The idea behind substitution
Substitution in an indefinite integral can look like a change of symbols: choose an inner expression, call it \(u\), integrate with respect to \(u\), and then replace \(u\) by the original expression. The reason this procedure works is not a separate integration trick. It is the Chain Rule applied to a proposed antiderivative.
If an antiderivative \(A\) has derivative \(F\), then composing \(A\) with a differentiable function \(g\) gives the function \(A(g(x))\). The Chain Rule tells us exactly what its derivative is: \(F(g(x))g'(x)\). That identity is the mathematical basis of substitution.
Definitions and conditions
A function \(A\) is an antiderivative of a function \(F\) on an interval \(J\) if \(A\) is differentiable on \(J\) and \(A'(u)=F(u)\) for every \(u\in J\). The letter \(u\) here is simply the input variable for the function \(A\).
To apply the result, we need \(A'(u)=F(u)\) wherever \(g(x)\) takes its values, and we need \(g\) to be differentiable at the \(x\)-values under consideration. In particular, the composition \(A(g(x))\) must be defined. These conditions allow the Chain Rule to be used at every point of \(I\).
The conclusion is a derivative identity on an interval:
Therefore the function \(A(g(x))\) is an antiderivative of the product on the right. No assumption that \(g\) is one-to-one or increasing is needed for this identity. The argument only requires differentiability and the stated domain conditions.
Why the formula works
The essential point is that differentiating a composition has two parts. First, differentiate the outer function \(A\), then evaluate that derivative at the inner expression \(g(x)\). Next, multiply by the derivative of the inner function. Since \(A'=F\), the first part becomes \(F(g(x))\); the second part is \(g'(x)\).
Assume \(A'(u)=F(u)\) for \(u\in J\).
Form \(A(g(x))\), which is defined because \(g(x)\in J\).
By the Chain Rule, its derivative is \(A'(g(x))g'(x)\).
Substitute \(A'(g(x))=F(g(x))\). The derivative is \(F(g(x))g'(x)\), as required.
This is a complete proof: the definition of antiderivative asks us to verify the derivative, and the Chain Rule gives exactly the required expression. In integral notation, the same result is written
Here \(C\) is an arbitrary constant. For any fixed constant \(C\), differentiating \(A(g(x))+C\) gives the same derivative because the derivative of a constant is zero. Thus every expression on the right is an antiderivative.
Worked examples
Worked Example: A power of a composite expression
Find an antiderivative of \(6x(3x^2+2)^4\). Let \(g(x)=3x^2+2\). Then \(g'(x)=6x\). For the outer function, \(A(u)=u^5/5\), since
Thus \(F(u)=u^4\), and \(F(g(x))g'(x)=(3x^2+2)^4(6x)\), which is the given integrand. The result is
Check by differentiating the proposed antiderivative:
The derivative matches the integrand, so the antiderivative is correct.
Worked Example: An exponential of a trigonometric expression
Find an antiderivative of \(e^{\sin x}\cos x\). Set \(g(x)=\sin x\), so \(g'(x)=\cos x\). An antiderivative of \(F(u)=e^u\) is \(A(u)=e^u\), because
Composing \(A\) with \(g\) gives \(A(g(x))=e^{\sin x}\). Therefore
The check uses the Chain Rule:
This verifies the proposed antiderivative.
Worked Example: A logarithm with a positive inner function
Find an antiderivative of \(\frac{2x}{x^2+5}\). Choose \(g(x)=x^2+5\), for which \(g'(x)=2x\). On the real line, \(x^2+5\geq 5>0\). An antiderivative of \(F(u)=1/u\) on the positive real numbers is \(A(u)=\ln u\), because \(A'(u)=1/u\) there. Hence
Differentiate to check:
The result differentiates to the original integrand. Positivity of \(x^2+5\) also ensures that the logarithm is defined for every real \(x\).
What substitution notation records
The common instruction “let \(u=g(x)\)” records the choice of inner function. It also signals that the derivative factor \(g'(x)\) must be accounted for. In differential notation, the relation \(du=g'(x)\,dx\) is a compact way of keeping track of that factor; the justification for the antiderivative remains the Chain Rule identity above.
For example, in the first worked example, \(u=3x^2+2\) gives \(du=6x\,dx\). The integral can then be recognized as \(\int u^4\,du\), whose antiderivative is \(u^5/5+C\). Replacing \(u\) by \(3x^2+2\) gives the displayed answer. The proof explains why that replacement is valid: differentiating the resulting composition restores both the outer derivative and the inner derivative.
This viewpoint also clarifies the role of a constant multiple. If the integrand contains \(k g'(x)\) alongside \(F(g(x))\), where \(k\) is constant, then an antiderivative is \(kA(g(x))+C\), since
A common mistake: leaving out the inner derivative
A frequent error is to replace \(u\) by \(g(x)\) in an antiderivative without checking whether the original integrand contains the corresponding factor \(g'(x)\). For instance, it would be incorrect to claim
The proposed answer differentiates to
which has the extra factor \(6x\). It is therefore not an antiderivative of \((3x^2+2)^4\). The formula from the Chain Rule applies directly to an integrand of the form \(F(g(x))g'(x)\), not to \(F(g(x))\) alone. If the inner derivative is absent, this particular substitution formula does not establish the claimed answer.
A reliable check is to differentiate the proposed antiderivative and compare the result with the entire original integrand, including every factor. This catches missing inner derivatives and other algebraic errors.
Why the result is useful
Substitution is often introduced as a practical way to recognize an integral. Its deeper value is that it lets us build antiderivatives systematically from known ones. Whenever an integrand is a composite \(F(g(x))\) multiplied by \(g'(x)\), a known antiderivative \(A\) of \(F\) immediately supplies the antiderivative \(A(g(x))\).
The result is also useful for checking an answer independently. Rather than relying on the substitution steps, differentiate the final expression. If the Chain Rule produces the original integrand at every point in the interval, the expression is an antiderivative there. No inverse of \(g\) is needed: the proof only differentiates the composition.
Check Your Understanding
Use the Chain Rule, not just a pattern match, to answer these questions.
- If \(A'(u)=F(u)\), what is the derivative of \(A(g(x))\)?
- What differentiability and domain conditions are needed for the composition result?
- Why does the integrand need the factor \(g'(x)\) in the substitution formula?
- Differentiate \(\frac{(x^2+1)^3}{3}\) and identify the integrand for which it is an antiderivative.
- Does the proof require \(g\) to be one-to-one? Explain briefly.
