Integrating Using Integration by Parts (BC ONLY) - AP Calculus BC Study Guide

The Core Idea: Integrating Using Integration by Parts (BC ONLY)

Integration by Parts is a fundamental technique used to find the antiderivative of a function that can be expressed as the product of two other functions. While the power rule, substitution, and other basic integration rules handle many forms, they are often insufficient for integrands involving products like $x\cos (x)$ or $x^2{e}^x$. This method provides a systematic way to tackle such integrals.

The entire technique is derived directly from the product rule for differentiation, $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. By integrating both sides of the product rule and rearranging the terms, we arrive at the integration by parts formula. In essence, integration by parts is the "inverse" of the product rule, allowing us to transform a difficult product integral into a potentially simpler one. The success of the method hinges on a strategic choice of which part of the product to differentiate and which part to integrate.

Key Formulas

The integration by parts method is based on a single core formula, which has forms for both indefinite and definite integrals.

Indefinite Integrals

The standard formula for integration by parts is expressed as:

$$\int udv=uv-\int vdu$$

• $\int udv$ represents the original integral you are trying to solve.

• $u$ is a part of the original integrand that you will differentiate to find $du$.

• $dv$ is the remaining part of the original integrand (including $dx$) that you will integrate to find $v$.

• $\int vdu$ is the new integral that results from the process, which should ideally be simpler to evaluate than the original.

Definite Integrals

When applying integration by parts to a definite integral from $a$ to $b$, the formula is adjusted to account for the bounds of integration:

$${\int }_a^{b}udv=[uv{]}_a^b-{\int }_a^{b}vdu$$

This is equivalent to:

$${\int }_a^{b}udv=(u(b)v(b)-u(a)v(a))-{\int }_a^{b}vdu$$

The term $[uv{]}_a^b$ is evaluated using the Fundamental Theorem of Calculus before the new integral ${\int }_a^{b}vdu$ is solved.

Understanding the Choice of $u$ and $dv$

The effectiveness of the integration by parts method depends entirely on the strategic selection of $u$ and $dv$ from the original integrand. An incorrect choice can lead to a new integral that is more complex than the original, defeating the purpose of the technique.

The primary goal is to choose $u$ and $dv$ such that the new integral, $\int vdu$, is easier to solve. This is generally achieved by following two principles:

1. Choosing $u$: The function chosen for $u$ should become simpler (or at least no more complex) after differentiation. Polynomials are excellent choices for $u$ because their degree decreases with each differentiation. For example, $x^3$ becomes $3x^2$, then $6x$, then $6$, and finally $0$.

2. Choosing $dv$: The expression chosen for $dv$ must be a function that you can readily integrate to find $v$. For example, if the integrand is $x\ln (x)$, choosing $dv=\ln (x)dx$ would be difficult, as the integral of $\ln (x)$ is not immediately obvious. However, choosing $dv=xdx$ is straightforward.

The process is a trade-off: you differentiate $u$ to get $du$ and integrate $dv$ to get $v$. The hope is that the product $vdu$ is a simpler expression to integrate than the original product $udv$.

Core Concepts & Rules

• Purpose: Integration by parts is a technique specifically used to find antiderivatives of functions that are expressed as a product.

• Origin: The formula is not arbitrary; it is derived by integrating the product rule for derivatives, $d(uv)=udv+vdu$.

• The Formula: The operational formula for the method is $\int udv=uv-\int vdu$.

• Strategic Choice: The success of the method is not automatic. It depends critically on the choice of which part of the integrand is designated as $u$ and which is $dv$. The goal is to make the new integral, $\int vdu$, simpler than the original.

• Repeated Application: For some functions, a single application of integration by parts is not sufficient to solve the integral. The process may need to be applied multiple times. This is common when one of the functions in the product is a polynomial of degree greater than one.

Step-by-Step Example 1: Basic Application

Problem: Find the indefinite integral $\int x\sin (x)dx$.

This integral involves a product of two functions, $x$ and $\sin (x)$, making it a prime candidate for integration by parts.

Step 1: Choose $u$ and $dv$

We need to select $u$ and $dv$. Let's consider our options based on the principle of simplifying the new integral.

• Option A: Let $u=x$ and $dv=\sin (x)dx$. Differentiating $u$ gives $du=dx$, which is simpler. Integrating $dv$ gives $v=-\cos (x)$, which is no more complex. The new integral will involve $(-\cos (x))(dx)$, which is simple.

• Option B: Let $u=\sin (x)$ and $dv=xdx$. Differentiating $u$ gives $du=\cos (x)dx$. Integrating $dv$ gives $v=\frac{1}{2}{x}^2$. The new integral would involve $(\frac{1}{2}{x}^2)(\cos (x)dx)$, which is more complex than the original.

Option A is the correct strategic choice.

• Let $u=x$

• Let $dv=\sin (x)dx$

Step 2: Find $du$ and $v$

• Differentiate $u$ to find $du$:

  $u=x⟹du=1\cdot dx=dx$

• Integrate $dv$ to find $v$:

  $dv=\sin (x)dx⟹v=\int \sin (x)dx=-\cos (x)$

Step 3: Apply the Integration by Parts Formula

Substitute $u,v,du,dv$ into the formula $\int udv=uv-\int vdu$.

$$\int x\sin (x)dx=(x)(-\cos (x))-\int (-\cos (x))dx$$

Step 4: Simplify and Solve the New Integral

Simplify the expression and evaluate the remaining integral.

$$\int x\sin (x)dx=-x\cos (x)-(-\int \cos (x)dx)$$

$$=-x\cos (x)+\int \cos (x)dx$$

The new integral is a basic one:

$$\int \cos (x)dx=\sin (x)+C$$

Step 5: Write the Final Answer

Substitute the result of the new integral back into the expression and include the constant of integration.

$$\int x\sin (x)dx=-x\cos (x)+\sin (x)+C$$

Step-by-Step Example 2: Repeated Application

Problem: Evaluate the definite integral ${\int }_0^1{x}^2{e}^{x}dx$.

