The substitution rule for integrals

Statement

Let \( \alpha, \beta \in \mathbb{R} \), let \( \varphi : [\alpha,\beta] \to \mathbb{R} \) be continuously differentiable, and let \( I \subseteq \mathbb{R} \) be an interval with \( \varphi([\alpha,\beta]) \subseteq I \) on which \( f : I \to \mathbb{R} \) is continuous. Then \( s \mapsto f(\varphi(s))\,\varphi'(s) \) is continuous on \( [\alpha,\beta] \), \( f \) is Riemann integrable between \( \varphi(\alpha) \) and \( \varphi(\beta) \), and \[ \int_{\alpha}^{\beta} f\bigl(\varphi(s)\bigr)\,\varphi'(s)\,\mathrm{d}s \;=\; \int_{\varphi(\alpha)}^{\varphi(\beta)} f(x)\,\mathrm{d}x. \] Neither injectivity nor monotonicity of \( \varphi \) is required: the two limits carry the orientation, and the oriented convention \( \int_p^q = -\int_q^p \) is in force. If \( \varphi \) is in addition strictly monotone, with \( \varphi(\alpha) = a \) and \( \varphi(\beta) = b \), the identity read from right to left evaluates \( \int_a^b f \) by inverse substitution, and it may then be restated in unsigned form as \( \int_{\varphi([\alpha,\beta])} f(x)\,\mathrm{d}x = \int_{\alpha}^{\beta} f(\varphi(s))\,\lvert \varphi'(s)\rvert\,\mathrm{d}s \).

Why it matters

Substitution is the chain rule read backwards, and with integration by parts it is one of only two general-purpose techniques for evaluating integrals in closed form. Every named method downstream of it — trigonometric substitution, the Weierstrass half-angle substitution \( t = \tan(s/2) \), hyperbolic substitution, partial-fraction shifts, separation of variables in a first-order ordinary differential equation, the \( u = g(x) \) step in every table of integrals — is this one theorem applied with a particular \( \varphi \). It is also what makes Leibniz's differential notation earn its keep: the mnemonic “put \( x = \varphi(s) \), then \( \mathrm{d}x = \varphi'(s)\,\mathrm{d}s \)” is not a formal manipulation of infinitesimals but a licensed abbreviation of the identity above.

Structurally the rule says that the integral is natural with respect to reparametrisation: it does not depend on how a path is traversed, only on where it starts, where it ends, and how the integrand is transported. That is why the same statement reappears, with the derivative replaced by a Jacobian determinant, as the change-of-variables theorem in \( \mathbb{R}^n \); with the derivative replaced by a Radon–Nikodym density, as the transformation law for probability densities; and with no correction factor at all, as the abstract image-measure formula \( \int g\,\mathrm{d}(\varphi_{*}\mu) = \int (g\circ\varphi)\,\mathrm{d}\mu \). One-dimensional substitution is the smallest case of a very large theorem.

Hypotheses

Proof

The whole content is: build an antiderivative for \( f \), compose it with \( \varphi \), and evaluate both sides by the fundamental theorem of calculus. Steps 1–3 set up the objects, Steps 4–7 give the identity, Steps 8–9 record the monotone variants.

