From textbook:
$$\int f(x)\,dx = \int f(g(t))g'(t)\,dt$$ This kind of substitution is called inverse substitution.
We can make this inverse substitution $x = a\sin \theta$ provided that it defines a one-to-one function. This can be accomplished by restricting $\theta$ to lie in the interval $[-\pi/2, \pi/2]$.
Image.
$\int f(x)\,dx$
$x = g(t) \longleftarrow$ Why does this need to be a one-to-one function?
$\frac{dx}{dt} = g'(t)$
$dx = g'(t)\,dt$
$\int f(g(t))g'(t)\,dt$
I would assume $g(t)$ would only need to be differentiable.
One way to think about it, is that the trig substitution is a transformation of the coordinate system. If the the transformation is not $1-1$, then you have some part of the $x-$axis (and the $xy-$ plane) that is folded over on itself.
A less abstract way of thinking about it is to consider:
$\int_a^b f(x) dx$ where $f(x)>0 \;\forall x\in (a,b)$
Clearly, $\int_a^b f(x) dx>0$
But if our substition is not $1-1$ there exists the possibility that $g^{-1}(b) = g^{-1}(a),$ in which case if we proceed with the substitution.
$x = g(t), dx = g'(t) dt\\ \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t))g'(t) dt = \int_{g^{-1}(a)}^{g^{-1}(a)} f(g(t))g'(t) dt = 0$
Creating a contradiction.