# Math, how do we know if a substitution is true?

For instance, in calculus we often do u-substitutions. Quite often, we do trignometric substitutions to solve integrals. For instance, if we have the following relation

$y=\sqrt{1-x^2}$

And we substitute $x = \sin u$, for $x \in [-1, 1]$; how do we know that our substitution is correct? If I graph $\sin u$ and $x$, it is clear these functions are very different.

How is there any relation at all? Why are we allowed to do substitutions like this?

• But you also need to care about $dx$ and $du$, not just $x$ and $u$ across such intervals. – Xoque55 Jan 9 '15 at 2:44
• In any part of mathematics, you can always invent a new variable, and substitute some function of it for some old variable, so long as you are consistent and make the same substitution everywhere in the problem. – Gerry Myerson Jan 9 '15 at 2:53

In the case of $u$-substitution you are setting $\sin(u)$ equal to $x$, forcing it to be true. The graph of $\sin(u)$ and $x$ look different because you aren't really graphing $\sin(u)$, you are graphing $\sin(x)$ instead (It is the $x$-axis). $\sin(x)=x$ is clearly a false identity. If you want to graph $\sin(u)$ on an $x$-axis, you can substitute '$u$' with $\arcsin(x)$, and then clearly the graph of $\sin(\arcsin(x))$ is just $x$ (For $[-1,1]$ at least). In this case, by defining $u$ to follow the relationship of $\sin(u)=x$, you can also substitute the $dx$ and make an integral easier to integrate. As for "how do we know that our substitution is correct?", the equal sign in this case does not signify "Are these things equal" as in solving systems of equations, but rather the equal sign signifies "I am defining $u$ such that this equation is true". In programming you can differentiate between the two types of equality with := (definition) and == (asking if they are equal).

Suppose we need to calculate the Riemann integral $$I=\int_{a}^{b}f(x)dx$$ where $f\in{\cal{R}}[a,b]$, id est, Riemann integrable on $[a,b].$
and there is a function $\phi(t)$ which is well defined on $[\alpha,\beta]$ such that:
1). $\phi([\alpha,\beta])=[a,b]$
2). $\phi(t)$ is always monotonic for all $t\in[\alpha,\beta]$
3). $\phi'(t)\in\cal{R}[\alpha,\beta]$
Hence we can substitute $x$ with $\phi(t)$ and the integral can be calculated as follows:
$$I=\int_{a}^{b}f(x)dx=\int_{\alpha}^{\beta}(f\circ\phi)(t)\phi'(t)dt$$ Only when three two requisites are met can we substitute $x$ with $\phi(t)$.The first two requisites make sure that $\phi:t\rightarrow x$ is one-on-one mapping, and the third guarantees the Riemann integrability of the new function $(f\circ\phi)(t)\phi'(t)$.

• You forgot $\phi$ differentiable, I think. – GPerez Jan 9 '15 at 4:20
• @GPerez Oh I'm sorry, I'll add it in my edit. Thanks for reminding – Vim Jan 9 '15 at 5:54

In your example we also have $u = \arcsin(x)$ on this domain. Really what we are saying by the substitution is that there is a way to get the graph of $y = x$ and squash or stretch the $x$ axis in a smooth way until it looks like the graph of $\sin$. But of course it's not $\sin(x)$ anymore, it's $\sin$ of something else, where the variable we now have is some transformed version of $x$. But I said what it was in my first sentence, it's $\sin(u)$ where $u = \arcsin(x)$.

How does this apply to calculus? The point is, if we want to integrate something by hand (i.e. analytically), it's convenient to have the integrand in a form we can recognise and maybe apply known rules to, so we make the change of variables. It's not strictly necessary to do so in order to solve the integral, but if we are careful we can do it and still get the right answer. Here, 'careful' means making sure to also transform how we measure the length element - originally, we wanted to use the measure of length $\mathrm{d}x$, but since we squashed the $x$ axis this is no longer relevant, and we need to use the length element $\mathrm{d}u$, which is relatable to $\mathrm{d}x$ by the smooth squashing we did.

Does this make sense? There is never really a claim that $x$ and $\sin{u}$ 'look like' the same graph.

In general, you're allowed to do substitutions whenever you're able to undo them. Precisely speaking, the substitution $$y = g(x)$$

has to be such that $g$ is a bijective function. What this means is that:

1. Two values $x_1\neq x_2$ cannot be mapped to the same value, i.e. $f(x_1) \neq f(x_2)$.
2. Every $y$ has to be the image under $f$ of a certain $x$.

Now for (definite) integrals, the same has to hold, but you need to have the additional hypothesis that $f$ is differentiable. I'll note that this is more of a sketch than anything, because we'd need to discuss the relative domains and codomains where they apply. I've skipped it to give a more concise idea.

Lets look at why your substitution is correct then (and in this case we will be obligated to consider the domain). Suppose your integral is $$\int_{-1}^1\sqrt{1-y^2} \,dy$$

Now, you say that $$y = \sin x\,, x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right)$$ I've chosen the domain where $x$ lies so that we do in fact get every $y$ in $[-1,1)$ (integration tip: it doesn't matter if we integrate in $[a,b]$ or $(a,b)$), and $\sin x$ does not repeat values here (you can verify these affirmations). Additionally, $\sin$ is differentiable, so here, the substitution is valid.

Actually, proving that we can do this does take some effort, so it's not as if we're dealing with some very evident fact. There are theorems to prove in the process.

One way of thinking about it is that the substitution is really just there to help us with the algebra. For example, when we find $\int \sqrt{1-x^2} dx$, we prove, using only algebra and trigonometry, that

$$\sqrt{1-x^2} = \frac{\cos(\arcsin(x))^2}{\sqrt{1-x^2}}.$$

We prove this in particular because $\frac{\cos(\arcsin(x))^2}{\sqrt{1-x^2}}$ can be written as $g'(h(x)) h'(x)$, where $g'(y)=\cos(y)^2$ and $h(x)=\arcsin(x)$. This lets us use the chain rule: we get $f(x)=\frac{d}{dx} g(h(x))$. Having written $f$ as the derivative of something, we can now use the FTC to say that $\int f(x) dx = g(h(x)) + C$. This reduces the problem to finding $g$, which is easier.

The notation $u=\arcsin(x),du=\frac{dx}{\sqrt{1-x^2}}$, etc. is really just a convenient shorthand for reasoning like the above.

• But how is it so that such a statement like $x=\sin u$ is valid? This seems like magic to me. – Jason Jan 9 '15 at 2:54
• @Jason Nope, that is only valid if we can show that it is bijective on the domain of $x$. Then we can have $h(x)=\sin^{-1}(x)$. – user1537366 Jan 9 '15 at 3:08

Jason, when we make a substitution such as $u = sin(x)$, we are not really comparing two things and stating their equality, we are declaring that $u$ is $\textbf{another name}$ for $sin(x)$. The symbol $u$ has no inherent meaning, we are defining it to be a representation of the function $sin(x)$. It is just a symbol.