Result

Corollaries & converses

• Indefinite form. If \( F' = f \) on \( I \) then \( \int f(\varphi(s))\,\varphi'(s)\,\mathrm{d}s = F(\varphi(s)) + C \). This is Step 4 alone; the definite statement is that plus the fundamental theorem.
• Logarithmic derivative. Taking \( f(x) = 1/x \) and \( \varphi = g \) with \( g \) nowhere zero on \( [\alpha,\beta] \): \( \int_{\alpha}^{\beta} \dfrac{g'(s)}{g(s)}\,\mathrm{d}s = \ln\lvert g(\beta)\rvert - \ln\lvert g(\alpha)\rvert \). The “nowhere zero” is the image condition of Step 1, not a technicality.
• Translation and dilation. \( \varphi(s) = s + h \) gives \( \int_{\alpha}^{\beta} f(s+h)\,\mathrm{d}s = \int_{\alpha+h}^{\beta+h} f \); \( \varphi(s) = \lambda s \) with \( \lambda \neq 0 \) gives \( \int_{\alpha}^{\beta} f(\lambda s)\,\mathrm{d}s = \lambda^{-1}\int_{\lambda\alpha}^{\lambda\beta} f \). Translation invariance of Lebesgue measure is this corollary in disguise.
• Reflection. \( \varphi(s) = a + b - s \) gives \( \int_a^b f(x)\,\mathrm{d}x = \int_a^b f(a+b-x)\,\mathrm{d}x \). The companion case \( \varphi(s) = -s \) gives \( \int_{-a}^{0} f(x)\,\mathrm{d}x = \int_{0}^{a} f(-s)\,\mathrm{d}s \), whence \( \int_{-a}^{a} f = 0 \) for odd \( f \) and \( \int_{-a}^{a} f = 2\int_0^a f \) for even \( f \).
• Periodicity. If \( f \) is continuous with period \( T \gt 0 \) then \( \int_{a}^{a+T} f \) is independent of \( a \). Substituting \( \varphi(s) = s+T \) on the interval with endpoints \( 0 \) and \( a \), and using \( f(s+T)=f(s) \), gives \( \int_{T}^{a+T} f = \int_{0}^{a} f \); splitting \( \int_a^{a+T} = \int_a^0 + \int_0^T + \int_T^{a+T} \) then cancels the outer two terms and leaves \( \int_0^T f \). Substituting on \( [a,a+T] \) instead proves only the weaker statement that \( a \mapsto a+T \) leaves the value unchanged.
• Trigonometric and Weierstrass substitutions. \( x = \sin\theta \) on \( [-\pi/2,\pi/2] \), \( x = \tan\theta \) on \( (-\pi/2,\pi/2) \), \( x = \sec\theta \) on \( [0,\pi/2) \), and \( t = \tan(s/2) \) turning any rational function of \( \sin s,\cos s \) into a rational function of \( t \). Each is Step 8 with an explicitly monotone \( \varphi \); the restricted range is what supplies injectivity and fixes the sign of the surd.
• Higher dimensions. The Jacobian change-of-variables theorem reduces to this one when \( n = 1 \), with \( \lvert\det D\varphi\rvert = \lvert\varphi'\rvert \) — matching the unsigned form of Step 9, not the oriented form, which is why the \( n \)-dimensional statement carries an injectivity hypothesis that the \( n=1 \) oriented statement does not.
• The converse fails. The identity \( \int_{\alpha}^{\beta} f(\varphi(s))\psi(s)\,\mathrm{d}s = \int_{\varphi(\alpha)}^{\varphi(\beta)} f \) holding for one particular \( f \) does not force \( \psi = \varphi' \): with \( f \equiv 1 \) it only says \( \int_\alpha^\beta \psi = \varphi(\beta)-\varphi(\alpha) \), which countless \( \psi \) satisfy. Demanding it for every continuous \( f \) is still not enough when \( \varphi \) folds: for \( \varphi(s)=s^2 \) on \( [-1,1] \) the right-hand side is \( \int_1^1 f = 0 \), while \( s \mapsto f(s^2)\psi(s) \) is odd, hence has integral \( 0 \), for every odd \( \psi \) — so \( \psi(s) = 3s \) serves as well as \( \varphi'(s) = 2s \). What does pin \( \psi \) down is “for every continuous \( f \)” together with \( \varphi' \gt 0 \) throughout: substituting \( x = \varphi(s) \) turns \( \int_{\alpha}^{\beta} f(\varphi(s))\bigl(\psi-\varphi'\bigr)(s)\,\mathrm{d}s = 0 \) into \( \int_a^b f(x)\,h(x)\,\mathrm{d}x = 0 \) for the continuous \( h = \bigl((\psi-\varphi')\circ\varphi^{-1}\bigr)\cdot(\varphi^{-1})' \), and a continuous function orthogonal to every continuous function is identically zero; since \( (\varphi^{-1})' \gt 0 \), this forces \( \psi = \varphi' \).

