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Integration by parts

Integration by parts

Integration by parts is a technique used in calculus to integrate the product of two functions. It is based on the product rule for differentiation and can be expressed by the formula:

udv=uvvdu\int u \, dv = uv - \int v \, du

Here, uu and dvdv are chosen such that:

  • uu is a differentiable function that you can easily differentiate to find dudu.
  • dvdv is a function that can be easily integrated to find vv.

The process involves:

  1. Identifying parts of the integrand to designate as uu and dvdv.
  2. Computing dudu (the derivative of uu) and vv (the integral of dvdv).
  3. Substituting these into the integration by parts formula.
  4. Simplifying the resulting integral if possible and solving.

This method is particularly useful when dealing with integrals that are products of polynomials, logarithmic functions, and exponential or trigonometric functions.

Part 1: Integration by parts intro

This video explains integration by parts, a technique for finding antiderivatives. It starts with the product rule for derivatives, then takes the antiderivative of both sides. By rearranging the equation, we get the formula for integration by parts. It helps simplify complex antiderivatives.

Here are the key points for studying "Integration by Parts":

  1. Formula: Understand the integration by parts formula:

    udv=uvvdu\int u \, dv = uv - \int v \, du

    where uu and dvdv are chosen parts of the integrand.

  2. Choosing uu and dvdv: Learn how to select uu and dvdv effectively. A common strategy is to choose uu as a function that simplifies upon differentiation and dvdv as a function that is easy to integrate.

  3. Differentiation and Integration: Familiarize yourself with how to differentiate uu to find dudu and integrate dvdv to find vv.

  4. Iterative Process: Recognize that integration by parts may require multiple applications, especially when the resulting integral is still complex.

  5. Practice Problems: Work on a variety of problems to gain proficiency in applying the method, including polynomial, logarithmic, and trigonometric integrands.

  6. Examples: Learn from examples that illustrate the process, including scenarios where integration by parts simplifies the calculation significantly.

  7. Special Cases: Be aware of special cases where integration by parts can be used to derive standard integral formulas or solve specific types of integrals efficiently.

Remember to practice regularly to solidify these concepts and improve your integration skills.

Part 2: Integration by parts: ∫x⋅cos(x)dx

This video shows how to find the antiderivative of x*cos(x) using integration by parts. It assigns f(x)=x and g'(x)=cos(x), making f'(x)=1 and g(x)=sin(x). The formula becomes x*sin(x) - ∫sin(x)dx, which simplifies to x*sin(x) + cos(x) + C.

When studying "Integration by parts" using the example ∫x⋅cos(x)dx, focus on the following key points:

  1. Integration by Parts Formula: The formula is ∫u dv = uv - ∫v du, where you need to identify parts of the integrand as u and dv.

  2. Choosing u and dv:

    • Select u as a function that simplifies upon differentiation (in this case, u = x).
    • Let dv be the remaining portion of the integrand (here, dv = cos(x)dx).
  3. Compute du and v:

    • Differentiate u to find du (du = dx).
    • Integrate dv to find v (v = ∫cos(x)dx = sin(x)).
  4. Substitute into the Formula: Plug u, du, v, and dv into the integration by parts formula.

  5. Evaluate the Integral: Simplify and evaluate the remaining integral ∫sin(x)dx.

  6. Combine Results: After integrating, combine the results from your substitution to find the final answer.

  7. Don't Forget the Constant of Integration: Always add the constant C at the end of your result.

By focusing on these key steps, you can effectively apply the integration by parts technique to similar problems.

Part 3: Integration by parts: ∫ln(x)dx

This video shows how to find the antiderivative of the natural log of x using integration by parts. We rewrite the integral as ln(x) times 1dx, then choose f(x) = ln(x) and g'(x) = 1. The antiderivative is xln(x) - x + C.

Key Points for "Integration by Parts: ∫ln(x)dx"

  1. Integration by Parts Formula:

    udv=uvvdu\int u \, dv = uv - \int v \, du
  2. Choosing uu and dvdv:

    • Set u=ln(x)u = \ln(x) (because its derivative simplifies the integral).
    • Then, du=1xdxdu = \frac{1}{x} \, dx.
    • Choose dv=dxdv = dx (which is straightforward to integrate).
  3. Finding vv:

    • Integrate dvdv: v=xv = x.
  4. Apply Integration by Parts:

    ln(x)dx=xln(x)x1xdx\int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx

    Simplifying gives:

    ln(x)dx=xln(x)1dx\int \ln(x) \, dx = x \ln(x) - \int 1 \, dx
  5. Integrate the Remaining Term:

    1dx=x\int 1 \, dx = x
  6. Combine Results:

    ln(x)dx=xln(x)x+C\int \ln(x) \, dx = x \ln(x) - x + C

    where CC is the constant of integration.

Main Result:

ln(x)dx=xln(x)x+C\int \ln(x) \, dx = x \ln(x) - x + C

Part 4: Integration by parts: ∫x²⋅𝑒ˣdx

Integration by parts helps find antiderivatives of products of functions. We assign f(x) and g'(x) to parts of the product. Then, we find f'(x) and g(x). The formula is ∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx. Sometimes, we use integration by parts twice!

