Lecture Notes on Integration Techniques

Integral of Polynomial Functions

  • We are trying to find the integral of x \cdot \cos(x) dx.
  • Recognize that there is a product of two functions, x and \cos(x).
  • The problem suggests using u-substitution as the primary method for solving integrals.

U-Substitution

  • The method applies to integrals such as \int x \cos(x) dx or \int \sin(t^2) \cos(t^2) t dt.

Example 1: Integral of \sin(t) \cos(t)

  • Consider the integral \int \sin(t) \cos(t) dt.
  • Here you have a function and its derivative.
  • Let u = \sin(t). Then \frac{du}{dt} = \cos(t), so du = \cos(t) dt.
  • The integral becomes \int u du = \frac{1}{2}u^2 + c = \frac{1}{2} \sin^2(t) + c.
  • The integral of \sin(t) is -\cos(t) + c.

Example 2: Integral of \sin(t^2) \cos(t^2) t dt

  • Consider the integral \int \sin(t^2)\cos(t^2)t dt.
  • Let u = \sin(t^2).
  • Then \frac{du}{dt} = \cos(t^2) \cdot 2t, so \frac{1}{2} du = t \cos(t^2) dt.
  • The integral becomes \int \frac{1}{2} \sin(t^2) du = \frac{1}{2} \int u du = \frac{1}{2} \cdot \frac{u^2}{2} + c = \frac{1}{4}u^2 + c.
  • Substituting back, we get \frac{1}{4}\sin^2(t^2) + c.
  • In this example, we can use u-substitution because the derivative of \sin(t^2) is present in the integral.

Derivatives of Composite Functions

  • The derivative of \sin(t^2) is \cos(t^2) \cdot 2t.
  • Using the power rule and chain rule, the derivative of \sin^2(t^2) is 2 \sin(t^2) \cdot \cos(t^2) \cdot 2t = 4t \sin(t^2) \cos(t^2).

Example 3: Integral of t \sin(t^2) \cos(t^2) dt

  • We need to solve \int t \sin(t^2) \cos(t^2) dt.
  • Let u = \sin(t^2). Then \frac{du}{dt} = \cos(t^2) \cdot 2t, so du = 2t \cos(t^2) dt.
  • Solve for dt to get \frac{du}{2t \cos(t^2)} = dt.
  • Substituting, we have \int t u \cos(t^2) \frac{du}{2t \cos(t^2)}.
  • Which simplifies to \int \frac{1}{2} u du = \frac{1}{2} \frac{u^2}{2} + c = \frac{1}{4} u^2 + c = \frac{1}{4} \sin^2(t^2) + c .

Practice Problems

  • The lecture mentions providing 10 practice problems.
  • Example problem: \int e^x \sqrt{1 + e^x} dx
  • Let u = 1 + e^x. Then du = e^x dx.
  • So the integral becomes \int \sqrt{u} du.
  • Another example: \int e^{x^2} x dx
  • Let u = x^2. So du = 2x dx.
  • This implies \frac{1}{2} du = x dx.
  • Replace x^2 with u in the exponential, then replace x dx with \frac{1}{2} du .
  • New integral: \int e^u \frac{1}{2} du.

Importance of U-Substitution from Previous Classes

  • U-substitution is a fundamental concept.
  • Advanced integration techniques may require multiple pages to solve, while u-substitution provides a more straightforward approach.
  • Understanding derivative rules is essential for mastering anti-derivatives.

Dealing with Derivatives and Anti-derivatives

  • Review derivative rules to better understand anti-derivatives.
  • The lecture states that you need to understand the derivative rules in order to apply anti derivative rules.

Formula Sheet

  • A formula sheet may not be provided for the exam.

Example 4: Integral of e^{\frac{1}{x}} \frac{1}{x^2} dx

  • Let u = \frac{1}{x}, so u = x^{-1}.
  • \frac{du}{dx} = -1 \cdot x^{-2} = -\frac{1}{x^2}.
  • du = -\frac{1}{x^2} dx\rightarrow -du = \frac{1}{x^2} dx .
  • Substitute back into the integral: \int e^u (-1) du = - \int e^u du = -e^u + c.
  • Substituting back for u: -e^{\frac{1}{x}} + c.
  • To verify, find the derivative of f(x) = -e^{\frac{1}{x}} + c.
  • f'(x) = -e^{\frac{1}{x}} \cdot (-\frac{1}{x^2}) + 0 = \frac{e^{\frac{1}{x}}}{x^2}, which confirms the anti-derivative.