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The substitution rule replaces a composite inner function with a new variable and its matching differential.
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Formula
Mathematical variables; for definite integrals, transform the bounds with u=g(x).
| Symbol | Meaning | Unit |
|---|---|---|
| Substituted inner function | dependent on the units of g | |
| g′(x) dx | Differential factor of the inner function | unit of u |
| Outer integrand after substitution | integrand |
\int f(g(x))g'(x)\,dx=\int f(u)\,du,\quad u=g(x)The rule reverses the chain rule: if F′=f, then (F∘g)′=f(g(x))g′(x). Set u=g(x) and replace du with g′(x)dx. For a definite integral, transform the bounds from x to u, or return to x before evaluating. A substitution is useful only when the matching differential factor is present.
Set u=g(x) when the factor g′(x) is present in the integrand. For definite integrals, transform the bounds; verify the resulting antiderivative by differentiation.
Substitution applies the chain rule in reverse.
Definite integral with transformed bounds
Each new bound is the image of the old bound under u=g(x).
Evaluate ∫01 2x e^(x2) dx using u=x2.
Set u=x2 and du=2x dx; the bounds remain 0 and 1.
Thus ∫01eᵘdu=e−1≈1.718.
Check: an antiderivative in x is e^(x2), whose endpoint difference is also e−1.
Setting u=g(x) but failing to replace g′(x)dx consistently.
Write du=g′(x)dx explicitly before rewriting the integrand.
Formula cluster
The reverse chain rule complements basic antiderivatives and integration by parts.
Compute ∫2x cos(x2) dx. Set u=x2 and du=2x dx. Then ∫cos(u) du=sin(u)+C=sin(x2)+C. Check: differentiating sin(x2) gives 2x cos(x2), the original integrand.
Simplify composite integrands, calculate areas, and derive antiderivatives systematically from the chain rule.
Here is how to work through a typical Integration by substitution (∫f(g(x))g′(x) dx = ∫f(u) du) task step by step:
Task
Evaluate ∫01 2x e^(x2) dx using u=x2.
Solution path
Set u=x2 and du=2x dx; the bounds remain 0 and 1.
Thus ∫01eᵘdu=e−1≈1.718.
Check: an antiderivative in x is e^(x2), whose endpoint difference is also e−1.
Look for an inner function g(x) whose derivative appears as a factor. In 2x cos(x2), x2 is the inner function and 2x its differential factor. The integrand becomes cos(u)du. If that factor is absent or cannot be matched, a simple substitution does not directly solve the integral; another method may be needed. For a definite integral, transformed bounds are part of the variable change. A quick differentiation check prevents sign and factor errors. The rule is especially useful when an inner derivative appears alongside the composite expression.
The chain rule differentiates F(g(x)) into F′(g(x))g′(x). In integration, recognize f(g(x))g′(x) as that derivative when F′=f. Therefore an antiderivative is F(g(x)). The check is direct: differentiate the result and confirm that both the outer function and factor g′(x) reappear. For a definite integral, transformed bounds are part of the variable change. A quick differentiation check prevents sign and factor errors. The rule is especially useful when an inner derivative appears alongside the composite expression. For example, ∫cos(u)du=sin(u)+C; back-substitution then replaces u with g(x).
Apply u=g(x) to the bounds too: x=a becomes u=g(a), and x=b becomes u=g(b). The integral is then entirely in u and can be evaluated without substituting back. Alternatively, return to x after integration and use the original bounds. Never mix a u-expression with x-bounds. For a definite integral, transformed bounds are part of the variable change. A quick differentiation check prevents sign and factor errors. The rule is especially useful when an inner derivative appears alongside the composite expression. This is especially convenient when g(a) and g(b) are easy to evaluate.
The replacement du=g′(x)dx transfers the inner function’s rate of change to the new variable. Without it, area scaling is wrong. In ∫2x cos(x2)dx, 2x dx becomes du. Replacing only x2 by u while leaving dx would not produce an integral in u and would give an incorrect factor. For a definite integral, transformed bounds are part of the variable change. A quick differentiation check prevents sign and factor errors. The rule is especially useful when an inner derivative appears alongside the composite expression.
Differentiate the antiderivative and compare every factor with the original integrand. The derivative of e^(x2) is 2x e^(x2), so it solves ∫2x e^(x2)dx. For a definite integral, also check the endpoint evaluation. These checks often catch a missing inner derivative or forgotten transformed bounds. For a definite integral, transformed bounds are part of the variable change. A quick differentiation check prevents sign and factor errors. The rule is especially useful when an inner derivative appears alongside the composite expression. For an indefinite integral, keep the constant of integration in the final antiderivative.
Quanta exam set
Curated exam set for "Integration by substitution":
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Which formula describes Integration by substitution?
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How do you rearrange Integration by substitution for Definite integral with transformed bounds?
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Which common mistake happens with Integration by substitution?
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