How to prove the Inverse Substitution Rule used in integration?

Question

Integration by substitution is a handy procedure used for solving integrals. Proofs of its correctness are readily available. Almost all the proofs, and much of the math.SE discussion, end up stating "From this, we show the substitution rule is correct. Don't confuse the substitution rule itself as the proof - it's just a mnemonic."

Yet, the "inverse substitution" (used e.g. to solve $\int \sqrt{1-x^2}$) is generally not proven, and just assumed from using the substition rule in reverse. Stewart omits the proof. Even Spivak (Ch.19) omits the proof, and just uses it as a technique!

This is especially troubling as the proofs I've seen for the substitution rule most definitely do not justify the inverse substitution rule.

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Update

Ninad Munshi and peek-a-boo comment that the inverse substitution rule doesn't need a proof, because it's a special case of the standard substitution rule. I'm having trouble seeing that. The standard substitution rule is if $p' = f$, then $$\int (f \circ u) \cdot u' = p \circ u,$$ a direct consequence of the chain rule $[p \circ u]' = (p' \circ u) \cdot u'$.

But I don't see how to make the inverse rule derive from that identity. If I write $u = x \mapsto \arcsin x$ (I'm deliberately using that notation to make it clear that $u$ is a function, not a number), and try $$\begin{align*}\int \sqrt{1 - x^2} \, dx &= \int \sqrt{1 - \sin^2 \arcsin x} \, dx\\ &= \int \cos \arcsin x \, dx\\ &= \int \cos \arcsin x \frac {\sqrt {1 - x^2}} {\sqrt {1 - x^2}} \, dx\\ &= \int (\cos \circ \arcsin) \cdot \arcsin' \cdot (x \mapsto \sqrt {1 - x^2}) \end{align*}$$ which is almost, but not quite, what I need to use the substitution rule to finish the problem.

So, an alternate way of phrasing the original question is: How can I finish the above to use the (already proven) substitution rule in the (inverse) manner (as required in this particular case)?

Answer 1

You need to re-read some of the things I wrote. I already wrote the formula \begin{align} \text{prim}(\phi) &= \text{prim}((\phi \circ \psi) \cdot \psi') \circ \psi^{-1}. \end{align} Apply this with $\phi(x)=\sqrt{1-x^2}$, and $\psi=\sin|_{[-\pi/2,\pi/2]}$. The choice of $\phi$ is obvious, while the choice of $\psi$ is what the entire subject of integration by trigonometric substitution is about, so you just have to recognize it.

Then, the function inside the brackets on the RHS is (for each $t\in [-\pi/2,\pi/2]$) \begin{align} [(\phi\circ\psi)\cdot \psi’](t)&=(\phi\circ\psi)(t)\cdot\psi’(t)\\ &=\sqrt{1-\sin^2t}\cdot\cos t\\ &=\cos^2t\\ &=\frac{1+\cos(2t)}{2}. \end{align} Be sure you know how to jusitfy the simplification of the square root. The primitive of this function is easy to compute, it is the function $f(t)=\frac{t}{2}+\frac{\sin(2t)}{4}$ (and I don’t care about the $+C$). Looking back at the formula for the primitive of $\phi$, we see that it is simply $f\circ \psi^{-1}$, i.e $f\circ \arcsin$, so (for each $x\in (-1,1)$), \begin{align} \text{prim}(\phi)(x)&=(f\circ\arcsin)(x)=\frac{\arcsin(x)}{2}+\frac{\sin(2\arcsin(x))}{4}. \end{align} I’ll let you simplify the trigonometry here (use the double angle formula).

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This should make sense because the usual presentation goes as follows:

Let $x=\sin t$, then $dx=\cos t\,dt$, and so \begin{align} \int\sqrt{1-x^2}\,dx&=\int\sqrt{1-\sin^2t}\cdot\cos t\,dt\\ &=\int \cos^2t\,dt\\ &=\int\frac{1+\cos(2t)}{2}\,dt\\ &=\frac{t}{2}+\frac{\sin(2t)}{4}\\ &=\frac{\arcsin(x)}{2}+\frac{\sin(2\arcsin(x))}{4} \end{align}