^{\alpha=2\pi}cos^2(\alpha)d\alpha = \int_{\alpha=0}^{\alpha=2\pi}\dfrac{1+cos(2\alpha)}{2}d\alpha \\\\ = \dfrac{1}{2}\int_{\alpha=0}^{\alpha=2\pi}(1+cos(2\alpha))d\alpha = \dfrac{1}{2} \left [\alpha+\dfrac{sin(2\alpha)}{2}\right ]_0^{2\pi} = \dfrac{1}{2} (2\pi) = \pi .[/tex]

3)

[tex]V_{eff} = \sqrt{\dfrac{u_0^2}{2\pi}\pi} = \sqrt{\dfrac{u_0^2}{2}} =\dfrac{u_0}{ \sqrt{2} } = \dfrac{ \sqrt{2} }{2}u_0 .[/tex]