Integration by parts (BC only) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The product rule reversal integration formula, LIATE mnemonic for u/dv selection, repeated integration by parts, tabular method, definite integral integration by parts, and solving for unknown integrals in cyclic cases.

You should already know: The product rule for differentiation, Antiderivatives of basic polynomial, exponential, and trigonometric functions, u-substitution for integration.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Integration by parts (BC only)?

Integration by parts is the core integration technique that reverses the product rule for differentiation, designed to integrate products of distinct function types that cannot be solved with u-substitution alone. Per the AP Calculus BC Course and Exam Description (CED), this topic falls within Unit 6: Integration and Accumulation of Change, and accounts for approximately 1-3% of the total AP exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often combined with other integration techniques (like partial fractions or u-substitution) in multi-step FRQ problems. Unlike u-substitution, which simplifies integrals from the chain rule, integration by parts splits a complex original integral into a simpler product term and a new, easier-to-evaluate integral. All BC candidates are expected to master all variations of this technique, from routine single-step problems to more complex cyclic and repeated applications that are exclusive to the BC curriculum.

2. The Core Integration by Parts Formula

Integration by parts is derived directly from the product rule for differentiation. For differentiable functions $u(x)$ and $v(x)$, the product rule states: $$\frac{d}{dx}\left[u(x)v(x)\right]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$ Integrate both sides with respect to $x$, and simplify the left-hand side via the Fundamental Theorem of Calculus: $$u(x)v(x)=\int u^{\prime }(x)v(x)dx+\int u(x)v^{\prime }(x)dx$$ Rearrange to isolate the integral of the product $u(x)v^{\prime }(x)$, which is our original integral: $$\int udv=uv-\int vdu$$ where $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$. For definite integrals over $[a,b]$, the formula becomes: $${\int }_a^{b}udv={\left[uv\right]}_a^b-{\int }_a^{b}vdu$$ The most critical step in using this formula is correctly choosing which function is $u$ and which is $dv$. The LIATE mnemonic helps: Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential. The function that comes first in this list is always $u$, and the other is $dv$. This works because we differentiate $u$, so we want to reduce complex functions like logarithms (which are hard to integrate, easy to differentiate) to simpler functions.

Worked Example

Evaluate the indefinite integral $\int 3x\ln xdx$.

1. By LIATE, logarithmic comes before algebraic, so set $u=\ln x$, $dv=3xdx$.
2. Compute $du$ and $v$: $du=\frac{1}{x}dx$, $v=\int 3xdx=\frac{3}{2}{x}^2$.
3. Substitute into the core formula: $\int 3x\ln xdx=uv-\int vdu=\frac{3}{2}{x}^2\ln x-\int \frac{3}{2}{x}^2\cdot \frac{1}{x}dx$.
4. Simplify and integrate the remaining term: $\frac{3}{2}{x}^2\ln x-\frac{3}{2}\int xdx=\frac{3}{2}{x}^2\ln x-\frac{3}{4}{x}^2+C$.
5. Factor for a cleaner final result: $\frac{3x^2}{4}\left(2\ln x-1\right)+C$.

Exam tip: When integrating a single function like $\ln x$ or $\arctan x$ (not an explicit product), write it as $f(x)\cdot 1dx$ and set $u=f(x)$, $dv=1dx$ — this converts it to a standard integration by parts problem automatically.

3. Repeated and Cyclic Integration by Parts

Most integrals involving higher-degree polynomials, or products of exponentials and trigonometric functions, require more than one application of integration by parts. For integrals with a polynomial $u$, each application reduces the degree of $u$ by 1, so an $n$-th degree polynomial requires $n$ applications to reduce $u$ to a constant. A special, BC-exclusive case is cyclic integration by parts, which occurs when integrating products of the form $e^{ax}\sin (bx)$ or $e^{ax}\cos (bx)$. After two applications of integration by parts, the original integral reappears on the right-hand side of the equation, allowing you to solve for it algebraically. This is a very common tested problem on the BC exam, so mastering the cyclic method is critical.

Worked Example

Evaluate $\int e^{3x}\cos 2xdx$.

