May 12, 2026Calculus II: Definite Integrals and Advanced Integration by Parts (Concise)

Definite Integrals and Integration by Parts

  • The formula for definite integrals by parts is: ∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du.
  • Definite integrals result in a numerical value representing area rather than a function with a constant of integration.
  • Standard example: ∫0π/2xcos⁡(x) dx=π2−1\int_0^{\pi/2} x \cos(x)\,dx = \frac{\pi}{2} - 1.

Integration of Logarithmic Functions

  • For functions like ln⁡(x)\ln(x) where the elementary anti-derivative is unknown, use integration by parts with u=ln⁡(x)u = \ln(x) and dv=1 dxdv = 1\,dx.
  • Logarithmic power rules (e.g., ln⁡(x2)=2ln⁡(x)\ln(x^2) = 2\ln(x)) should be applied before integration to simplify the expression.
  • Example calculation: ∫1eln⁡(x) dx=1\int_1^e \ln(x)\,dx = 1.

Repeated Integration by Parts and Symmetry

  • Integrals such as ∫0π/2x2sin⁡(x) dx\int_0^{\pi/2} x^2 \sin(x)\,dx require applying the parts formula twice to reduce the power of xx to zero.
  • Even and Odd Functions: If a function is even (like the product of two odd functions xsin⁡(x)x \sin(x)), its integral from −a-a to aa is twice the integral from 00 to aa.
  • The integral ∫−ππxsin⁡(x) dx\int_{-\pi}^{\pi} x \sin(x)\,dx results in 2π2\pi.

Algebraic Manipulation in Integration

  • Complex expressions like ∫ln⁡(x2+1) dx\int \ln(x^2 + 1)\,dx result in internal fractions such as x2x2+1\frac{x^2}{x^2 + 1}.
  • Techniques like polynomial long division or the "adding and subtracting one" trick are used to simplify the integrand: x2x2+1=1−1x2+1\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}.
  • The term 1x2+1\frac{1}{x^2 + 1} integrates to arctan⁡(x)\arctan(x).

Recursive Integration Techniques

  • Some integrals (e.g., ∫excos⁡(ln⁡(x)) dx\int e^x \cos(\ln(x))\,dx) reappear in their original form during integration by parts.
  • Method: Treat the repeating integral as a variable, move it to the same side of the equation, and divide by the resulting coefficient to isolate the solution.

Introduction to Trigonometric Integrals

  • Always check for uu-substitution possibilities before attempting integration by parts for products of sines and cosines.
  • For integrals with powers of sine and cosine: If one power is odd, peel off one factor for the derivative term and convert the remaining even powers using the Pythagorean identity cos⁡2(x)+sin⁡2(x)=1\cos^2(x) + \sin^2(x) = 1.
  • Example: ∫sin⁡3(x)cos⁡2(x) dx\int \sin^3(x) \cos^2(x)\,dx can be converted by replacing sin⁡2(x)\sin^2(x) with 1−cos⁡2(x)1 - \cos^2(x).

Questions & Discussion

  • Question regarding ∫−ππxsin⁡(x) dx\int_{-\pi}^{\pi} x \sin(x)\,dx: Why is the total area not zero even though the bounds are symmetric?
  • Response: Because both xx and sin⁡(x)\sin(x) are odd functions, their product results in an even function. This creates identical areas above the x-axis on both sides of the y-axis, summing to 2π2\pi rather than canceling out.