Fails without

Common errors

• “Substitute the integrand, keep the limits.” Writing \( \int_0^1 2s\,e^{s^2}\,\mathrm{d}s = \int_0^1 e^{x}\,\mathrm{d}x \). The limits are \( \varphi(0)=0 \) and \( \varphi(1)=1 \) here only by coincidence of \( \varphi(s)=s^2 \); for \( \varphi(s)=s^2+3 \) the limits become \( 3 \) and \( 4 \). Either transform the limits or convert back to \( s \) before evaluating — never mix.
• Losing \( \varphi' \). \( \int f(\varphi(s))\,\mathrm{d}s \neq \int f(x)\,\mathrm{d}x \) in general. The derivative factor is the whole content of the rule; without it the identity is simply false, e.g. \( \int_0^1 e^{s^2}\mathrm{d}s \approx 1.4627 \) while \( \int_0^1 e^{x}\mathrm{d}x = e-1 \approx 1.7183 \).
• “\( \varphi \) must be monotone.” Not for the oriented form: \( \varphi(s) = s^2 - s \) on \( [0,1] \) is not monotone, yet \( \int_0^1 f(s^2-s)(2s-1)\,\mathrm{d}s = \int_0^0 f = 0 \) for every continuous \( f \) on \( [-1/4, 0] \), which is correct. Monotonicity is needed to run the substitution backwards, and for the \( \lvert\varphi'\rvert \) form.
• Sign of a surd. Putting \( x = \sin\theta \) and writing \( \sqrt{1-x^2} = \cos\theta \). True only when \( \cos\theta \ge 0 \), i.e. \( \theta \in [-\pi/2,\pi/2] \); in general \( \sqrt{1-\sin^2\theta} = \lvert\cos\theta\rvert \). The same trap is \( \sqrt{x^2} = \lvert x\rvert \), not \( x \).
• Forgetting to restore the original variable. An indefinite integral answered as \( \tfrac{1}{2}u^2 + C \) is not an answer; it must be \( \tfrac{1}{2}\varphi(s)^2 + C \).
• Substituting inside an improper or singular integral without checking. Any substitution that maps a bounded parameter interval onto an unbounded one, or that passes through a pole of \( f \), needs a separate limiting argument — the compact-interval theorem does not cover it.
• “\( \mathrm{d}x = \varphi'(s)\,\mathrm{d}s \) is an equation about infinitesimals.” It is an abbreviation for the theorem, licensed by it. The notation is reliable because the theorem is true, not the other way round.

Discussion

The rule is older than its proof. Leibniz's 1670s calculus was built on the differential \( \mathrm{d}x \), and substitution was, for him, nothing more than the observation that \( \mathrm{d}x \) obeys the same transformation law as the derivative — a piece of notation so well designed that it computes correctly by itself. Euler used it with complete freedom throughout the Institutiones calculi integralis (1768–1770). The modern justification arrived only with Cauchy's definition of the integral as a limit of sums (1823) and its refinement by Riemann (1854), at which point the hypotheses became visible: continuity of \( f \) on the image, and continuous differentiability of \( \varphi \). Everything difficult about the theorem lives in those hypotheses, since the proof itself is four lines once the fundamental theorem is available.

It is worth being clear about the direction of logical dependence. Substitution does not stand beside the chain rule and the fundamental theorem as a third principle; it is their composite, and the proof above is little more than a statement of that fact. This also explains the asymmetry between differentiation and integration in practice. The chain rule is an algorithm: given \( F \circ \varphi \), differentiate mechanically. Substitution is a search: given an integrand, guess a factorisation of the form \( f(\varphi(s))\varphi'(s) \). No such guess is guaranteed to exist within the elementary functions — Liouville's theory of integration in finite terms, made algorithmic by Risch (1969), decides exactly when it does, and shows that \( \int e^{-s^2}\mathrm{d}s \) has no elementary antiderivative. Substitution is therefore a powerful heuristic attached to a very sharp theorem, and the two should not be confused.

