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Math101learn.math101.caIntegration by Parts
A rigorous, example-driven guide to integration by parts, including hypotheses, method choice, verification, and practice.
The central idea
The product rule rearranges to integration by parts: $\int u\,dv=uv-\int v\,du$. For definite integrals, evaluate $uv$ at both bounds and subtract the remaining definite integral. The method is useful when differentiating one factor simplifies it while the other factor is readily integrated.
Definitions, hypotheses, and notation
A useful preference list—logarithmic, inverse trigonometric, algebraic, trigonometric, exponential—can suggest $u$, but it is a heuristic, not a theorem. The real test is whether $du$ simplifies and $dv$ integrates. For polynomial times exponential or sine, repeated integration by parts eventually differentiates the polynomial to zero.
For integrals such as $\int e^x\cos xdx$, applying parts twice reproduces the original integral. Collecting that term algebraically then solves the equation. Constants of integration should be added only after the indefinite equation is resolved.
Conceptual meaning
Integration by parts transfers a derivative from one factor to another. It does not eliminate complexity automatically; the choices of $u$ and $dv$ should make the new integral simpler or create an equation involving the original integral.
A dependable method and decision rule
- Factor the integrand conceptually into a choice of $u$ and $dv$.
- Prefer a $u$ that simplifies under differentiation and a $dv$ with known antiderivative.
- Compute $du$ and $v$ explicitly.
- Substitute into $uv-\int vdu$ with the minus sign visible.
- Repeat, solve algebraically, or apply bounds as the resulting structure requires.
Fully worked example
Graphical or geometric meaning
The formula balances two product-rule contributions. The product differential is $d(uv)=u\,dv+v\,du$; moving the second accumulated contribution to the other side creates the subtraction in integration by parts.
Common mistakes and why they fail
Verification and reasonableness checks
- Differentiate the final antiderivative.
- Compare complexity before and after the transfer.
- For definite integrals, estimate sign and apply all bounds consistently.
Keep the product differential visible
The identity follows from $d(uv)=u\,dv+v\,du$, so $\int u\,dv=uv-\int v\,du$. For definite integrals, write $[uv]_a^b-\int_a^b v\,du$ so the boundary term remains visible. A mnemonic may suggest $u$, but the real criterion is whether differentiating $u$ and integrating $dv$ make the new integral simpler. Polynomial factors work well because repeated differentiation eventually reaches zero. Cyclic examples such as $\int e^x\cos x\,dx$ require applying the rule twice and solving algebraically for the original integral. Differentiate the final result to check the minus sign, product term, and constant factors together. Add the integration constant only after resolving any equation involving the original integral.
Practice
- Evaluate $\int x\cos x dx$.
- Evaluate $\int\ln x dx$.
- Which factor is a natural $u$ in $\int x^2e^x dx$?
Answers and brief solutions
- $x\sin x+\cos x+C$.
- $x\ln x-x+C$.
- $x^2$.
Connections and next steps
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Which antiderivative equals ∫x e^x dx?
- du=dx and v=e^x.
- ∫u dv=uv−∫v du=xe^x−e^x+C.
- Factor to get e^x(x−1)+C.
End of lesson
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