According to the properties of the Dirac Delta function $\int\delta(t)\cos(t)dt = \cos(0) = 1$.
I tried proving this using integration by parts but got stumped:
$
\require{cancel}
\int_{-\infty}^{\infty}\delta(t)\cos(t)dt = H(t)\cos(t)\big|_{-\infty}^{\infty} + \int_{-\infty}^{\infty}H(t)\sin(t)dt = \cancel{0 \cdot \cos(t)|_{-\infty}^{0}} + 1 \cdot \cos(t)\big|_{0}^{\infty} + \cancel{ \int_{-\infty}^{0}0 \cdot \sin(t)dt } + \int_{0}^{\infty}1 \cdot \sin(t)dt = \cos(t)\big|_0^{\infty} – \cos(t)|_0^{\infty} = 0
$
$H$ is the step function.
Best Answer
The integral of $H(t) \sin(t)$ over an infinite domain is obviously not convergent; it oscillates between $-1$ and $1$ as the upper limit grows.
Instead, you need to e.g. choose some $a>0$ and compute $$ \begin{align} \int_{-a}^a \delta(t) \cos(t) \ {\rm d}t &= \left[H(t) \cos(t) \right]_{-a}^a + \int_{-a}^a H(t) \sin(t) \ {\rm d}t \\ &= \cos(a) + \int_0^a \sin(t) \ {\rm d}t \\ &= \cos(a) - \left[\cos(a) - \cos(0)\right] \\ &= 1 \end{align} $$ Now you can take the limit $a \to \infty$ safely. (The oscillating $\cos(a)$ terms have cancelled.)
Edit: In some sense your calculation would still have "given you the right answer" if you had been slightly more careful with your evaluation of $\left[H(t) \cos(t) \right]_{-\infty}^\infty$. For some reason you split it up into $[0,\infty)$ and $(-\infty,0)$ and then in one of these you chose $H(0)=0$ and $H(0)=1$. This isn't at all consistent! The more consistent thing would be to directly evaluate $H(\infty) = 1$ and $H(-\infty) = 0$ and write "$\cos(\infty)$"... but obviously this now makes clear why you should be careful with $\infty$s! Had you ploughed ahead and written $\cos(\infty) - \left[\cos(t) \right]_0^\infty = \cos(0) = 1$ you would have at least got the right answer however. Then you might realize that you should go back and replace $\infty \to a$ throughout to be more precise.