Answer
$$\dfrac{1}{2}$$
Work Step by Step
Let us consider that $I=\int_{0}^{\infty} e^{-x} \cos x dx$
Plug in $a=e^{-x} \implies da=\cos x \ dx$
Integrate the integral by using limits.
$I=\lim\limits_{z \to \infty} \int_{0}^{z} e^{-x} \cos x \ dx \\=\lim\limits_{z \to \infty} [ e^{-x} \sin x ]_0^z-\lim\limits_{z \to \infty}\int_0^b -\sin x (-e^{-x} \ dx \\=0-\lim\limits_{z \to \infty}[ -e^{-x} \cos x ]_0^z- \lim\limits_{z \to \infty}\int_0^z e^{-x} \cos x \ dx\\=\dfrac{1}{2}(0+1)\\=\dfrac{1}{2}$