When studying "Integration by Parts" for the integral ∫x²⋅eˣ dx, focus on the following key points:

  1. Integration by Parts Formula:

    • The formula is derived from the product rule of differentiation:
      udv=uvvdu\int u \, dv = uv - \int v \, du
    • Choose uu and dvdv strategically to simplify the integral.
  2. Choosing uu and dvdv:

    • Typically, uu is chosen as the polynomial part (x²) since its derivative will simplify the integral.
    • dvdv is chosen as the remaining part (eˣ dx).
  3. Differentiation and Integration:

    • Differentiate uu to find dudu and integrate dvdv to find vv:
      • u=x2u = x²du=2xdxdu = 2x \, dx
      • dv=exdxdv = eˣ dxv=exv = eˣ
  4. Applying the Formula:

    • Substitute into the integration by parts formula:
      x2exdx=x2exex(2x)dx\int x² eˣ \, dx = x² eˣ - \int eˣ (2x) \, dx
  5. Second Integration by Parts:

    • Now, you may need to apply integration by parts again on the integral 2xexdx\int 2x eˣ \, dx using the same strategy.
    • Choose u=2xu = 2x and dv=exdxdv = eˣ dx.
  6. Final Combination:

    • After solving the second integral, combine all parts to get the final result, maintaining any necessary constants of integration.
  7. Check Your Work:

    • Differentiate your final result to confirm it matches the original integrand x² eˣ.

By mastering these points, you will develop a solid understanding of how to use integration by parts effectively, especially with integrals involving products of polynomials and exponential functions.

Part 5: Integration by parts: ∫𝑒ˣ⋅cos(x)dx

In the video, we learn about integration by parts to find the antiderivative of e^x * cos(x). We assign f(x) = e^x and g'(x) = cos(x), then apply integration by parts twice. The result is the antiderivative e^x * sin(x) + e^x * cos(x) / 2 + C.

To study the integration of the function excos(x)dx\int e^x \cos(x) \, dx using integration by parts, focus on the following key points:

  1. Integration by Parts Formula: Understand the formula:

    udv=uvvdu\int u \, dv = uv - \int v \, du

    where you will need to choose uu and dvdv appropriately.

  2. Choosing uu and dvdv:

    • Set u=exu = e^x and dv=cos(x)dxdv = \cos(x) \, dx.
    • Differentiate uu to find dudu and integrate dvdv to find vv.
  3. Calculate dudu and vv:

    • du=exdxdu = e^x \, dx
    • v=sin(x)v = \sin(x)
  4. Apply the formula: Substitute into the integration by parts formula:

    excos(x)dx=exsin(x)sin(x)exdx\int e^x \cos(x) \, dx = e^x \sin(x) - \int \sin(x) e^x \, dx
  5. Second Integral: Now, focus on exsin(x)dx\int e^x \sin(x) \, dx and repeat the integration by parts.

  6. Using Integration by Parts Again: Choose u=exu = e^x and dv=sin(x)dxdv = \sin(x) \, dx. Determine dudu and vv again.

  7. Set up the new equation: After applying integration by parts again, you get another equation that will lead you back to the original integral.

  8. Solve for the original integral: Combine the results to isolate the original integral and solve for it.

  9. Final Result: Write the solution in terms of the original integral, leading to a formula involving exe^x, sin(x)\sin(x), and cos(x)\cos(x).

  10. Conclusion: Use the derived equation to express excos(x)dx\int e^x \cos(x) \, dx in a closed form.

These key points will guide you through the process of solving excos(x)dx\int e^x \cos(x) \, dx effectively using integration by parts.

Part 6: Integration by parts: definite integrals

When finding a definite integral using integration by parts, we should first find the antiderivative (as we do with indefinite integrals), but then we should also evaluate the antiderivative at the boundaries and subtract.

When studying "Integration by Parts: Definite Integrals," focus on the following key points:

  1. Formula: Understand the integration by parts formula:

    udv=uvvdu\int u \, dv = uv - \int v \, du

    where uu and dvdv must be chosen appropriately.

  2. Choosing uu and dvdv: Apply the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to help select uu and dvdv.

  3. Differentiation and Integration: Compute dudu (the derivative of uu) and vv (the integral of dvdv) after making your selections.

  4. Bounds Adjustment: For definite integrals, evaluate the expression uvuv at the bounds, and subtract the integral evaluated at the same bounds:

    abudv=[uv]ababvdu\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du
  5. Iterative Application: Sometimes, integration by parts may need to be applied more than once if the resulting integral is still complex.

  6. Simplifying the Integral: Focus on simplifying the integral vdu\int v \, du. If it becomes simpler, take advantage of this to obtain the final result.

  7. Examples Practice: Work through several examples to reinforce understanding and become adept at selecting uu and dvdv.

  8. Recognize Patterns: After more practice, recognize integral patterns that can simplify the process, especially with common functions.

By focusing on these key elements, you can effectively master the application of integration by parts to definite integrals.