--- title: "AP Calculus BC 6.11: Integration by Parts" description: "Review AP Calculus BC integration by parts, including the formula, LIATE strategy, repeated and cyclic integrals, definite integrals, and exam setup." canonical: "https://fiveable.me/ap-calc/unit-6/integrating-using-integration-by-parts/study-guide/O4P3LchNoZnWElf8zETV" type: "study-guide" subject: "AP Calculus AB/BC" unit: "Unit 6 – Integration and Accumulation of Change" lastUpdated: "2026-06-09" --- # AP Calculus BC 6.11: Integration by Parts ## Summary Review AP Calculus BC integration by parts, including the formula, LIATE strategy, repeated and cyclic integrals, definite integrals, and exam setup. ## Guide Integration by parts is a BC-only technique for integrating a product of two [functions](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink"). You rewrite the integral using the formula $\int u\,dv = uv - \int v\,du$, picking $u$ and $dv$ so the new integral is easier than the one you started with. For AP Calculus BC, track the minus sign and your choice of $u$ and $dv$ carefully. ## Why This Matters for the AP Calculus Exam Integration by parts shows up only on the AP Calculus BC exam, not AB. It is one of several antidifferentiation tools you are expected to recognize and apply, and BC questions often mix techniques so you have to decide which one fits an [integrand](/ap-calc/key-terms/integrand "fv-autolink"). This skill supports both finding [indefinite integrals](/ap-calc/key-terms/indefinite-integrals "fv-autolink") and evaluating definite integrals, and it connects to later BC topics like improper integrals. On both multiple-choice and free-response work, you will need to recognize when a product of an algebraic factor and a transcendental factor (like $x e^x$ or $x^2\cos x$) calls for integration by parts, set it up cleanly, and carry signs correctly through repeated steps. ## Key Takeaways - The formula is $\int u\,dv = uv - \int v\,du$, and it comes from reversing [the product rule](/ap-calc/unit-2/product-rule/study-guide/qQXYTmpHvjsqAWOVzcEw "fv-autolink"). - Choose $u$ and $dv$ so the new integral $\int v\,du$ is simpler than the original. - Use LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to pick $u$. - Some integrals need integration by parts more than once; track signs carefully each time. - For cyclic integrals (like $\int e^x\cos x\,dx$), apply the method twice, get the original integral back, then solve algebraically. - Include $+C$ for indefinite integrals, and for definite integrals evaluate the $uv$ boundary term at the [limits](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/defining-convergent-divergent-infinite-series/study-guide/CIVFHStGQM90EJ4GtIDB "fv-autolink") too. ## Integration by Parts Basics Take a look at the following integral: $$ \int x^2 \sin(x)\,dx $$ You cannot finish this with the tools you already have: [substitution](/ap-calc/unit-6/integrating-functions-using-long-division-completing-square/study-guide/ju79RFY6f5aKWjFK "fv-autolink") fails, and you only know how to integrate $x^2$ and $\sin(x)$ separately. Integration by parts is built for exactly this situation. The method comes from reversing the product rule for differentiation. Here is the product rule, with $u$ and $v$ as two functions: $$ \frac{d}{dx} uv = \textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\textcolor{red}{ v}\cdot \textcolor{teal}{u'} $$ If you integrate both sides, you get: $$ uv = \int\textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\int\textcolor{red}{ v}\cdot \textcolor{teal}{u'} $$ Rearranging gives the integration by parts formula: $$ \int \textcolor{blue}{u} \, \textcolor{pink}{dv} = uv - \int \textcolor{red}{v} \, \textcolor{teal}{du} $$ Where: - $\textcolor{blue}{u}$ and $\textcolor{pink}{dv}$ are selected parts of the integrand. - $\textcolor{teal}{du}$ is the [derivative](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/finding-taylor-polynomial-approximations-functions/study-guide/LszguYzKz0M6GdqTRSr6 "fv-autolink") of $\textcolor{blue}{u}$ with respect to the variable of integration. - $\textcolor{red}{v}$ is the [antiderivative](/ap-calc/key-terms/antiderivative "fv-autolink") of $\textcolor{pink}{dv}$. ## Application and Strategy To apply integration by parts, follow these steps: 1. **Select $u$ and $dv$ carefully.** Choose them so the resulting integral is easier to handle. The LIATE mnemonic helps you pick $u$ quickly. 