
\begin{enumerate}
\item $I_0=\dint_0^{\frac{\pi}{2}} \sin x \dx=\left[-\cos x \right]_0^{\frac{\pi}{2}}=\cos 0 - \left(-\cos\frac{\pi}{2}\right)=1$.

$J_0=\dint_0^{\frac{\pi}{2}} \cos x \dx=\left[\sin x \right]_0^{\frac{\pi}{2}}=\sin 0 - \sin\frac{\pi}{2}=1$.

\item

\begin{enumerate}
\item $F_n^\prime(x)=n\e^{-nx}\sin x - \e^{-nx}\cos x$.

Ainsi,
\begin{align*}
\int_0^{\frac{\pi}{2}} F_n^\prime(x)\dx & = n\int_0^{\frac{\pi}{2}} \e^{-nx}\sin x\dx - \e^{-nx}\cos x\dx\\[8pt]
\Aboxed{-\e^{-n\frac{\pi}{2}} & = nI_n-J_n}
\end{align*}

\item $G_n^\prime(x)=n\e^{-nx}\cos x + \e^{-nx}\sin x$.

Ainsi,

\begin{align*}
\int_0^{\frac{\pi}{2}} G_n^\prime(x)\dx & = n\int_0^{\frac{\pi}{2}}\e^{-nx}\cos x\dx + \int_0^{\frac{\pi}{2}}\e^{-nx}\sin x\dx\\
\Aboxed{1 & = nJ_n+I_n}
\end{align*}

\item On a :
\[
\left\lbrace
\begin{array}{cccccc}
I_n & + & nJ_n & = & 1 & (E_1)\\
-nI_n & + & J_n & = & \e^{-n\frac{\pi}{2}} & (E_2)
\end{array}
\right.
\]

En faisant $(E_1)-n(E_2)$, on a :

\[ (1+n^2)I_n = 1-n\e^{-n\frac{\pi}{2}} \]
D'où :
\[ \boxed{I_n = \frac{1-n\e^{-n\frac{\pi}{2}}}{1+n^2}} \]

De plus, en faisant $(E_2)+n(E_1)$, on a :
\[ (1+n^2)J_n=\e^{-n\frac{\pi}{2}}+n\;,\]
d'où :
\[ \boxed{J_n = \frac{n+e^{-n\frac{\pi}{2}}}{1+n^2}} \]
\end{enumerate}

\item $\lim\limits_{n \rightarrow +\infty}\left( -n\frac{\pi}{2}\e^{-n\frac{\pi}{2}} \right)=0$ donc $\lim\limits_{n \rightarrow +\infty} I_n=\lim\limits_{n \rightarrow +\infty}\frac{1}{1+n^2}=0$.

\medskip

De même, $\lim\limits_{n \rightarrow +\infty} J_n=\lim\limits_{n \rightarrow +\infty} \frac{n}{1+n^2}=0$.
\end{enumerate}