This integral requires integration by parts. Because the polynomial part ($x^2$) has a degree of 2, we anticipate needing to apply the method twice.

Step 1: First Application of Integration by Parts

• Choose $u$ and $dv$: Let $u=x^2$ (since it simplifies upon differentiation) and $dv=e^{x}dx$.

• Find $du$ and $v$:

  • $u=x^2⟹du=2xdx$

  • $dv=e^{x}dx⟹v=\int e^{x}dx=e^x$

• Apply the definite integral formula:${\int }_a^{b}udv=[uv{]}_a^b-{\int }_a^{b}vdu$

$${\int }_0^1{x}^2{e}^{x}dx=[x^2{e}^x{]}_0^1-{\int }_0^1(e^x)(2xdx)$$

Step 2: Simplify and Prepare for the Second Application

First, evaluate the $[uv]$ term. Then, simplify the new integral.

$$[x^2{e}^x{]}_0^1=(1^2{e}^1)-(0^2{e}^0)=e-0=e$$

The new integral is:

$${\int }_0^{1}2x{e}^{x}dx$$

So our expression is now:

$${\int }_0^1{x}^2{e}^{x}dx=e-{\int }_0^{1}2x{e}^{x}dx$$

The new integral still contains a product, so we must apply integration by parts again.

Step 3: Second Application of Integration by Parts

Focus on ${\int }_0^{1}2x{e}^{x}dx$.

• Choose $u$ and $dv$: Let $u=2x$ and $dv=e^{x}dx$.

• Find $du$ and $v$:

  • $u=2x⟹du=2dx$

  • $dv=e^{x}dx⟹v=e^x$

• Apply the formula:

$${\int }_0^{1}2x{e}^{x}dx=[2x{e}^x{]}_0^1-{\int }_0^1(e^x)(2dx)$$

Step 4: Solve the Final Integral and Substitute Back

Evaluate the parts of the second application.

• $[2x{e}^x{]}_0^1=(2(1)e^1)-(2(0)e^0)=2e-0=2e$

• ${\int }_0^{1}2e^{x}dx=[2e^x{]}_0^1=(2e^1)-(2e^0)=2e-2$

Putting the second application together:

$${\int }_0^{1}2x{e}^{x}dx=2e-(2e-2)=2$$

Step 5: Combine Results for the Final Answer

Substitute the result from Step 4 back into the expression from the end of Step 2.

$${\int }_0^1{x}^2{e}^{x}dx=e-\left({\int }_0^{1}2x{e}^{x}dx\right)$$

$$=e-(2)$$

$$=e-2$$

Using Your Calculator

Integration by parts is a purely analytical technique for finding an antiderivative. A graphing calculator cannot perform the symbolic steps of choosing $u$ and $dv$ or applying the formula $\int udv=uv-\int vdu$. You must be able to perform the entire procedure by hand.

However, the calculator is an excellent tool for checking your work on definite integrals.

To verify the result of Example 2, ${\int }_0^1{x}^2{e}^{x}dx=e-2$:

1. Calculate the exact answer: Find a decimal approximation for your analytical answer.

   $e-2\approx 2.71828-2=0.71828$

2. Use the calculator's numerical integration feature: On a TI-84 style calculator, this is typically found under the MATH menu as fnInt( or by pressing ALPHAWINDOW and selecting $\text{intf}(x)dx$.

3. Enter the original integral:

   • fnInt(X^2 * e^(X), X, 0, 1)

   • The calculator will return a numerical approximation, which should be very close to $0.71828$.

4. Compare: If the calculator's result matches your analytical result's approximation, you can be confident in your by-hand calculation. If they differ, re-check your work for errors in differentiation, integration, or algebra.

AP Exam Quick Hit

Common Question Types

• Standard Indefinite Integral: You will be asked to find the antiderivative of a product of functions that requires a single application of integration by parts.

  • Example: Find $\int 5x{\sec }^2(x)dx$.

• Definite Integral with Repeated Application: You will be asked to evaluate a definite integral where the integrand requires applying the integration by parts formula more than once.

  • Example: Evaluate ${\int }_1^e(\ln x{)}^{2}dx$.

• Integration by Parts with Transcendental Functions: Questions often involve products of polynomial, logarithmic, exponential, and trigonometric functions.

  • Example: Find the antiderivative of $f(x)=e^{2x}\cos (x)$.

Common Mistakes

• Incorrect Choice of $u$ and $dv$: Choosing $u$ and $dv$ such that the resulting integral $\int vdu$ is more complicated than the original. For $\int x{e}^{x}dx$, choosing $u=e^x$ and $dv=xdx$ is a classic mistake that leads to a more difficult integral.

• Sign Errors in the Formula: Forgetting the minus sign in the formula and writing $uv+\int vdu$ instead of the correct $uv-\int vdu$.

• Forgetting + C: Omitting the constant of integration + C on all indefinite integral problems.

• Errors with Definite Integrals: Incorrectly applying the bounds. A common error is to only evaluate $uv$ at the upper bound $b$ instead of correctly calculating $[uv{]}_a^b=u(b)v(b)-u(a)v(a)$. Another is forgetting to carry the bounds over to the new integral, ${\int }_a^{b}vdu$.

• Algebraic Errors in Substitution: When performing repeated integration by parts, it is easy to make a sign error or drop a coefficient when substituting the result of the second integration back into the expression for the first. For example, forgetting to distribute the negative sign in an expression like $e-(2e-2)$.