Determine the Radon transform of $\chi_{(-1/2,1/2)}$

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I need to find the radon transform of the following function. But I got stuck in finding this integral. Assume that $\delta$ is the Dirac delta distribution. Let $\chi = \chi_{(-1/2,1/2)}$ be given by

$$\chi(t) = \begin{cases} 1 & |t|< 1/2 \\ 0 & \text{otherwise}\end{cases}$$

What then does the following equal?

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \chi(x)\chi(y)\delta(s-x\cos\theta-y\sin\theta)\,dx\,dy$$

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What stand in front of $\delta$ in the integrand is unimportant. This is because $\delta (a - b x) = \delta(a/b -x) /|b|$ (provided, of course, that $b\neq 0$). Applying this twice will do the job. P.S. The formula above is proved by the change of variables $x\mapsto x/b.$