Why Back-Substitution Matters
A substitution can turn a complicated integrand into a familiar one. For a definite integral, there are two consistent ways to finish: change the limits to match the substituted variable, or return the antiderivative to the original variable and use the original limits. This tutorial focuses on the second approach.
The key is to keep the variable and the limits together. If the antiderivative is written in terms of \(x\), evaluate it at the original \(x\)-limits. Do not use the original limits on an expression that is still written in the substituted variable.
The Method and Its Justification
Suppose the integrand can be written as \(h(g(x))g'(x)\), and let \(H\) be an antiderivative of \(h\), so that \(H'(u)=h(u)\). The substitution \(u=g(x)\) turns the integrand into \(h(u)\,du\). An antiderivative in the new variable is \(H(u)\). To use the original limits, substitute \(g(x)\) back into this antiderivative, giving \(H(g(x))\).
Why does this give the correct antiderivative in \(x\)? The chain rule gives
Thus \(H(g(x))\) is an antiderivative of the original integrand. If the integrand is continuous on \([a,b]\) and \(g\) and \(H\) are differentiable as needed, the Fundamental Theorem of Calculus then gives the definite integral as the value of this antiderivative at \(b\) minus its value at \(a\).
This result follows from the chain rule and the Fundamental Theorem of Calculus, not from a special rule for guessing the new bounds. The substitution helps find the antiderivative; after returning to \(x\), the ordinary definite-integral evaluation rule applies.
A Reliable Sequence of Steps
Set \(u=g(x)\) so the integrand and its differential become simpler. Keep track of how \(du\) relates to the original differential.
Find an antiderivative \(H(u)\) of the resulting expression in \(u\).
Replace every \(u\) in \(H(u)\) with its expression in \(x\), giving \(H(g(x))\).
Evaluate \(H(g(x))\) at the original upper and lower \(x\)-limits, then subtract lower from upper.
At each stage, note which variable is being used. The differential \(du\) belongs to the substituted integral, and the original limits belong to an antiderivative written in \(x\). This bookkeeping prevents a common mismatch.
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^1 2x\cos(x^2)\,dx\) by back-substitution.
Step 1: Substitute. Let \(u=x^2\). Then \(du=2x\,dx\), so the integral becomes \(\int \cos u\,du\) while we find an antiderivative.
Step 2: Integrate in \(u\). Since the derivative of \(\sin u\) is \(\cos u\), an antiderivative is \(\sin u\).
Step 3: Substitute back. Since \(u=x^2\), the antiderivative in the original variable is \(\sin(x^2)\).
Step 4: Apply the original limits. The original interval is \(0\leq x\leq 1\), so evaluate at \(x=1\) and \(x=0\):
Check: Differentiating the returned antiderivative gives \(\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x=2x\cos(x^2)\), the original integrand. The limits were applied to the expression in \(x\), as required.
Worked Example: A Logarithmic Antiderivative
Evaluate \(\int_0^2 \frac{x}{x^2+1}\,dx\) by back-substitution.
Step 1: Substitute and account for the factor. Let \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\frac12du\). The transformed integrand for finding the antiderivative is \(\frac{1}{2u}\,du\).
Step 2: Integrate in \(u\). Because \(u=x^2+1\) is positive for every real \(x\), the logarithmic antiderivative can be written as \(\frac12\ln u\).
Step 3: Substitute back. Replace \(u\) by \(x^2+1\), giving \(\frac12\ln(x^2+1)\).
Step 4: Apply the original limits. Evaluate at \(x=2\) and \(x=0\):
Check: Differentiating the returned antiderivative gives \(\frac12\cdot\frac{2x}{x^2+1}=\frac{x}{x^2+1}\). This matches the integrand, and the evaluation uses the original endpoints \(0\) and \(2\).
Worked Example: A Decreasing Inner Function
Evaluate \(\int_0^{\pi/2}\sin x\,e^{\cos x}\,dx\) by back-substitution.
Step 1: Substitute. Let \(u=\cos x\). Then \(du=-\sin x\,dx\), or \(\sin x\,dx=-du\). The minus sign must be included, so the expression to integrate in \(u\) is \(-e^u\,du\).
Step 2: Integrate in \(u\). An antiderivative of \(e^u\) is \(e^u\), so an antiderivative of \(-e^u\) is \(-e^u\).
Step 3: Substitute back. Replacing \(u\) by \(\cos x\) gives \(-e^{\cos x}\).
Step 4: Apply the original limits. Use \(x=\pi/2\) for the upper limit and \(x=0\) for the lower limit:
Check: The derivative of the returned antiderivative is \(\frac{d}{dx}(-e^{\cos x})=-e^{\cos x}(-\sin x)=\sin x\,e^{\cos x}\), which is the original integrand. The negative sign from \(du=-\sin x\,dx\) is reflected in the antiderivative.
A Common Bounds Mistake
The substitution variable and the original variable have different limits. If you find \(H(u)\) and have not yet substituted back, its input is \(u\), not \(x\). Applying \(x\)-limits directly to \(H(u)\) is therefore a variable mismatch.
Worked Example: Correcting a Bounds Mismatch
Consider \(\int_1^2 2x\cos(x^2)\,dx\). Let \(u=x^2\), so \(du=2x\,dx\). Integrating in \(u\) gives \(\sin u\).
Incorrect calculation: Using the original endpoints \(1\) and \(2\) directly on \(\sin u\) gives \(\sin 2-\sin 1\). This treats the original \(x\)-limits as though they were \(u\)-values.
Correction by back-substitution: Replace \(u\) with \(x^2\), then apply the original \(x\)-limits:
Check: The derivative of \(\sin(x^2)\) is \(2x\cos(x^2)\). Also, the substituted endpoint values would be \(u=1^2=1\) and \(u=2^2=4\), consistent with the corrected result. The mistaken expression used \(2\) where the corresponding substituted endpoint is \(4\).
Why the Method Is Useful
Back-substitution lets you use an antiderivative found in a convenient variable while still evaluating the definite integral with its original limits. It is especially straightforward when the substitution includes a negative differential or a constant factor: those signs and factors are handled while finding the antiderivative, and then the original endpoints are applied to the complete expression in \(x\).
Before finishing a definite integral by substitution, check three things:
- Does the antiderivative contain only the original variable before the original limits are used?
- Was the full differential relation, including any negative sign or constant factor, included?
- Was the value at the lower limit subtracted from the value at the upper limit?
These checks keep the variable, differential, antiderivative, and limits consistent. The central idea is simple: integrate in whichever variable makes the work easier, but apply the original bounds only after the antiderivative has been returned to the original variable.
Check Your Understanding
For each question, keep track of which variable the antiderivative uses and which limits belong to it.
- For \(\int_0^1 6x^2 e^{2x^3}\,dx\), choose a substitution and state the antiderivative after returning to \(x\). What original limits will you apply?
- For \(\int_0^{\pi/2}\cos x\sin(\sin x)\,dx\), use \(u=\sin x\). Find an antiderivative in \(u\), substitute back, and evaluate with the original limits.
- In \(\int_1^3 \frac{2x}{x^2+4}\,dx\), why is it incorrect to use \(1\) and \(3\) as the endpoints of an antiderivative written in \(u=x^2+4\)?
- State the chain-rule calculation that explains why \(H(g(x))\) is an antiderivative of \(h(g(x))g'(x)\) when \(H'(u)=h(u)\).
- If \(u=g(x)\) and \(du=-f(x)\,dx\), what sign must be included when rewriting an integral containing \(f(x)\,dx\) in terms of \(du\)?
