= Solution
Put $u=x+y$. Then $u'=1+\cos u=2\cos^2(u/2)$. Along the branch through $u(0)=0$, the <chain rule> yields $[\tan(u/2)]'=1$, hence $\tan(u/2)=x$. Choose the continuous branch through zero to obtain
$$
\boxed{y(x)=2\arctan x-x,\qquad x\in\mathbb R.}
$$
Indeed $y'=(1-x^2)/(1+x^2)=\cos(2\arctan x)=\cos(x+y)$, and $y(0)=0$. The right-hand side has a bounded derivative with respect to $y$, so the usual <uniqueness theorem for ordinary differential equations> rules out another solution through the same point.
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