Suppose I have an integral to evaluate with respect to the column vector of $n$ variables, $x=[x_1,x_2,...x_n]^T$;
$\int f(x)dx$
And I make a change of variables to a new column vector, $y$, of length $n$, by setting $y=Ax$ where $A$ is an $(n\times n)$ orthogonal matrix. Is it true that the jacobian determinant of this transformation is $1$, if so why? I.e. does
$\int f(x)dx=\int f(y)dy$
Since $A$ is orthogonal, $\det(A) = \pm 1$. Why? By definition, $A A^T = I$, and taking determinant of both sides gives $$\det(A)^2 = \det(A)\det(A^T) = \det(A A^T) = \det(I) = 1. $$
Now as the change of coordinates $y = \phi(x) = Ax$ is linear, its derivative $(D\phi)(x)$ equals $A$. By change of variables, $$\int f(y) dy = \int f(Ax)\, |\det (D\phi)(x)| dx = \int f(Ax)\, |\det A| dx = \int f(Ax) dx. $$