2. **Differentiate and integrate.** Compute $du$ from $u$ and $v$ from $dv$. 3. **Apply the formula.** Plug $u$, $dv$, $v$, and $du$ into the integration by parts formula. 4. **Evaluate the new integral.** The integral on the right may still need work. Repeat integration by parts if necessary. 5. **Solve for the original integral.** If the same integral shows up on both sides, use algebra to solve for it. **LIATE Mnemonic for choosing $u$:** - **L**ogarithmic functions - **I**nverse [trigonometric functions](/ap-calc/key-terms/trigonometric-functions "fv-autolink") - **A**lgebraic functions - **T**rigonometric functions - **E**xponential functions Whatever appears earlier in LIATE usually becomes $u$, and the rest becomes $dv$. ## Integration by Parts Practice Problems ### Example 1 Evaluate using integration by parts: $$ \int xe^x \, dx $$ Go through LIATE to choose $u$. Algebraic comes before Exponential, so set $u = x$ and $dv = e^x\,dx$. Now collect the parts: - $u = x$ - $du = dx$ - $v = e^x$ - $dv = e^x \,dx$ Apply the formula: $$ \int x \cdot e^x \, dx = x \cdot e^x - \int e^x \, dx $$ Evaluate the new integral: $$ \int x \cdot e^x \, dx = x \cdot e^x - e^x + C $$ Therefore, $$ \int xe^x \, dx = x e^x - e^x + C $$ ### Example 2 Evaluate: $$ \int \ln(x)\,dx $$ This does not look like a product at first. Rewrite the integrand as $1 \cdot \ln(x)$ so you have two functions, then use LIATE. Logarithmic ranks ahead of Algebraic, so set $u = \ln(x)$ and $dv = 1\,dx$. - $u = \ln(x)$ - $du = \frac{1}{x} \,dx$ - $v = x$ - $dv = 1 \,dx$ Apply integration by parts: $$ \int \ln(x)\,dx = x\ln(x)- \int x\cdot \frac{1}{x}\,dx $$ The integral on the right simplifies: $$ x\ln(x)- \int dx $$ Since $\int dx = x$, $$ \int \ln(x)\,dx = x\ln(x)- x + C $$ ### Example 3 Evaluate: $$ \int x^2 \cos(x) \, dx $$ Pick $u$ and $dv$ with LIATE, then find $du$ and $v$. - $u = x^2$ - $du = 2x\,dx$ - $v = \sin x$ - $dv = \cos x\,dx$ Apply the formula: $$ \int x^2 \cos(x) \, dx = x^2 \cdot \sin(x) - \int 2x \cdot \sin(x) \, dx $$ The remaining integral $\int 2x \cdot \sin(x) \, dx$ needs integration by parts again. Choose: - $u = 2x$ - $du = 2\,dx$ - $v = -\cos x$ - $dv = \sin x\,dx$ So: $$ \int 2x \cdot \sin(x) \, dx = -2x \cdot \cos(x) - \int (-2) \cdot \cos(x) \, dx = -2x\cos(x) + 2\int \cos(x)\,dx $$ Now substitute back. Watch the sign: you are subtracting the whole expression above. $$ \int x^2 \cos(x) \, dx = x^2 \sin(x) - \left(-2x\cos(x) + 2\int \cos(x)\,dx\right) = x^2 \sin(x) + 2x\cos(x) - 2\int \cos(x)\,dx $$ Finally, $\int \cos(x)\,dx = \sin(x)$, so: $$ \int x^2 \cos(x) \, dx = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C $$ ### Challenge Question Evaluate: $$ \int e^x \cos x\,dx $$ Try it before checking the answer below. > Hint: Use integration by parts twice, keeping $e^x$ in the same role both times. $$ \int e^x \cos x\,dx = \frac{e^x\sin x + e^x\cos x}{2} + C $$ When you apply integration by parts twice without switching which factor is $u$, you get the original integral back on the right side. Move both copies of the integral to the same side and divide by two to finish. ## How to Use This on the AP Calculus Exam ### Problem Solving - Scan the integrand for a product of an algebraic factor and a transcendental factor (exponential, trig, log, or inverse trig). That pattern signals integration by parts. - Use LIATE to pick $u$ so that $du$ gets simpler. With polynomials times $e^x$, $\sin x$, or $\cos x$, the polynomial is usually $u$ because its derivative eventually drops to a constant. - Write out $u$, $du$, $v$, and $dv$ before plugging in. This keeps signs and terms organized when you repeat the process. ### Common Trap - For repeated integration by parts, the most common error is a dropped or flipped sign on the $uv$ term or the integral being subtracted. Carry the full minus sign through every step. - For cyclic integrals like $\int e^x\sin x\,dx$ or $\int e^x\cos x\,dx$, do not switch roles between the two rounds. If you do, you just undo your first step. ### Free Response - For a definite integral, evaluate the $uv$ boundary term at the upper and lower limits, not just the leftover integral. - Show a clear antiderivative and include $+C$ for indefinite integrals. Avoid stringing together equal signs between expressions that are not actually equal. Clean notation makes your work easier to follow. ## Common Misconceptions - **Integration by parts is not just "reverse product rule applied directly."