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Integration by Parts

HyperWrite's Integration by Parts Study Guide is your comprehensive resource for understanding and applying this essential calculus technique. This guide covers the key concepts, formulas, and strategies needed to solve a wide variety of integrals using Integration by Parts.

What is Integration by Parts?

Integration by Parts is a powerful technique used to evaluate integrals that cannot be solved using basic integration rules. It is based on the product rule for differentiation and allows you to transform a complex integral into a simpler one. The technique is particularly useful when the integrand is a product of two functions, one of which can be easily differentiated and the other easily integrated.

The Integration by Parts Formula

The Integration by Parts formula is derived from the product rule for differentiation:

∫u dv = uv - ∫v du

Where:

  • u is a function that can be easily differentiated
  • dv is a function that can be easily integrated
  • du is the derivative of u
  • v is the integral of dv
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Choosing u and dv

The success of Integration by Parts depends on choosing the appropriate functions for u and dv. A helpful mnemonic for selecting u is LIPET:

  1. Logarithmic functions
  2. Inverse trigonometric functions
  3. Polynomial functions
  4. Exponential functions
  5. Trigonometric functions

Choose u from the integrand based on this order of preference, and let dv be the remaining part of the integrand.

Applying Integration by Parts

To apply Integration by Parts, follow these steps:

  1. Identify the integrand and choose u and dv based on the LIPET rule.
  2. Differentiate u to find du and integrate dv to find v.
  3. Substitute u, du, v, and dv into the Integration by Parts formula.
  4. Simplify the resulting expression and evaluate the new integral, if possible.
  5. If the new integral is still too complex, repeat the process until you obtain a solvable integral.

Examples of Integration by Parts

Example 1: Evaluate ∫x cos(x) dx

Let u = x and dv = cos(x) dx. Then, du = dx and v = sin(x).

∫x cos(x) dx = x sin(x) - ∫sin(x) dx = x sin(x) + cos(x) + C

Example 2: Evaluate ∫ln(x) dx

Let u = ln(x) and dv = dx. Then, du = (1/x) dx and v = x.

∫ln(x) dx = x ln(x) - ∫x (1/x) dx = x ln(x) - ∫1 dx = x ln(x) - x + C

Common Questions and Answers

When should I use Integration by Parts?

Integration by Parts is most effective when the integrand is a product of two functions, one of which can be easily differentiated (u) and the other easily integrated (dv). It is particularly useful when the integrand involves logarithmic, inverse trigonometric, polynomial, exponential, or trigonometric functions.

Can Integration by Parts be used more than once in a single problem?

Yes, sometimes you may need to apply Integration by Parts multiple times to solve a complex integral. This process is called repeated Integration by Parts or recursive Integration by Parts.

What if I can't solve the new integral after applying Integration by Parts?

If the new integral is still too complex to solve directly, you can try applying Integration by Parts again or consider using other integration techniques, such as trigonometric substitution, partial fraction decomposition, or tabular integration.

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Conclusion

Integration by Parts is a versatile and essential technique for solving complex integrals in calculus. By understanding the formula, selecting appropriate functions for u and dv, and following the step-by-step process, you can tackle a wide range of integration problems. With practice and exposure to various examples, you will develop the skills and intuition needed to apply Integration by Parts effectively.

Integration by Parts
Master the technique of Integration by Parts to solve complex integrals
How do I choose which function to use as u in Integration by Parts?
Use the LIPET rule to choose u: Logarithmic, Inverse trigonometric, Polynomial, Exponential, and Trigonometric functions, in that order of preference.

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