For a random vector $\mathbf{X} = [x_1,x_2,...,x_n] \in \mathbb{R}^n$ uniformly distributed on the surface of the unit sphere, the PDF is the inverse of the surface $$f_\mathbf{X}(\mathbf{x}) = 2\pi^{-n/2}\Gamma(n/2) \delta(\left\lVert \mathbf{x}\right\rVert = 1) \tag{1}$$ where $\delta$ is the delta function.
I am interested in $$f_{X_1,...,X_{n-1}}(x_1,...,x_{n-1}) = \int_{-1}^1 f_\mathbf{X}(\mathbf{x}) dx_n \\= 2\pi^{-n/2}\Gamma(n/2) \int_{-1}^1 \delta(x_n^2 = 1 - x_1^2-...-x_{n-1}^2) dx_n = 2\pi^{-n/2}\Gamma(n/2) \times 2$$ because there is 2 points $x_n^2 = \pm\sqrt{1 - x_1^2-...-x_{n-1}^2}$.
I am not sure this is correct because if I continue like that, the marginal distribution $f_{X_1}(x_1)$ will be something does not depend on $x_1$ which is not true.
Thanks for hints.
One can think about this problem very geometrically: Let $F(r)$ be defined as the surface area of the boundary of a $n$ dimensional ball of radius $r$. Then the pdf of $X_1$ is $f_1(x)= \frac{F(r)}{\int_{-1}^{1} F(r) dx}$ where $r=\sqrt{1-x^2}$.