Why Nested Substitutions Need Care
A definite integral may contain several layers of composition. One substitution can expose a simpler integral, but that new integral may still be easier to evaluate after changing variables again. Each change is straightforward on its own; the challenge is keeping the integrand, differential, and bounds synchronized.
For example, if the original variable is \(x\), a first substitution might introduce \(u=g(x)\), and a second might introduce \(v=h(u)\). After the first change, the integral is written entirely in \(u\); after the second, it is written entirely in \(v\). The final evaluation must use the final integrand and the final endpoints together.
The Rule for Successive Changes of Variable
Suppose \(g\) is continuously differentiable on \([a,b]\), \(h\) is continuously differentiable on an interval containing \(g([a,b])\), and \(f\) is continuous on an interval containing \(h(g([a,b]))\). Then the definite integral below has a valid nested substitution:
The first change is \(u=g(x)\), so \(du=g'(x)\,dx\). Its endpoints are \(u=g(a)\) and \(u=g(b)\). The second change is \(v=h(u)\), so \(dv=h'(u)\,du\), with endpoints \(v=h(g(a))\) and \(v=h(g(b))\). The formula does not require either substitution to be increasing: if a change reverses endpoint order, the bounds remain in the order obtained by applying the substitution to the original lower and upper endpoints.
In practice, an integral may be presented in a form where the factors for the second substitution become visible only after the first change. In that case, write out the intermediate integral explicitly before making the next change. This makes it possible to check that the new differential really accounts for the factors being replaced.
Why the Nested Rule Works
The chain rule explains the factors in the rule, and the Fundamental Theorem of Calculus turns that derivative into the definite integral. To see this, choose an antiderivative \(F\) of \(f\) on the interval containing the values of \(h(g(x))\). Thus \(F'(v)=f(v)\). The chain rule gives
The composition \(F(h(g(x)))\) is continuously differentiable under the stated assumptions, so its derivative is continuous and the Fundamental Theorem of Calculus applies. Therefore,
Applying the same theorem to the transformed integrals gives
and
All three integrals therefore have the same value. The continuity assumptions matter: continuous differentiability of the substitutions ensures the displayed integrands are continuous, so these proper definite integrals and the applications of the Fundamental Theorem are justified.
Worked Example: An Exponential After Two Linear Changes
Worked Example: An Exponential After Two Linear Changes
Evaluate \(\displaystyle \int_0^1 2x e^{1+x^2}\,dx\) by making two successive substitutions.
First substitution: Let \(u=x^2\). Then \(du=2x\,dx\). Transform the endpoints: when \(x=0\), \(u=0^2=0\); when \(x=1\), \(u=1^2=1\). Thus
Second substitution: Let \(v=1+u\). Then \(dv=du\). The new endpoints are \(v=1+0=1\) and \(v=1+1=2\). Hence
Check: An antiderivative of the original integrand is \(e^{1+x^2}\), since
Evaluating it at the original endpoints gives \(e^{1+1^2}-e^{1+0^2}=e^2-e\), agreeing with the substitutions.
Worked Example: A Nonlinear Second Substitution
Worked Example: A Nonlinear Second Substitution
Evaluate \(\displaystyle \int_0^1 2x(1+x^2)^3\,dx\), using a second substitution after simplifying the inner expression.
Let \(u=1+x^2\), so \(du=2x\,dx\). When \(x=0\), \(u=1+0^2=1\); when \(x=1\), \(u=1+1^2=2\). The integral becomes
Now let \(v=u^2\). Then \(dv=2u\,du\), or \(u\,du=\frac12\,dv\). Since \(u^3\,du=u^2(u\,du)\), and \(u^2=v\), the transformed integrand is \(\frac{v}{2}\,dv\). The endpoints become \(v=1^2=1\) and \(v=2^2=4\). Therefore,
Check: An antiderivative in the original variable is \(\frac{(1+x^2)^4}{4}\), because
Its endpoint difference is \(\frac{(1+1^2)^4}{4}-\frac{(1+0^2)^4}{4}=\frac{16-1}{4}=\frac{15}{4}\), as required.
Worked Example: Tracking Reversed Endpoints
Worked Example: Tracking Reversed Endpoints
Evaluate \(\displaystyle \int_0^1 2(2-x)e^{(2-x)^2}\,dx\) by two substitutions. The first substitution is decreasing over the original interval, so its transformed endpoints will appear in reverse order.
Let \(u=2-x\), so \(du=-dx\). When \(x=0\), \(u=2\); when \(x=1\), \(u=1\). Substituting the integrand, differential, and endpoints gives
Now let \(v=u^2\), so \(dv=2u\,du\). At the lower \(u\)-endpoint \(2\), \(v=4\); at the upper \(u\)-endpoint \(1\), \(v=1\). Thus
The reversal is handled by the bounds, not by silently changing their order. Equivalently, reversing the final bounds would require changing the sign: \(-\int_4^1 e^v\,dv=\int_1^4 e^v\,dv\).
Check: The function \(-e^{(2-x)^2}\) is an antiderivative of the original integrand, since
Its endpoint difference is \(-e^{(2-1)^2}-(-e^{(2-0)^2})=-e+e^4=e^4-e\), which confirms the result.
A Common Mistake: Losing a Differential Factor
In the second worked example, after setting \(v=u^2\), it would be incorrect to claim that \(u^3\,du=v^3\,dv\). This replaces the power \(u^3\) with \(v^3\) but ignores the differential. In fact, \(dv=2u\,du\), so one factor of \(u\) belongs with \(du\):
The incorrect transformation would produce \(\int_1^4 v^3\,dv=\frac{4^4-1^4}{4}=\frac{255}{4}\), not the correct value \(\frac{15}{4}\). The bounds \(1\) and \(4\) are correct, but correct bounds cannot repair an incorrectly transformed differential. Each substitution must account for the entire product of the integrand and differential.
A Reliable Procedure
When a definite integral calls for successive substitutions, organize the work one variable at a time.
Write the substitution and its differential, then compute the images of both original endpoints.
Express the integrand, differential, and bounds using the new variable before proceeding.
Derive its differential from the new substitution, rewrite every factor, and transform the current bounds.
Use the final variable and its bounds throughout the evaluation, then differentiate an antiderivative in the original variable and compare endpoint values.
This procedure is useful because errors often arise from changing only part of an integral. A written intermediate line exposes missing factors, old variables left behind, or endpoints that were not updated. If an endpoint order reverses, retain that order and let the signed definite integral represent the orientation correctly.
Check Your Understanding
For each question, keep the variable, differential, and endpoints in view at every change.
- In the substitution \(u=x^2+1\) on \([0,2]\), what are the transformed endpoints?
- If \(v=u^3\), what differential relation connects \(dv\) and \(du\), and how would you transform \(u^2\,du\)?
- Why must the bounds be transformed again when a second substitution is made?
- Suppose a substitution sends the original lower endpoint to a larger new value than the original upper endpoint. What should be done with the transformed bounds?
- State the chain-rule identity that justifies the nested substitution formula when \(F'=f\).
