Đáp án:
$\displaystyle\int\limits_0^{\pi} e^x\cos xdx =-\dfrac{e^{\pi}+1}{2}$
Giải thích các bước giải:
$I =\displaystyle\int\limits_0^{\pi} e^x\cos xdx$
Đặt $\begin{cases}u = \cos x\\dv = e^xdx\end{cases}\longrightarrow \begin{cases}du = -\sin xdx\\v = e^x\end{cases}$
Ta được:
$\quad I = e^x\cos x\Bigg|_0^{\pi} +\displaystyle\int\limits_0^{\pi} e^x\sin xdx$
$\to I = -e^{\pi} -1 + \displaystyle\int\limits_0^{\pi}e^x\sin xdx$
Đặt $\begin{cases}t = \sin x\\dz = e^xdx\end{cases}\longrightarrow \begin{cases}dt = \cos xdx\\z = e^x\end{cases}$
Ta được: $I = -e^{\pi} - 1 + e^x\sin x\Bigg|_0^{\pi} - \displaystyle\int\limits_0^{\pi} e^x\cos xdx$
$\to I = -e^{\pi} - 1- \displaystyle\int\limits_0^{\pi} e^x\cos xdx$
$\to \displaystyle\int\limits_0^{\pi} e^x\cos xdx = -e^{\pi} -1 - \displaystyle\int\limits_0^{\pi} e^x\cos xdx$
$\to 2\displaystyle\int\limits_0^{\pi} e^x\cos xdx = -e^{\pi} -1$
$\to \displaystyle\int\limits_0^{\pi} e^x\cos xdx =-\dfrac{e^{\pi}+1}{2}$
