Justification for indefinite integration by substitution

Question

I know that if we want to compute a definite integral using substitution, we use $$\int_{a}^b f(\phi(t))\phi '(t)~dt = \int_{\phi(a)}^{\phi(b)}f(x)~dx\tag1$$ I know that this method can go from left to right or right to left.

If we want to compute an indefinite integral using substitution, we use $$\int f(t)~dt = \int g(\phi(t))\phi '(t)~dt = G(\phi(t)) + C, \tag{2}$$ where $G$ is an antiderivative of $g$ and $C$ is the constant of integration.

For example, for evaluating functions like $\dfrac{1}{\sqrt{4 - x^2}}$ we can make the substitution $x = 2\sin \theta$ and find $dx$ in terms of $d\theta$. Substituting its value in the integral, it becomes computable. However, pretending that $\dfrac{dx}{d\theta}$ is a fraction and substituting the value of $dx$ in the integral just because it "works" is not very rigorous. Is there a rigorous statement for computing indefinite integrals that can't be expressed in the aforementioned form using substitution, just like we have for definite integrals, or can they be computed using $(2)$ in a way that I can't figure out?

Answer 1

You ask a good question. In the example you mentioned, the first step is simply writing down an antiderivative in abstract form, which is

$$\tag 1 \int_0^x \frac{1}{\sqrt {4-t^2}}\,dt,\,\,|x|<2.$$

We know that $(1)$ is an antiderivative for the integrand by the FTC. Now consider the function $t=2\sin s.$ This function maps $(-\pi/2,\pi/2)$ to $(-2,2)$ nicely. By the first equation you wrote,

$$\tag 2 \int_0^x \frac{1}{\sqrt {4-t^2}}\,dt = \int_0^{\arcsin x/2} \frac{2\cos s}{\sqrt {4-(2\sin s)^2}}\,ds.$$

After we realize that the denominator on the right equals $2\cos s,$ we get $(1) = \arcsin x/2,$ and we're done.

This is the justification behind the scenes, although after a while you'll be cancelling $dx$'s with the best (worst) of them.

Answer 2

Define $y=f(x)$ and $x=s(t)$. Compute $\dfrac{dy}{dt}$ using chain rule: $$\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}$$ Now, to reverse the process, compute $$\int \frac{dy}{dx}\dfrac{dx}{dt}dt$$ which by our assumptions is $$y=f(x)$$ Now, you are given $\dfrac{dy}{dx}$ and you have defined $x$ as a function of $t$, so you can find $\dfrac{dx}{dt}$. Your problem is to find $y=f(x)$. That's the justification you are looking for. Introducing the new parameter $t$ is done just to simplify the integral, and all follows from the chain rule. You can try using your functions in the above equations.

NOTE: In the integral, you can hack the notation and say $$\int\dfrac{dy}{dx}\dfrac{dx}{dt}dt=\int\dfrac{dy}{dx}dx$$ Edit after the comment
Let me use your example here. You are given $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{4-x^2}}$ and you want to find $y=f(x)$, i.e the function whose derivative is this. You introduce $x=2\sin\theta$. So you know $\dfrac{dx}{d\theta}=2\cos\theta$. Take it from here.

Answer 3

$$\int_{a}^b f(\phi(t))\phi '(t)~dt =\int_{\phi(a)}^{\phi(b)}f(x)~dx\tag1$$

$$\int f(t)~dt = \int g(\phi(t))\phi '(t)~dt = G(\phi(t)) + C\tag2$$

pretending that $\dfrac{dx}{d\theta}$ is a fraction just because it "works" is not very rigorous. Is there a rigorous statement for computing indefinite integrals that can't be expressed in the aforementioned form using substitution, just like we have for definite integrals, or can they be computed using $(2)$ in a way that I can't figure out?

Your query sounds like the two theorems are not intrinsically the same, so let's specify them using matching notation, and in full:

1. If $g'$ is integrable on $[a,b]$ and $f$ is integrable, and has an antiderivative, on $g[a,b],$ then $$\int_a^bf\big(g(x)\big)\ g'(x)\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\,\mathrm{d}u.\tag{✓1}$$
2. If $g'$ has domain $D,$ $f$ has an antiderivative $F$ on $g[D]$ and $u=g(x),$ then $$\int f(g(x))\ g'(x)\,\mathrm dx=\int f(u)\,\mathrm du;$$ furthermore, on each interval $i$ of $D,$ $$\int f(g(x))\ g'(x)\,\mathrm dx=F(g(x))+C_i.\tag{✓2}$$

Sidestepping the informal notation, indefinite integration by subsitution ( $✓2$) is straightforward to rigorously derive.

Proof

By the chain rule, on $D,$ $$f(u)=F'(u)=F'(g(x))=f(g(x))g'(x)).$$ Then, noting that a function's indefinite integral is literally its family of antiderivatives, the first result of $(✓2)$ immediately follows.

On each interval $i$ of $D,$ $$\int f(u)\,\mathrm du=F(u)+C=F(g(x))+C_i,$$ and the second result of $(✓2)$ immediately follows.

$(✓1)$ follows from the Fundamental Theorem of Calculus.

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Example (with $g(x)=x^3):$ $$\int\frac{x^2}{x^6+1}\,\mathrm dx=\int\frac1{3\Big(\color{violet}{(x^3)}^2+1\Big)}\color{cyan}{(3x^2)}\,\mathrm dx,$$ so the final result is $$\frac13\arctan\color{violet}{(x^3)}+C$$ while the intermediate result is $$\int\frac1{3(u^2+1)}\,\mathrm du.$$

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Finally, a small disagreement with peek-a-boo's answer:

Often times you see the following statement instead: "Let $u = g(x).$ Then $$ \int f(g(x))\ g'(x)\,\mathrm dx = \int f(u) \,\mathrm du"\tag A$$

This statement strictly speaking isn't accurate because if we differentiate both sides we don't get the same result, because $(f \circ g)g'$ is certainly not equal to $f.$

Suppose we're asked to find $\displaystyle \int 2x e^{x^2}\,\mathrm dx.$ First set $u = x^2$ ....$$\int 2x e^{x^2}\,\mathrm dx = \int e^u \,\mathrm du."\tag B$$ This is the 'dubious equality' in the sense that these aren't strictly equal as functions. To see this, note that we can simply change the variable of integration from $u$ to $x$ on the RHS.

No: $u$ is a dummy (i.e., bound) variable in the integral $\displaystyle\int_3^7 e^u \,\mathrm du$ but not in an indefinite integral, like $(\mathrm B)$'s $\displaystyle\int e^u \,\mathrm du,$ as it does not disappear from the indefinite integral's result.

Similarly, in $(\mathrm A),$ $$\int f(u) \,\mathrm du\not\equiv \int f(x) \,\mathrm dx,\\\frac{\mathrm d}{\mathrm dx}\int f(u) \,\mathrm du\not\equiv f(x),$$ so differentiating both sides of $(\mathrm A)$ certainly does not return the generally false equation $\displaystyle f(g(x))\ g'(x)=f(x).$

As long as its differentiability and antidifferentiability conditions (see $(✓2)$) are satisfied, statement $(A)$ is perfectly accurate!