Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$
The differential equation of the family of circles with fixed radius $r$ units and centre on the line $y=3$, is
The degree of the differential equation
$$ x\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{5 / 3}+7 \frac{d y}{d x}+y=0 $$
The curve that satisfies the differential equation $x y d y-\left(1+y^2\right) d x=0$ passes through $(1,0)$ and intersects the curve $x^2+3 y^2=3$ at an angle $\theta$. Then, $\frac{2 \theta}{\pi}=$
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