1. By LIATE, trigonometric comes before exponential, so set $u=\cos 2x$, $dv=e^{3x}dx$. Compute $du=-2\sin 2xdx$, $v=\frac{1}{3}{e}^{3x}$.
2. First application: $\int e^{3x}\cos 2xdx=\frac{1}{3}{e}^{3x}\cos 2x-\int \frac{1}{3}{e}^{3x}(-2\sin 2x)dx=\frac{1}{3}{e}^{3x}\cos 2x+\frac{2}{3}\int e^{3x}\sin 2xdx$.
3. Apply integration by parts a second time to the new integral: set $u=\sin 2x$, $dv=e^{3x}dx$, so $du=2\cos 2xdx$, $v=\frac{1}{3}{e}^{3x}$. This gives: $\frac{2}{3}\left(\frac{1}{3}{e}^{3x}\sin 2x-\int \frac{1}{3}{e}^{3x}(2\cos 2x)dx\right)=\frac{2}{9}{e}^{3x}\sin 2x-\frac{4}{9}\int e^{3x}\cos 2xdx$.
4. Let $I=\int e^{3x}\cos 2xdx$ to simplify notation. Substitute back into the original equation: $I=\frac{1}{3}{e}^{3x}\cos 2x+\frac{2}{9}{e}^{3x}\sin 2x-\frac{4}{9}I$.
5. Collect terms with $I$ on the left: $I+\frac{4}{9}I=\frac{e^{3x}}{9}(3\cos 2x+2\sin 2x)$, so $\frac{13}{9}I=\frac{e^{3x}}{9}(3\cos 2x+2\sin 2x)$, and $I=\frac{e^{3x}(3\cos 2x+2\sin 2x)}{13}+C$.

Exam tip: Always add the constant of integration $C$ only after you have isolated the original integral $I$ on the left-hand side of the equation. Adding $C$ early will lead to an incorrect constant multiple in your final result.

4. Tabular Integration by Parts (DI Method)

Tabular integration is a time-saving shortcut for repeated integration by parts that works exclusively when one function (the $u$ term) differentiates to zero after a finite number of steps — this is almost always a polynomial of any degree. The method organizes work into two columns: one labeled $D$ (for differentiation of $u$) and one labeled $I$ (for integration of $dv$). You differentiate $u$ repeatedly until you get zero, and integrate $dv$ the same number of times. You then add alternating signs starting with a positive sign for the first term, multiply diagonally (each $D$ entry times the next $I$ entry in the next row), and sum all terms to get the antiderivative. This method eliminates the repeated algebraic write-out of step-by-step integration by parts, reducing the chance of sign and arithmetic errors, and is fully acceptable for use on the AP exam (even in FRQ, as long as your table is shown if requested).

Worked Example

Evaluate $\int (2x^2-5x)e^{2x}dx$.

1. By LIATE, algebraic comes before exponential, so assign $u=2x^2-5x$ to the $D$ column, $dv=e^{2x}$ to the $I$ column.
2. Differentiate $D$ until 0, integrate $I$ the same number of steps:
   • Row 1 (D): $2x^2-5x$; Row 1 (I): $e^{2x}$
   • Row 2 (D): $4x-5$; Row 2 (I): $\frac{1}{2}{e}^{2x}$
   • Row 3 (D): $4$; Row 3 (I): $\frac{1}{4}{e}^{2x}$
   • Row 4 (D): $0$; Row 4 (I): $\frac{1}{8}{e}^{2x}$
3. Add alternating signs starting with + for Row 1: $+,-,+$.
4. Multiply each D entry by the I entry from the next row, multiply by the sign, and sum: $+(2x^2-5x)\left(\frac{1}{2}{e}^{2x}\right)-(4x-5)\left(\frac{1}{4}{e}^{2x}\right)+4\left(\frac{1}{8}{e}^{2x}\right)+C$
5. Simplify by factoring out $\frac{1}{4}{e}^{2x}$: $\frac{e^{2x}}{4}\left(2(2x^2-5x)-(4x-5)+2\right)+C=\frac{e^{2x}}{4}(4x^2-14x+7)+C$.

Exam tip: Never use tabular integration for cyclic cases like $e^x\sin x$. You will never reach a zero derivative, so the method will not work and will lead to an incorrect result.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Choosing $u=x$ and $dv=\ln xdx$ when integrating $\int x\ln xdx$. Why: Students mix up LIATE order, forgetting that logarithms come before algebraic functions, so they end up needing to integrate $\ln x$ for $v$, which is unnecessary and complicates the problem. Correct move: Always follow LIATE: set $u=\ln x$, $dv=xdx$ for this type of integral.
• Wrong move: In cyclic integration by parts, after getting $I=...-kI$, the student leaves the answer as $I=...$ with the $-kI$ still on the right. Why: Students forget that the original integral reappears and needs to be solved for algebraically. Correct move: Assign the original integral the placeholder name $I$ at the start of the problem, so you can easily collect like terms.
• Wrong move: For definite integration by parts, the student leaves the boundary term $[uv{]}_a^b$ as a function of $x$ instead of evaluating it at the bounds. Why: Students focus on integrating the $\int vdu$ term and forget the boundary term entirely. Correct move: Always evaluate and simplify $[uv{]}_a^b$ first, before integrating the second term.
• Wrong move: In tabular integration, the student starts alternating signs with a negative first term. Why: Students confuse the sign flip from the $-\int vdu$ term in the core formula. Correct move: Write the sign explicitly next to each row of the table, starting with $+$ for the first row, to avoid this error.
• Wrong move: Claiming $\int \arctan xdx$ cannot be solved with integration by parts because it is not a product of two functions. Why: Students think integration by parts only works for explicit products. Correct move: Rewrite the integral as $\int 1\cdot \arctan xdx$, set $u=\arctan x$, $dv=1dx$, and proceed with the standard formula.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Evaluate ${\int }_0^{\pi }x\sin xdx$. Which of the following is the correct result? A) $-\pi$ B) $\pi$ C) $0$ D) $1$

