Đặt: $\ln x=t \Rightarrow x=e^t \Rightarrow dx=e^tdt$
Đổi cận: $x=1 \Rightarrow t=0$
$x=e^{\pi} \Rightarrow t=\pi$
Ta có:
$\int\limits_1^{e^{\pi}}\cos(\ln x)dx$
$=\int\limits_0^{\pi}\cos te^tdt$
$=\int\limits_0^{\pi}\cos td(e^t)$
$=\cos te^t\left|\begin{array}{l}\pi\\0\end{array}\right.-\int\limits_0^{\pi}e^td(\cos t)$
$=-e^{\pi}-1+\int\limits_0^{\pi}\sin te^tdt$
$=-e^{\pi}-1+\int\limits_0^{\pi}\sin td(e^t)$
$=-e^{\pi}-1+\sin te^t\left|\begin{array}{l}\pi\\0\end{array}\right.-\int\limits_0^{\pi}e^td(\sin t)$
$=-e^{\pi}-1-\int\limits_0^{\pi}\cos te^tdt$
$\Rightarrow \int\limits_0^{\pi}\cos te^tdt=\dfrac{-e^{\pi}-1}{2}$