The geometric content is naturality under reparametrisation. Write the integrand as the differential form \( \omega = f(x)\,\mathrm{d}x \) on \( I \); then \( \varphi^{*}\omega = f(\varphi(s))\varphi'(s)\,\mathrm{d}s \) is its pullback, and the theorem says \( \int_{[\alpha,\beta]} \varphi^{*}\omega = \int_{\varphi_{*}[\alpha,\beta]} \omega \), the right-hand side being the integral over the oriented image chain — the path traced by \( \varphi \), counted with multiplicity and sign — and not over the image set. In this language the absence of an injectivity hypothesis is no surprise: pullback is defined for any smooth map, and a path that doubles back simply contributes cancelling oriented pieces. The unsigned form of Step 9 is the statement about measure rather than about forms, and measure cannot cancel, which is why it alone needs injectivity. The same split explains why the \( n \)-dimensional Jacobian theorem, which is a measure statement, carries \( \lvert\det D\varphi\rvert \) and demands a diffeomorphism, while its differential-form counterpart carries \( \det D\varphi \) and demands only orientation.

In the Lebesgue theory the clean general statement drops the correction factor altogether. For a measurable \( \varphi : (\Omega,\mathcal{A},\mu) \to (\Omega',\mathcal{A}') \), the image measure \( \varphi_{*}\mu = \mu \circ \varphi^{-1} \) satisfies \( \int_{\Omega'} g\,\mathrm{d}(\varphi_{*}\mu) = \int_{\Omega} (g\circ\varphi)\,\mathrm{d}\mu \) for every non-negative measurable \( g \) — proved by the usual bootstrap from indicators through simple functions and monotone convergence, with no smoothness anywhere. The Jacobian \( \lvert\varphi'\rvert \) is not part of that theorem; it appears only when one insists on writing \( \varphi_{*}\mu \) back in terms of Lebesgue measure, at which point \( \lvert\varphi'\rvert \) is the Radon–Nikodym derivative \( \mathrm{d}(\varphi_{*}^{-1}\lambda)/\mathrm{d}\lambda \), and its existence is exactly absolute continuity. The sharpest one-dimensional real-variable version says: if \( \varphi : [\alpha,\beta] \to \mathbb{R} \) is absolutely continuous and \( f \) is Borel with \( (f\circ\varphi)\varphi' \) Lebesgue integrable, then the identity holds — \( C^1 \) is comfortably sufficient, the Cantor function is exactly what absolute continuity excludes, and the probabilistic density-transformation rule \( p_Y(y) = p_X(\varphi^{-1}(y))\,\lvert(\varphi^{-1})'(y)\rvert \) is this same theorem read once more, with \( \varphi \) required injective because a density is a measure and measures do not cancel.

Common misconceptions. Substitution is not “change of variables in the integrand”: it changes the integrand, the measure, and the limits together, and any two out of three is wrong. It does not require \( \varphi \) to be invertible, monotone, or even locally injective — only that it be \( C^1 \) and land inside the region where \( f \) is continuous. Conversely it gives no licence to substitute through a point where \( f \) or \( \varphi \) misbehaves, and the resulting fallacies (\( \pi/2 = 0 \) above) look entirely routine on the page. Finally, the appearance of \( \lvert\varphi'\rvert \) rather than \( \varphi' \) is not a matter of taste: the absolute value belongs to the unsigned, set-theoretic reading and comes packaged with injectivity, while the signed \( \varphi' \) belongs to the oriented reading and comes free.

Worked examples

Example 1. Evaluate \( \displaystyle\int_{0}^{4} \frac{\mathrm{d}x}{1 + \sqrt{x}} \), justifying every hypothesis. The integrand is elementary but has no obvious antiderivative; the substitution \( x = \varphi(s) = s^2 \) rationalises it, and is used in the inverse direction (Step 8), so monotonicity must be checked.

Example 2. A bead of mass \( m = 0.250\ \mathrm{kg} \) is pushed along a straight rail by the model force \( F(x) = \dfrac{k\,x}{(x^2+a^2)^{3/2}} \), where \( k = 2.00\ \mathrm{N\,m^2} \) and \( a = 0.100\ \mathrm{m} \) is a fixed offset. Find the work done as the bead moves from \( x_1 = 0 \) to \( x_2 = 0.200\ \mathrm{m} \), and the resulting speed if it starts from rest.

