Calculus 2 Lecture 7.1: Integration By Parts

Section 7.1: Introduction to New Integration Techniques

Two Main Types of Integration:

1. Direct Integration: If a function fits the integration table, it can be integrated directly without any modifications. This is often the simplest approach and involves recognizing standard integral forms.

2. Substitution: If a function does not align with those found in the integration table, substitution techniques are employed to transform the function into a form that can be integrated directly. This often involves identifying a part of the function as a new variable to simplify the integral.

Substitution and the Chain Rule

• Understanding Substitution: The process of substitution is essentially the inverse of the chain rule used in differentiation. In calculus, when we have a composite function, the derivative is found through the chain rule.

• Key Note: Unlike derivatives, there isn't a straightforward, analogous product rule for integrals. This makes the understanding of each integral unique.

Investigating the Product Rule

• Basic Product Rule: The product rule states that if f(x) and g(x) are functions, then:

[ f'(x) \cdot g(x) + f(x) \cdot g'(x) ]

• Integrating Both Sides: When we integrate both sides of the product rule, we can investigate the associated relationships.

[ \int (f'(x) \cdot g(x) + f(x) \cdot g'(x)) , dx = f(x) \cdot g(x) ]

• Resulting Relationship: This leads us to:

[ \int f(x) \cdot g'(x) , dx = f(x) \cdot g(x) - \int g(x) \cdot f'(x) , dx ]

Integration by Parts

• Deriving Integration by Parts: This method is derived from the product rule and allows for integrating products of functions.

• Formulation:

[ \int u , dv = u \cdot v - \int v , du ]

• Choosing u and v is crucial for simplifying the integral.

Guidelines for Choosing (u) and (v)

• Manageable Integrals: Choose u such that the integral of v is manageable and straightforward.

• Simplification: Ensure that the derivative du (the derivative of u) further simplifies the integral problem, making the calculation easier.

• Example: For the integral ( \int x e^x , dx ):

  • Let u = x (hence, du = dx) to reduce complexity.

  • Then let v = e^x \, dx.

  • The result after integration leads to:

[ x \cdot e^x - \int e^x , dx = x \cdot e^x - e^x + C ]

More Complex Integrations

• Assessment Steps: Not all integrals will lend themselves to simple substitutions or be found in standard tables. The process to approach complex integrals includes:

  1. Check if the integral appears in the table.

  2. Assess if substitution techniques are applicable.

  3. If neither of the above works, opt for integration by parts.

Reduction Formula for Sine

• For integrals involving powers of sine (( \sin^n(x) )), where ( n \geq 2 ), a reduction formula simplifies the integral:

[ \int \sin^n(x) , dx = -\frac{1}{n} \sin^{n-1}(x) \cdot \cos(x) + \frac{n-1}{n} \int \sin^{n-2}(x) , dx ]

Example of Reducing Power of Sine

• To compute ( \int \sin^4(x) , dx ), you would apply the formula:

  • [ -\frac{1}{4} \sin^3(x) \cdot \cos(x) + \frac{3}{4} \int \sin^2(x) , dx ]

  • Continue solving until the integral can be evaluated completely.

Important Reminders

• Constant of Integration: Always include ( + C ) when performing indefinite integrals to indicate the constant of integration.

• Complex Integrals: When approaching integrals that seem complex, it's often beneficial to break the task into smaller, more familiar parts and re-apply previously learned techniques such as substitution and the product rule as necessary.

Conclusion

Understanding integration techniques is crucial for solving various mathematical problems, particularly in calculus where they enable the evaluation of areas under curves and other complex functions. Mastery of substitution, integration by parts, and reduction formulas empowers students to tackle integrals with confidence and skill.