Worked Solution: Use integration by parts with $u=x$, $dv=\sin xdx$, so $du=dx$ and $v=-\cos x$. Apply the definite integral formula: ${\int }_0^{\pi }x\sin xdx={\left[-x\cos x\right]}_0^{\pi }-{\int }_0^{\pi }(-\cos x)dx=\left(-\pi \cos \pi -0\right)+{\int }_0^{\pi }\cos xdx=(\pi )+{\left[\sin x\right]}_0^{\pi }=\pi +0=\pi$. The correct answer is B.

Question 2 (Free Response)

Consider the function $f(x)=x^2\cos x$ on the interval $[0,2\pi ]$. (a) Evaluate the indefinite integral $\int x^2\cos xdx$. (b) Evaluate the definite integral ${\int }_0^{2\pi }{x}^2\cos xdx$, the net area under $f(x)$ on $[0,2\pi ]$. (c) The region bounded by $y=f(x)$, the x-axis, $x=0$, and $x=2\pi$ is rotated around the x-axis. Set up (but do not evaluate) the integral for the volume of the resulting solid.

Worked Solution: (a) Use tabular integration: $u=x^2$, $dv=\cos xdx$. D column: $x^2,2x,2,0$; I column: $\cos x,\sin x,-\cos x,-\sin x$; signs: $+,-,+$. Sum terms: $+x^2(\sin x)-2x(-\cos x)+2(-\sin x)+C=x^2\sin x+2x\cos x-2\sin x+C$. (b) Evaluate the antiderivative from (a) at the bounds: ${\left[x^2\sin x+2x\cos x-2\sin x\right]}_0^{2\pi }=\left(0+2(2\pi )(1)-0\right)-\left(0+0-0\right)=4\pi$. (c) By the disk method for volume of revolution, $V=\pi {\int }_0^{2\pi }{\left[f(x)\right]}^{2}dx=\pi {\int }_0^{2\pi }{x}^4{\cos }^{2}xdx$.

Question 3 (Application / Real-World Style)

The marginal revenue of a company producing $x$ thousand units of a good is given by $R^{\prime }(x)=x{e}^{-x/3}$ thousand dollars per thousand units, for $0\le x\le 6$. Find the total revenue earned from producing the first 6 thousand units, rounded to the nearest thousand dollars.

Worked Solution: Total revenue $R={\int }_0^6{R}^{\prime }(x)dx={\int }_0^{6}x{e}^{-x/3}dx$. Use integration by parts: $u=x$, $dv=e^{-x/3}dx$, so $du=dx$, $v=-3e^{-x/3}$. Apply the formula: $R={\left[-3x{e}^{-x/3}\right]}_0^6-{\int }_0^6(-3e^{-x/3})dx=-18e^{-2}+3{\int }_0^6{e}^{-x/3}dx=-18e^{-2}+3{\left[-3e^{-x/3}\right]}_0^6=-18e^{-2}-9(e^{-2}-1)=-27e^{-2}+9\approx -27(0.1353)+9\approx -3.65+9=5.35\approx 5$. The total revenue from producing the first 6 thousand units is approximately 5 thousand dollars.

7. Quick Reference Cheatsheet

8. What's Next

Integration by parts is a foundational technique for all advanced integration methods you will learn next in Unit 6, including integration by partial fractions and integration of improper integrals. Many multi-step AP exam problems require combining integration by parts with u-substitution or partial fractions, so mastering u/dv selection and avoiding common sign errors is critical to solving these problems efficiently and correctly. Beyond integration, integration by parts is also used to derive reduction formulas for power integrals, which occasionally appear on MCQ, and it is used in solving many applied differential equation problems for BC. Without a solid command of this technique, you will struggle with the more complex integration problems that make up a significant portion of the BC exam.

• Integration by partial fractions
• Improper integrals
• Volume of revolution
• First-order differential equations