Why does Trig Substitution need the substitution to be one-to-one?

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.

• I suppose the extreme example would be the substitution $x=0$. – Rene Schipperus Oct 14 '16 at 22:36
• Please learn MathJax and type out all images. This helps others find this post. – Em. Oct 14 '16 at 22:39
• @Max Oh I didn't think about that from a search perspective, I will next time. – A_for_ Abacus Oct 14 '16 at 22:44
• @ReneSchipperus Heh, that's real funny XD – Simply Beautiful Art Oct 14 '16 at 23:01
• @SimpleArt Actually it was meant seriously, see my comment to the answer. – Rene Schipperus Oct 14 '16 at 23:06

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.

$x = g(t), dx = g'(t) dt\\ \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t))g'(t) dt = \int_{g^{-1}(a)}^{g^{-1}(a)} f(g(t))g'(t) dt = 0$

• Of course if $g^{\prime}(c)\neq 0$ then $g$ will be injective in a nbh of $c$ so can be used to find the indefinite integral. – Rene Schipperus Oct 14 '16 at 22:51
• How did you turn $g^{-1}(b)$ into $g^{-1}(a)$ in the integral? – A_for_ Abacus Oct 14 '16 at 23:09
• @A_for_Abacus if it is not $1$ to $1$, then there may exist some $a$ such that $g^{-1}(b)=g^{-1}(a)$. – Simply Beautiful Art Oct 14 '16 at 23:12