** You are reorganizing the product rule into a usable formula, then choosing $u$ and $dv$ strategically. Picking them poorly can make the integral harder. - **LIATE chooses $u$, not $dv$.** The function earlier in LIATE becomes $u$; everything else (including $dx$) becomes $dv$. - **Signs matter when you repeat the method.** When you substitute a second integration by parts result back in, you are subtracting the entire expression, so distribute the minus sign across every term. - **You still need $+C$.** Indefinite integrals require a [constant of integration](/ap-calc/key-terms/constant-of-integration "fv-autolink") even after multiple steps. - **Cyclic integrals are solved with algebra, not endless repetition.** Once the original integral reappears, stop and solve for it instead of applying the method again. - **This is BC-only material.** Integration by parts is assessed on the AP Calculus BC exam, so AB students will not see it tested. ## Related AP Calculus Guides - [Unit 6 Overview: Integration and Accumulation of Change](/ap-calc/unit-6/review/study-guide/GcRakhgwYqAcIAg0y7WB) - [6.1 Integration and Accumulation of Change](/ap-calc/unit-6/integration-accumulation-change/study-guide/NWRV9MaRJO4Eno32l5Xp) - [6.12 Integrating Using Linear Partial Fractions](/ap-calc/unit-6/using-linear-partial-fractions/study-guide/VNjHMatlmaZoFyt7yLPH) - [6.5 Interpreting the Behavior of Accumulation Functions Involving Area](/ap-calc/unit-6/interpreting-behavior-accumulation-functions-involving-area/study-guide/uVNoabC47nUgOdjr) - [6.4 The Fundamental Theorem of Calculus and Accumulation Functions](/ap-calc/unit-6/fundamental-theorem-calculus-accumulation-functions/study-guide/TyDEzN9M5wieA4Kw) - [6.13 Evaluating Improper Integrals](/ap-calc/unit-6/evaluation-improper-integrals/study-guide/DRGur03jVQG3WXLiNyuz) ## Vocabulary - **antiderivative**: Functions whose derivative equals a given function; the reverse process of differentiation. - **definite integral**: The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis. - **indefinite integral**: Antiderivatives of a function, represented as ∫f(x)dx = F(x) + C, where C is an arbitrary constant. - **integrands**: The function being integrated in an integral expression. - **integration by parts**: A technique for finding antiderivatives of products of functions, based on the product rule for derivatives. ## FAQs ### What is integration by parts? Integration by parts is a Calculus BC integration technique for products of functions. It rewrites an integral using the reverse of the product rule. ### What is the integration by parts formula? The formula is integral u dv equals uv minus integral v du. You choose u and dv from the original integrand. ### How do I choose u for integration by parts? Use LIATE as a guide: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. The earlier type usually works best as u. ### When do I need integration by parts more than once? Use it more than once when the new integral is still a product that can be simplified by repeating the method, such as powers of x times trig or exponential functions. ### What is a cyclic integration by parts problem? A cyclic problem is one where applying integration by parts twice brings back the original integral, so you solve algebraically for that original integral. ### Is integration by parts on AP Calculus AB? No. Integration by parts is tested on AP Calculus BC, not AP Calculus AB. ## Structured Data ```json {"@context":"https://schema.org","@type":"FAQPage","inLanguage":"en","mainEntity":[{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-6/integrating-using-integration-by-parts/study-guide/O4P3LchNoZnWElf8zETV#what-is-integration-by-parts","name":"What is integration by parts?","acceptedAnswer":{"@type":"Answer","text":"Integration by parts is a Calculus BC integration technique for products of functions. It rewrites an integral using the reverse of the product rule."}},{"@type":"Question","@id":"https://fiveable.me/ap-calc/unit-6/integrating-using-integration-by-parts/study-guide/O4P3LchNoZnWElf8zETV#what-is-the-integration-by-parts-formula","name":"What is the integration by parts formula?","acceptedAnswer":{"@type":"Answer","text":"The formula is integral u dv equals uv minus integral v du. 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