Question: Let $W$ be a Brownian motion with $W(0) = 0$. Determine $E[\cos{W(t)}+\sin{W(t)}]$.
I let
$Y=\cos{W(t)}$
$dY=-\sin{W(t)}dW(t)-\frac{1}{2}\cos{W(t)}(dW(t))^2=-\sin{W(t)}dW(t)-\frac{1}{2}\cos{W(t)}dt$
and similarly,
$Z=\sin{W(t)}$
$dZ=\cos{W(t)}dW(t)-\frac{1}{2}\sin{W(t)}dt$
But I am not sure how to continue now.
$E[Y(t)]=\cos{W(0)}-\int^{t}_{0}\frac{1}{2}\cos{W(s)}ds$
and
$E[Z(t)]=\sin{W(0)}-\int^{t}_{0}\frac{1}{2}\sin{W(s)}ds$, am I right?
What we need to compute $E[\cos{W(t)}+\sin{W(t)}]$ is the fact that $$ W(t) \sim \mathcal{N}(0,t). $$ By definition, it holds $$ E[\cos{W(t)}+\sin{W(t)}]= \int_\mathbb{R} (\cos x+\sin x)\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} dx, $$ where $\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}$ is the probability density function of $W(t)$. Since we have characteristic function of normal distributions $$ \int_\mathbb{R} e^{i\xi x}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} dx = \int_\mathbb{R} (\cos \xi x +i\sin \xi x)\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} dx =e^{-\frac{t\xi^2}{2}}, \quad\forall \xi\in\mathbb{R}, $$ by comparing the real and imaginary part on both sides, it follows that $$ E[\cos{W(t)}+\sin{W(t)}] = e^{-\frac{t}{2}}+0=e^{-\frac{t}{2}}. $$