For every natural number $n$, we consider the following integrals:

$$I_n = \int_0^{\pi} \mathrm{e}^{-nx} \sin(x) \mathrm{d}x, \quad J_n = \int_0^{\pi} \mathrm{e}^{-nx} \cos(x) \mathrm{d}x$$

\begin{enumerate}
  \item Calculate $I_0$.
  \item a. Justify that, for every natural number $n$, we have $I_n \geqslant 0$.\\
b. Show that, for every natural number $n$, we have $I_{n+1} - I_n \leqslant 0$.\\
c. Deduce from the two previous questions that the sequence $(I_n)$ converges.
  \item a. Show that, for every natural number $n$, we have:
$$I_n \leqslant \int_0^{\pi} \mathrm{e}^{-nx} \mathrm{d}x$$
b. Show that, for every natural number $n \geqslant 1$, we have:
$$\int_0^{\pi} \mathrm{e}^{-nx} \mathrm{d}x = \frac{1 - \mathrm{e}^{-n\pi}}{n}.$$
c. Deduce from the two previous questions the limit of the sequence $(I_n)$.
  \item a. By integrating by parts the integral $I_n$ in two different ways, establish the two following relations, for every natural number $n \geqslant 1$:
$$I_n = 1 + \mathrm{e}^{-n\pi} - nJ_n \quad \text{and} \quad I_n = \frac{1}{n}J_n$$
b. Deduce that, for every natural number $n \geqslant 1$, we have
$$I_n = \frac{1 + \mathrm{e}^{-n\pi}}{n^2 + 1}$$
  \item It is desired to obtain the rank $n$ from which the sequence $(I_n)$ becomes less than $0.1$. Copy and complete the fifth line of the Python script below with the appropriate command.
\begin{verbatim}
from math import *
def seuil() :
    n = 0
    I = 2
    ...
        n=n+1
        I =(1+exp(-n*pi))/(n*n+1)
return n
\end{verbatim}
\end{enumerate}