Problems

1. Evaluate \( \displaystyle\int_{0}^{1} x\,e^{-x^2}\,\mathrm{d}x \), stating which \( \varphi \) and which \( f \) you are using and checking the hypotheses.
   Solution

   Take \( \varphi(x) = -x^2 \) and \( f(u) = -\tfrac{1}{2}e^{u} \). Then \( \varphi \in C^{\infty}[0,1] \), \( \varphi'(x) = -2x \), and \[ f\bigl(\varphi(x)\bigr)\varphi'(x) = -\tfrac{1}{2}e^{-x^2}\cdot(-2x) = x\,e^{-x^2}, \] which is the integrand. The image is \( \varphi([0,1]) = [-1,0] \), and \( f \) is continuous on all of \( \mathbb{R} \), so certainly on that image. The theorem gives \[ \int_0^1 x e^{-x^2}\,\mathrm{d}x = \int_{\varphi(0)}^{\varphi(1)} f(u)\,\mathrm{d}u = \int_{0}^{-1} \left(-\tfrac{1}{2}e^{u}\right)\mathrm{d}u = \tfrac{1}{2}\int_{-1}^{0} e^{u}\,\mathrm{d}u, \] using the orientation convention to flip the limits. Hence \[ \int_0^1 x e^{-x^2}\,\mathrm{d}x = \tfrac{1}{2}\Bigl[e^{u}\Bigr]_{-1}^{0} = \tfrac{1}{2}\bigl(1 - e^{-1}\bigr) = \tfrac{1}{2}(1 - 0.367879\ldots) = 0.316060\ldots \approx 0.3161 . \] Equivalently, in the everyday notation, put \( u = x^2 \), \( \mathrm{d}u = 2x\,\mathrm{d}x \), limits \( 0 \to 1 \), giving \( \tfrac{1}{2}\int_0^1 e^{-u}\mathrm{d}u \) — the same number.

2. Evaluate \( \displaystyle\int_{-1}^{1} \sqrt{1-x^2}\,\mathrm{d}x \) by trigonometric substitution, and say precisely where the choice of range for \( \theta \) is used.
   Solution

   Put \( \varphi(\theta) = \sin\theta \) on \( [\alpha,\beta] = [-\pi/2,\ \pi/2] \). Then \( \varphi \in C^{\infty} \), \( \varphi'(\theta) = \cos\theta \), \( \varphi(-\pi/2) = -1 \), \( \varphi(\pi/2) = 1 \), and \( \varphi \) is strictly increasing on this interval (its derivative is positive on the interior), so it is a bijection onto \( [-1,1] \) — this is what licenses the inverse-substitution direction. With \( f(x) = \sqrt{1-x^2} \), continuous on \( [-1,1] \), \[ \int_{-1}^{1}\sqrt{1-x^2}\,\mathrm{d}x = \int_{-\pi/2}^{\pi/2} \sqrt{1-\sin^2\theta}\;\cos\theta\,\mathrm{d}\theta = \int_{-\pi/2}^{\pi/2} \lvert\cos\theta\rvert\,\cos\theta\,\mathrm{d}\theta. \] Here the range is spent: on \( [-\pi/2,\pi/2] \) we have \( \cos\theta \ge 0 \), so \( \lvert\cos\theta\rvert = \cos\theta \) and the integrand is \( \cos^2\theta \). (On \( [0,\pi] \), say, this identification would be false on \( (\pi/2,\pi] \) and would produce a sign error.) Using \( \cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta) \), \[ \int_{-\pi/2}^{\pi/2}\cos^2\theta\,\mathrm{d}\theta = \tfrac{1}{2}\Bigl[\theta + \tfrac{1}{2}\sin 2\theta\Bigr]_{-\pi/2}^{\pi/2} = \tfrac{1}{2}\Bigl[\bigl(\tfrac{\pi}{2}+0\bigr) - \bigl(-\tfrac{\pi}{2}+0\bigr)\Bigr] = \frac{\pi}{2} \approx 1.5708 . \] This is the area of a unit half-disc, as it must be.

