In 7.12 Other sampling distribution
For example, if we know that the errors are being generated by the unavoidable and uncontrollable rotation of some small object, in such a way that when it is at angle $\theta$, e = $\alpha cos\theta$ but the actual angle is unknown, a little analysis shows that the prior probability assignment $p(e\mid I) = (\pi \sqrt{\alpha^2-e^2})^{-1} \;, e^2 < \alpha^2$ correctly describes our state of knowledge about the error.
I want to verify the $p(e\mid I) =(\pi \sqrt{\alpha^2-e^2})^{-1} $ via $\int_{0^{+}}^{\pi^{}-}p(e\mid I) d\theta = 1 \tag{1}$ (1) is wrong.
The right integral is:
$$\int_{-e^{+}}^{+e^{-}}p(e\mid I)de = \frac{1}{\alpha\pi}\int_{\pi^{-}}^{0^{+}}\frac{1}{|sin\theta|}d\alpha cos\theta =\frac{1}{\pi}\int_{0^{+}}^{\pi^{-}}\frac{sin\theta}{|sin\theta|}d\theta = 1$$ Any help will be appreciated, thanks.
$$\int_{-e^{+}}^{+e^{-}}p(e\mid I)de = \frac{1}{\alpha\pi}\int_{\pi^{-}}^{0^{+}}\frac{1}{|sin\theta|}d\alpha cos\theta =\frac{1}{\pi}\int_{0^{+}}^{\pi^{-}}\frac{sin\theta}{|sin\theta|}d\theta = 1$$