Topic 6: Integration by Substitution

Introduction to Integration

• Topic 6 focuses on integration.

• Integration is the reverse process of differentiation. Integrating a function and then differentiating the result yields the original function.

• Example: Integrating acceleration over time gives velocity; differentiating velocity gives acceleration.

Indefinite vs. Definite Integrals

• Indefinite Integrals: Integrals without limits. This lecture focuses on indefinite integrals, which result in a function plus a constant.

• Definite Integrals: Integrals with defined limits (A to B), representing the area under the curve between those limits.

Antiderivatives and the Constant of Integration

• Antiderivative: Another term for an integral. If integrating f(x) results in F(x), then F(x) is the antiderivative of f(x).

• Antiderivatives are not unique; they include an arbitrary constant C.

  • For example, \int 2x \, dx = x^2 + C; possible antiderivatives include x^2 + 5 and x^2 - 32.

Notation for Integration

• The integral symbol is an elongated "S," representing summation.

  • \int f(x) \, dx = F(x) + C

  • f(x) is the integrand, and C is the arbitrary constant.

• Today's focus remains on indefinite integrals without limits.

Textbook Integrals and Formula Sheet

• A formula sheet will be provided during the exam.

• Integrals on the formula sheet may be presented as derivatives. For example:

  • The formula sheet might provide \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}.

  • Understanding this is equivalent to knowing that \int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C.

Key Properties of Integration

• Constants within an integral can be factored out:

  • \int k \cdot f(x) \, dx = k \int f(x) \, dx

• Integrals of sums can be separated:

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

Exercise 1: Warm-up Examples

Part A: \int (\frac{1}{2x} - \frac{1}{2x^2}) \, dx

1. Separate the integral and factor out constants:

   • \int (\frac{1}{2x} - \frac{1}{2x^2}) \, dx = \frac{1}{2} \int \frac{1}{x} \, dx - \frac{1}{2} \int \frac{1}{x^2} \, dx

2. Rewrite the second term using exponents: \frac{1}{2} \int \frac{1}{x} \, dx - \frac{1}{4} \int x^{-2} \, dx

3. Integrate each term:

   • \frac{1}{2} \int \frac{1}{x} \, dx = \frac{1}{2} \ln |x| + C_1

   • \frac{1}{4} \int x^{-2} \, dx = \frac{1}{4} * \frac{x^{-1}}{-1} = - \frac{1}{4x} + C_2

4. Combine constants and simplify:

   • \frac{1}{2} \ln |x| + \frac{1}{4x} + C \text{, where } C = C1 + C2

Part B: \int (\frac{e^x}{3} + 2\cos x) \, dx

1. Separate the integral and factor out constants:

   • \int (\frac{e^x}{3} + 2\cos x) \, dx = \frac{1}{3} \int e^x \, dx + 2 \int \cos x \, dx

2. Integrate each term:

   • \frac{1}{3} \int e^x \, dx = \frac{1}{3}e^x

   • 2 \int \cos x \, dx = 2\sin x

3. Add the arbitrary constant:

   • \frac{1}{3}e^x + 2\sin x + C

Part C: \int (\frac{\sqrt[3]{x^2}}{5} - \frac{7}{2\sqrt{x}} + 6x) \, dx

1. Rewrite terms with fractional exponents:

   • \int (\frac{x^{\frac{2}{3}}}{5} - \frac{7}{2}x^{-\frac{1}{2}} + 6x) \, dx

2. Separate the integral and factor out constants:

   • \frac{1}{5} \int x^{\frac{2}{3}} \, dx - \frac{7}{2} \int x^{-\frac{1}{2}} \, dx + 6 \int x \, dx

3. Integrate each term:

   • \frac{1}{5} \int x^{\frac{2}{3}} \, dx = \frac{1}{5}*\frac{x^{\frac{5}{3}}}{\frac{5}{3}} = \frac{3}{25} x^{\frac{5}{3}}

   • -\frac{7}{2} \int x^{-\frac{1}{2}} \, dx = -\frac{7}{2}*\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = -7x^{\frac{1}{2}}

   • 6 \int x \, dx = 3x^2

4. Add the constant of integration and simplify:

   • \frac{3}{25}x^{\frac{5}{3}} - 7\sqrt{x} + 3x^2 + C

Part D (Skipped)

Integration by Substitution

• Integration is generally more complex than differentiation.

• Some functions cannot be integrated analytically.

• Integration by substitution is related to the chain rule in differentiation.

• General form: \int f(g(x)) \cdot g'(x) \, dx

Exercise 2: Integration by Substitution

Part A: \int (x^2 + 1)^{50} \cdot 2x \, dx

1. Substitution: Let u = x^2 + 1.

2. Differentiate: \frac{du}{dx} = 2x \Rightarrow du = 2x \, dx

3. Rewrite Integral: \int u^{50} \, du

4. Integrate: \frac{u^{51}}{51} + C

5. Substitute Back: \frac{(x^2 + 1)^{51}}{51} + C

6. Check Answer: Differentiate the result to see if it matches the original integrand.

Part B: \int x^2 \sqrt{x^3 - 1} \, dx

1. Substitution: Let u = x^3 - 1.

2. Differentiate: \frac{du}{dx} = 3x^2 \Rightarrow dx = \frac{1}{3x^2} \, du

3. Rewrite Integral: \int x^2 \sqrt{u} \cdot \frac{1}{3x^2} \, du = \frac{1}{3} \int u^{\frac{1}{2}} \, du

4. Integrate: \frac{1}{3} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{9}u^{\frac{3}{2}} + C

5. Substitute Back: \frac{2}{9}(x^3 - 1)^{\frac{3}{2}} + C

Part C: \int xe^{x^2} \, dx

1. Substitution: Let u = x^2.

2. Differentiate: \frac{du}{dx} = 2x \Rightarrow dx = \frac{1}{2x} \, du

3. Rewrite Integral: \int xe^u \cdot \frac{1}{2x} \, du = \frac{1}{2} \int e^u \, du

4. Integrate: \frac{1}{2}e^u + C

5. Substitute Back: \frac{1}{2}e^{x^2} + C

Part D: \int \frac{3x^2}{x^3 + 5} \, dx

1. Substitution: Let u = x^3 + 5.

2. Differentiate: \frac{du}{dx} = 3x^2 \Rightarrow dx = \frac{1}{3x^2} \, du

3. Rewrite Integral: \int \frac{3x^2}{u} \cdot \frac{1}{3x^2} \, du = \int \frac{1}{u} \, du

4. Integrate: \ln |u| + C

5. Substitute Back: \ln |x^3 + 5| + C

• General Rule: If the integral is in the form \int \frac{f'(x)}{f(x)} \, dx, the result is \ln |f(x)| + C. In this case, f(x) = x^3 + 5 and f'(x) = 3x^2.