Let's rewrite this as : $\displaystyle\int_0^{\infty} e^{-\tau}\left(1 + \frac{1}{2}\cos\left(400\pi (t-\tau)\right)\right)d\tau$
In exponential form this becomes : $\displaystyle\int_0^{\infty} e^{-\tau} + \frac14 e^{i 400\pi (t -\tau)-\tau}+ \frac14 e^{-i 400\pi (t -\tau)-\tau}d\tau$
$\displaystyle =\left[e^{-\tau} - \frac{e^{i 400\pi t -\tau(1+i400\pi)}}{4(1+i400\pi)}- \frac{e^{-i 400\pi t -\tau(1-i400\pi)}}{4(1-i400\pi)}\right]_0^{\infty}$ $\displaystyle =1+\frac{e^{i 400\pi t}}{4(1+i400\pi)}+\frac{e^{-i 400\pi t}}{4(1-i400\pi)}$
EDIT Let's rewrite this in standard trigonometric form :
$\displaystyle =1+\frac{(1-i400\pi)e^{i 400\pi t}+ (1+i400\pi)e^{-i 400\pi t}}{4(1+(400\pi)^2)}$
$\displaystyle =1+\frac{\cos(400\pi t) + 400\pi\sin(400\pi t)}{2(1+(400\pi)^2)}$