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The cross product of the two in-plane directions is $\mathbf{u}\times \mathbf{v}=(-5,1,2)$, with Euclidean norm $\sqrt{30}$. Either sign gives a unit normal vector:The plane equation is $\hat{\mathbf{n}}\cdot (\mathbf{r}-\mathbf{A})=0$, so the distance from a point to a plane gives

On the specified domain $\sin x\ne 0$, so multiplying by $\sin x$ is legitimate. Use the double-angle identity to findThe first factor gives $x=\pm \pi /2$. The second gives ${\sin }^{2}x=1/2$, yielding four more values in the interval. ThusEvery listed value has nonzero sine and satisfies the original equation. Dividing by $\cos x$ would incorrectly lose the two solutions $\pm \pi /2$.

For the indefinite integral, substitute $u=x^2$, so $x^{3}dx=udu/2$. An integration by parts then givesFor the definite integral, the original PDF's upper limit is $\sqrt{\pi /2}$, so the corresponding upper value of $u$ is $\pi /2$. ThereforeBoth requested integrals are thus evaluated with the same antiderivative.

Use completing the square to obtainThe locus is therefore a circle with centre $(1,0)$ and radius $2$. It meets the horizontal axis at $(-1,0)$ and $(3,0)$ and the vertical axis at $(0,\pm \sqrt{3})$; its highest and lowest points are $(1,\pm 2)$.