Explain integration by parts.
Understand the Problem
The user is asking about integration by parts, which is a calculus technique used to integrate the product of two functions. This is a request for information, not a specific problem to solve.
Answer
The integration by parts formula is $\int u \, dv = uv - \int v \, du$
Answer for screen readers
The integration by parts formula is: $$ \int u , dv = uv - \int v , du $$ Examples:
- $\int x \cos(x) , dx = x \sin(x) + \cos(x) + C$
- $\int x e^x , dx = x e^x - e^x + C$
- $\int \ln(x) , dx = x \ln(x) - x + C$
- $\int x^2 e^x , dx = x^2 e^x - 2x e^x + 2 e^x + C$
Steps to Solve
- Introduce Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It's used when integrating a product of two functions. The formula is: $$ \int u , dv = uv - \int v , du $$ where $u$ and $v$ are functions of $x$.
- Choosing $u$ and $dv$
The key to using integration by parts is selecting appropriate $u$ and $dv$. A helpful guideline is the acronym LIATE:
- Logarithmic functions
- Inverse trigonometric functions
- Algebraic functions
- Trigonometric functions
- Exponential functions
This order suggests which function should be $u$. Choose $u$ based on what comes first in the list; the rest becomes $dv$. The goal is to choose $u$ such that its derivative is simpler than $u$ itself, and $dv$ such that it's easy to integrate.
- Example 1: $\int x \cos(x) , dx$
Let $u = x$ and $dv = \cos(x) , dx$. Then, $du = dx$ and $v = \int \cos(x) , dx = \sin(x)$. Applying the integration by parts formula:
$$ \int x \cos(x) , dx = x \sin(x) - \int \sin(x) , dx = x \sin(x) + \cos(x) + C $$
- Example 2: $\int x e^x , dx$
Let $u = x$ and $dv = e^x , dx$. Then, $du = dx$ and $v = \int e^x , dx = e^x$. Applying the integration by parts formula:
$$ \int x e^x , dx = x e^x - \int e^x , dx = x e^x - e^x + C $$
- Example 3: $\int \ln(x) , dx$
Let $u = \ln(x)$ and $dv = dx$. Then, $du = \frac{1}{x} dx$ and $v = \int dx = x$. Applying the integration by parts formula:
$$ \int \ln(x) , dx = x \ln(x) - \int x \cdot \frac{1}{x} , dx = x \ln(x) - \int 1 , dx = x \ln(x) - x + C $$
- Repeated Integration by Parts
Sometimes, you may need to apply integration by parts more than once. For example, consider $\int x^2 e^x , dx$. First, let $u = x^2$ and $dv = e^x , dx$. Then $du = 2x , dx$ and $v = e^x$.
$$ \int x^2 e^x , dx = x^2 e^x - \int 2x e^x , dx $$
Now we need to integrate $\int 2x e^x , dx$. Let $u = 2x$ and $dv = e^x , dx$. Then $du = 2 , dx$ and $v = e^x$.
$$ \int 2x e^x , dx = 2x e^x - \int 2 e^x , dx = 2x e^x - 2 e^x + C_1$$
Substitute back: $$ \int x^2 e^x , dx = x^2 e^x - (2x e^x - 2 e^x) + C = x^2 e^x - 2x e^x + 2 e^x + C $$
The integration by parts formula is: $$ \int u , dv = uv - \int v , du $$ Examples:
- $\int x \cos(x) , dx = x \sin(x) + \cos(x) + C$
- $\int x e^x , dx = x e^x - e^x + C$
- $\int \ln(x) , dx = x \ln(x) - x + C$
- $\int x^2 e^x , dx = x^2 e^x - 2x e^x + 2 e^x + C$
More Information
Integration by parts is a powerful technique for integrating products of functions, especially when simpler methods don't work. The LIATE rule is a useful guide for selecting $u$ and $dv$, but practice is key to mastering this technique. Sometimes one must apply integration by parts multiple times to arrive at the final answer.
Tips
- Incorrectly choosing $u$ and $dv$: This is the most common mistake. Using LIATE helps, but understanding which choice leads to simpler integrals is crucial.
- Forgetting the constant of integration: Always remember to add $+ C$ after performing indefinite integration.
- Incorrectly calculating derivatives or integrals: Errors in finding $du$ or $v$ will lead to an incorrect result.
- Sign errors: Be careful with signs, especially when substituting back into the integration by parts formula.
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