3. Prove the reflection corollary \( \displaystyle\int_{0}^{a} f(x)\,\mathrm{d}x = \int_{0}^{a} f(a-x)\,\mathrm{d}x \) for continuous \( f \) and \( a \gt 0 \), and use it to evaluate \( \displaystyle J = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,\mathrm{d}x \).
   Solution

   The corollary. Let \( \varphi(s) = a - s \) on \( [0,a] \). It is \( C^{\infty} \) with \( \varphi'(s) = -1 \), \( \varphi(0) = a \), \( \varphi(a) = 0 \), and \( \varphi([0,a]) = [0,a] \), on which \( f \) is continuous. The theorem gives \[ \int_{0}^{a} f(a-s)\,(-1)\,\mathrm{d}s = \int_{a}^{0} f(x)\,\mathrm{d}x = -\int_{0}^{a} f(x)\,\mathrm{d}x, \] and multiplying by \( -1 \) yields \( \int_0^a f(a-s)\,\mathrm{d}s = \int_0^a f(x)\,\mathrm{d}x \), as claimed.

   The integral. Apply this with \( a = \pi/2 \). The integrand is continuous on \( [0,\pi/2] \): on \( [0,\pi/2] \) both \( \sin x \) and \( \cos x \) are non-negative, so the square roots are real, and \[ \bigl(\sqrt{\sin x}+\sqrt{\cos x}\bigr)^2 = \sin x + \cos x + 2\sqrt{\sin x \cos x} \ \ge\ \sin x + \cos x \ \ge\ 1, \] the last inequality because \( (\sin x + \cos x)^2 = 1 + \sin 2x \ge 1 \) and \( \sin x + \cos x \ge 0 \) there. So the denominator is at least \( 1 \) and never vanishes. Using \( \sin(\pi/2 - x) = \cos x \) and \( \cos(\pi/2 - x) = \sin x \), \[ J = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,\mathrm{d}x . \] Adding the two expressions for \( J \), \[ 2J = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,\mathrm{d}x = \int_{0}^{\pi/2} 1\,\mathrm{d}x = \frac{\pi}{2}, \] so \( J = \pi/4 \approx 0.7854 \). Notice that no antiderivative of the integrand was ever found; the substitution was used purely as a symmetry.

4. Evaluate \( \displaystyle I = \int_{0}^{\pi} \frac{x\,\sin x}{1 + \cos^2 x}\,\mathrm{d}x \).
   Solution

   The integrand is continuous on \( [0,\pi] \) since \( 1+\cos^2 x \ge 1 \). Substitute \( \varphi(t) = \pi - t \) on \( [0,\pi] \), which is \( C^{\infty} \), maps \( [0,\pi] \) onto itself, and has \( \varphi' = -1 \). With \( \sin(\pi-t) = \sin t \) and \( \cos^2(\pi-t) = \cos^2 t \), the reflection corollary of Problem 3 (with \( a = \pi \)) gives \[ I = \int_{0}^{\pi} \frac{(\pi - t)\sin t}{1+\cos^2 t}\,\mathrm{d}t = \pi\int_{0}^{\pi} \frac{\sin t}{1+\cos^2 t}\,\mathrm{d}t \;-\; I . \] Hence \( 2I = \pi K \) where \( K = \int_0^{\pi} \dfrac{\sin t}{1+\cos^2 t}\,\mathrm{d}t \). For \( K \), substitute \( \psi(t) = \cos t \) on \( [0,\pi] \): \( \psi \in C^{\infty} \), \( \psi'(t) = -\sin t \), \( \psi(0) = 1 \), \( \psi(\pi) = -1 \), and \( \psi([0,\pi]) = [-1,1] \), on which \( u \mapsto (1+u^2)^{-1} \) is continuous. Then \[ K = -\int_{0}^{\pi} \frac{\psi'(t)}{1+\psi(t)^2}\,\mathrm{d}t = -\int_{1}^{-1}\frac{\mathrm{d}u}{1+u^2} = \int_{-1}^{1}\frac{\mathrm{d}u}{1+u^2} = \arctan(1) - \arctan(-1) = \frac{\pi}{2}. \] Therefore \( 2I = \pi \cdot \dfrac{\pi}{2} \) and \[ I = \frac{\pi^2}{4} \approx 2.4674 . \] Both substitutions were applied in the oriented direction of Step 7, so no inverse function was ever needed; \( \varphi(t) = \pi - t \) served only as a symmetry, and \( \psi = \cos \) was used to push the limits from \( [0,\pi] \) to \( [1,-1] \), the reversal of orientation supplying the sign that cancels the minus sign of \( \psi' \).

