5. Transform the left side of the equation $(\sin x+\cos y)(\cos x-\sin y)=$ $=\sin x \cos x+\cos y \cos x-\sin x \sin y-\cos y \sin y=$ $=\cos (x+y)+\frac{1}{2}(\sin 2 x-\sin 2 y)=\cos (x+y)+\sin (x-y) \cos (x+y)$.

Substitute the obtained expression into the original equation

$$
\cos (x+y)+\sin (x-y) \cos (x+y)=1+\sin (x-y) \cos (x+y)
$$

From this, we get $\cos (x+y)=1$. Solving this equation, we find

$$
x+y=2 \pi k, k \in Z
$$

Therefore, the smallest positive value for $x+y$ is $2 \pi$.
Answer: $(x+y)_{\min }=2 \pi$.