
\medskip

$S = \dint_0^{\frac{\pi}{4}} \sin(x)\e^{x}\dx$. Posons: $\begin{aligned}[t]
u(x) = \sin(x) & \qquad  u'(x) = \cos(x)\\
v'(x) = \e^x & \qquad v(x) = \e^x
\end{aligned}$

Alors,
\[
S = \big[ \sin(x)\e^x \big]_0^{\frac{\pi}{4}} - \underbrace{\dint_0^{\frac{\pi}{4}}\cos(x)\e^x\dx}_{=I}.
\]
Calculons alors $I$ en posant: $\begin{aligned}[t]
u(x) = \cos(x) & \qquad  u'(x) = -\sin(x)\\
v'(x) = \e^x & \qquad v(x) = \e^x
\end{aligned}$
\begin{align*}
I & = \big[ \cos(x)\e^x \big]_0^{\frac{\pi}{4}} - \dint_0^{\frac{\pi}{4}} -\sin(x)\e^x\dx\\
& = \big[ \cos(x)\e^x \big]_0^{\frac{\pi}{4}} + \dint_0^{\frac{\pi}{4}} \sin(x)\e^x\dx\\
& = \big[ \cos(x)\e^x \big]_0^{\frac{\pi}{4}} +S.
\end{align*}

Ainsi,
\begin{align*}
S & = \big[ \sin(x)\e^x \big]_0^{\frac{\pi}{4}} - \left( \big[ \cos(x)\e^x \big]_0^{\frac{\pi}{4}} +S \right)\\
S & = \big[ \sin(x)\e^x \big]_0^{\frac{\pi}{4}} - \big[ \cos(x)\e^x \big]_0^{\frac{\pi}{4}} - S \\
2S & = \big[ \sin(x)\e^x - \cos(x)\e^x \big]_0^{\frac{\pi}{4}}\\
S & = \dfrac{1}{2}\Big[ \big(\sin(x) - \cos(x)\big)\e^x \Big]_0^{\frac{\pi}{4}}\\
S & = \dfrac{1}{2}\Big( \underbrace{\big(\sin\frac{\pi}{4}-\cos\frac{\pi}{4}\big)}_{=0}\e^{\frac{\pi}{4}} - \underbrace{\big(\sin0-\cos0)}_{=-1}\e^0  \Big)\\
\Aboxed{S & = \dfrac{1}{2}}
\end{align*}