5. The following argument concludes that \( \displaystyle I = \int_{-1}^{1}\frac{\mathrm{d}x}{1+x^2} \) is zero. Substituting \( x = 1/u \) gives \( \mathrm{d}x = -u^{-2}\,\mathrm{d}u \) and \( (1+x^2)^{-1} = u^2/(1+u^2) \); the limits \( x = -1 \) and \( x = 1 \) correspond to \( u = -1 \) and \( u = 1 \); hence \( I = \int_{-1}^{1} \frac{u^2}{1+u^2}\cdot\left(-\frac{1}{u^2}\right)\mathrm{d}u = -I \), so \( I = 0 \). Locate the error precisely, and give the correct value.
   Solution

   The correct value. \( x \mapsto (1+x^2)^{-1} \) is continuous on \( [-1,1] \) and strictly positive, so \( I \gt 0 \) immediately; by Part II of the fundamental theorem with the antiderivative \( \arctan \), \[ I = \arctan(1) - \arctan(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2} \approx 1.5708 . \] So the argument is certainly wrong.

   The error. The substituting map is \( \varphi(u) = 1/u \), and the theorem demands \( \varphi \in C^1[\alpha,\beta] \) with \( \varphi([\alpha,\beta]) \) inside the domain of continuity of \( f \). Here the proposed parameter interval is \( [\alpha,\beta] = [-1,1] \), and \( \varphi \) is not defined at \( u = 0 \), let alone continuously differentiable there. Step 3 of the proof — forming \( G = F\circ\varphi \) on the closed parameter interval — therefore never gets off the ground, and neither does the application of Part II in Step 6, which needs \( G \) continuous on the whole of \( [\alpha,\beta] \).

   What is actually true. On \( [\alpha,\beta] = [1,R] \) with \( R \gt 1 \) the substitution is perfectly legal, and it gives \[ \int_{1/R}^{1}\frac{\mathrm{d}x}{1+x^2} = \int_{1}^{R}\frac{\mathrm{d}u}{1+u^2}, \] a correct identity (both sides equal \( \arctan 1 - \arctan(1/R) = \arctan R - \arctan 1 \), using \( \arctan(1/t) = \pi/2 - \arctan t \) for \( t \gt 0 \)). Letting \( R \to \infty \) recovers \( \int_0^1 = \int_1^{\infty} = \pi/4 \), and the mirror computation on \( [-R,-1] \) gives \( \int_{-1}^{-1/R} \mathrm{d}x/(1+x^2) = \int_{-R}^{-1}\mathrm{d}u/(1+u^2) \). So the parameter set that \( u \mapsto 1/u \) really carries onto \( [-1,1]\setminus\{0\} \) is \( \lvert u\rvert \ge 1 \), not \( [-1,1] \): the fallacy has quietly swapped the true, unbounded \( u \)-domain for its complement. Nothing diverges here — both regions carry the same finite integral, \( \int_{\lvert u\rvert \ge 1} \mathrm{d}u/(1+u^2) = \pi/2 = \int_{-1}^{1}\mathrm{d}u/(1+u^2) \) — which is precisely why the arithmetic looks self-consistent and only the stray minus sign gives it away. This is the singularity-crossing failure mode: everything on the page looks routine, and only the hypothesis check catches it.