Why reduction formulas are useful
A reduction formula replaces an integral in a family with one or more integrals of lower complexity. Rather than treating each integral as unrelated, we identify a parameter—such as a power \(n\)—and derive a rule that connects the case \(n\) to a simpler case. Repeating the rule eventually reaches an integral that can be evaluated directly.
This workshop compares three families: powers of a logarithm, products of a polynomial and an exponential, and powers of sine. Their recurrences look different because the factors behave differently under differentiation and integration. The shared strategy is to choose integration by parts so that one part becomes simpler, state the parameter conditions, reduce to a base case, and check the result.
We use the integration-by-parts identity \(\int u\,dv=uv-\int v\,du\). It requires differentiable \(u\) and an integrable differential \(dv\) on the interval under consideration. For indefinite integrals, the resulting antiderivatives are understood up to an additive constant.
Logarithmic powers
For \(x>0\) and an integer \(n\geq 0\), define \(L_n(x)=\int(\ln x)^n\,dx\). For \(n\geq1\), take \(u=(\ln x)^n\) and \(dv=dx\). Then \(du=n(\ln x)^{n-1}dx/x\) and \(v=x\). Substitution into integration by parts gives the recurrence.
The base case is \(L_0(x)=x+C\), since \((\ln x)^0=1\). To use the recurrence, apply it repeatedly until the power is zero. The condition \(x>0\) matters: the real logarithm \(\ln x\) is defined only there.
Worked Example: Integrating a Cubed Logarithm
Evaluate \(\int(\ln x)^3\,dx\) for \(x>0\). First reduce from \(n=3\) to \(n=2\):
Next reduce \(L_2\), then \(L_1\):
Substitute these expressions in order:
Check by differentiation. The derivative of the proposed expression is
This equals the integrand, so the antiderivative is correct on \(x>0\).
Polynomial-exponential products
For a fixed nonzero constant \(a\) and integer \(n\geq0\), define \(E_n(x)=\int x^n e^{ax}\,dx\). For \(n\geq1\), choose \(u=x^n\) and \(dv=e^{ax}dx\). Then \(du=nx^{n-1}dx\) and \(v=e^{ax}/a\), where \(a\ne0\) ensures this antiderivative formula applies.
The base case is \(E_0=e^{ax}/a+C\). Each use lowers the polynomial degree by one, so the recurrence terminates after finitely many steps. When \(a=0\), the integrand is \(x^n\), and this recurrence does not apply; the power rule is the direct method.
Worked Example: Integrating a Quadratic Times an Exponential
Evaluate \(\int x^2e^{2x}\,dx\). Here \(a=2\). Apply the recurrence at \(n=2\):
At \(n=1\),
Therefore \(E_1=xe^{2x}/2-e^{2x}/4\), and substitution into \(E_2\) gives
To check, let \(P(x)=x^2/2-x/2+1/4\). The product rule gives \(\frac{d}{dx}(e^{2x}P)=e^{2x}(2P+P')\), and
Thus the derivative is \(x^2e^{2x}\), as required.
Sine-power recurrence
For an integer \(n\geq2\), write \(S_n(x)=\int\sin^n x\,dx\). Separate one factor of sine and set \(u=\sin^{n-1}x\), \(dv=\sin x\,dx\). Then \(du=(n-1)\sin^{n-2}x\cos x\,dx\) and \(v=-\cos x\). Integration by parts yields
Now use \(\cos^2 x=1-\sin^2 x\). The remaining integral becomes \(S_{n-2}-S_n\), so
Collect the \(S_n\) terms and divide by \(n\), which is nonzero for \(n\geq2\). This proves the recurrence rather than merely suggesting it:
The two base cases are \(S_0=x+C\) and \(S_1=-\cos x+C\). The recurrence decreases the exponent by two, so the appropriate base case is determined by whether the starting exponent is even or odd.
Worked Example: Reducing a Fourth Power of Sine
Evaluate \(\int\sin^4 x\,dx=S_4\). Apply the recurrence with \(n=4\):
Apply it again with \(n=2\):
Substitute this into \(S_4\):
For the check, the derivative of \(\sin^3x\cos x\) is \(3\sin^2x\cos^2x-\sin^4x\), and the derivative of \(\sin x\cos x\) is \(\cos^2x-\sin^2x\). Therefore the derivative of the proposed answer is
Using \(\cos^2x=1-\sin^2x\), this simplifies to \(\sin^4x\). The recurrence-based answer passes the differentiation check.
Choosing and checking a recurrence
The three recurrences have a common logic: choose the factor to differentiate so it becomes simpler, and choose the other factor so it has a manageable antiderivative. In the logarithm family, differentiating the logarithmic power reduces its exponent. In the exponential family, differentiating \(x^n\) reduces polynomial degree. In the sine family, integration by parts produces a term involving \(\sin^{n-2}x\) after the identity \(\cos^2x=1-\sin^2x\) is used.
Define a symbol such as \(L_n\), \(E_n\), or \(S_n\), so the recurrence can be applied without losing track of the integral.
Confirm the domain and parameter restrictions: \(x>0\) for \(\ln x\), \(a\ne0\) for the stated exponential recurrence, and \(n\geq2\) for the sine recurrence.
Keep the endpoint or algebraic terms from every recurrence step; do not stop with a lower-index integral still unevaluated.
Use product and chain rules, simplify, and confirm that the derivative equals the original integrand.
A common sign error
A frequent error in the polynomial-exponential recurrence is to use a plus sign before the lower-degree integral. For \(E_1=\int xe^{2x}\,dx\), an incorrect calculation might say
This sign contradicts integration by parts: the formula is \(uv-\int v\,du\), so the correct recurrence has a minus sign. The incorrect proposed answer differentiates to
which is not \(xe^{2x}\). Correcting the sign gives \(E_1=e^{2x}(x/2-1/4)+C\), whose derivative is \(xe^{2x}\). This illustrates why retaining signs carefully and checking by differentiation are parts of the method, not optional finishing touches.
When reduction formulas matter
Reduction formulas are especially useful when a family contains many related integrals and direct integration by parts would otherwise be repeated without a clear plan. A recurrence makes the simplification visible: the logarithmic exponent decreases by one, the polynomial degree decreases by one, and the sine exponent decreases by two. It also exposes termination conditions and base cases, helping prevent an apparently successful calculation from circling back to the original integral.
These formulas are not interchangeable. The logarithmic and polynomial-exponential recurrences are indefinite-integral relations with constants absorbed into the result. Trigonometric recurrences also include an explicit \(x\)-dependent endpoint term. In all three settings, the formula depends on its stated conditions and on careful algebra. The final differentiation check provides a direct test that the complete antiderivative—not merely the recurrence step—has been evaluated correctly.
Check Your Understanding
Answer each question by writing the recurrence or reasoning, not just a final expression.
- State the recurrence for \(L_n=\int(\ln x)^n\,dx\), its domain condition, and its base case.
- Why does the polynomial-exponential recurrence require \(a\ne0\), and what happens to the integrand when \(a=0\)?
- Use the sine-power recurrence to express \(S_5\) in terms of \(S_3\) and then \(S_1\).
- In the derivation of the sine recurrence, which identity turns the remaining cosine factor into lower powers of sine?
- What does differentiating an alleged antiderivative establish, and why is a recurrence step alone not a complete check?
