L3-L4: Integration by Substitution

Lecture notes covering the substitution method (u-substitution) for evaluating integrals. This method is the reverse of the chain rule and depends on the idea of a differential.

1. Substitution Basics

In finding the antiderivative for some functions, many standard techniques fail. Substitution can sometimes remedy this problem.

Differential: If $u = f(x)$, then the differential of $u$, written $du$, is defined as: $$du = f'(x),dx$$

Example: If $u = 2x^3 + 1$, then $du = 6x^2,dx$.

2. Integration by Substitution (The Method)

Method of integration related to chain rule differentiation. If $u$ is a function of $x$, then we can use the formula: $$\int f,dx = \int \left(\frac{f}{du/dx}\right) du$$

More commonly, if $u = g(x)$ and $du = g'(x),dx$, then: $$\int f[g(x)]g'(x),dx = \int f(u),du = F(u) + C = F(g(x)) + C$$

The core process is: Pick $u$ → Compute $du$ → Substitute in → Integrate → Back-substitute.

3. Choosing $u$

In general, for the types of problems in this course, there are three primary cases for choosing $u$:

  1. The quantity under a root or raised to a power
  2. The quantity in the denominator
  3. The exponent on $e$

Some integrands may need to be rearranged to fit one of these cases. Also, remember that $du$ is the derivative of $u$.

4. Handling Missing Constant Factors

We haven't always needed to modify $du$ explicitly, because in some problems $du$ (or a constant multiple of it) already appears in the integrand. When it doesn't, we can multiply and divide by the needed constant.

Key idea: If $du = k \cdot \text{(something)},dx$ but the integrand only has $\text{(something)},dx$, multiply inside the integral by $k$ and counteract it by multiplying the entire integral by $1/k$.

flowchart TD
    A([Start: ∫ f(g(x)) g'(x) dx]) --> B{Pick u}
    B -->|"Case 1"| C1["Quantity under a root<br/>or raised to a power"]
    B -->|"Case 2"| C2["Quantity in the denominator"]
    B -->|"Case 3"| C3["Exponent on e"]
    C1 --> D[Compute du = g'(x) dx]
    C2 --> D
    C3 --> D
    D --> E{Does du match<br/>integrand exactly?}
    E -->|Yes| F[Substitute directly]
    E -->|No| G["Multiply / divide<br/>by constant k"]
    F --> H[Integrate ∫ f(u) du]
    G --> H
    H --> I[Back-substitute u = g(x)]
    I --> J([F(g(x)) + C])

5. Worked Examples by Category

5.1 Direct Substitution (Perfect $du$ Match)

Example 2.1.1: Find $\displaystyle\int (2x^3 + 1)^4 \cdot 6x^2,dx$

  • Let $u = 2x^3 + 1$, then $du = 6x^2,dx$
  • Substitute: $\displaystyle\int u^4,du = \frac{u^5}{5} + C$
  • Back-substitute: $\displaystyle\frac{(2x^3 + 1)^5}{5} + C$

Example 2.1.6: Find $\displaystyle\int 3x^2(x^3 - 5)^9,dx$

  • Let $u = x^3 - 5$, then $du = 3x^2,dx$
  • $\displaystyle\int u^9,du = \frac{u^{10}}{10} + C = \frac{(x^3 - 5)^{10}}{10} + C$

5.2 Substitution Requiring Constant Adjustment

Example 2.1.2 & 2.1.3: Find $\displaystyle\int x^2\sqrt{x^3 + 1},dx$

  • Let $u = x^3 + 1$, then $du = 3x^2,dx$
  • We have $x^2,dx$ but need $3x^2,dx$: $$\int x^2\sqrt{x^3 + 1},dx = \frac{1}{3}\int \sqrt{u},du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} + C = \frac{2}{9}(x^3 + 1)^{3/2} + C$$

Example 2.1.4: Find $\displaystyle\int \frac{x + 3}{(x^2 + 6x)^2},dx$

  • Let $u = x^2 + 6x$, then $du = (2x + 6),dx = 2(x + 3),dx$
  • Multiply inside by 2, outside by $1/2$: $$= \frac{1}{2}\int \frac{du}{u^2} = \frac{1}{2}\int u^{-2},du = \frac{1}{2} \cdot \frac{u^{-1}}{-1} + C = \frac{-1}{2(x^2 + 6x)} + C$$

5.3 Variable Substitution (Avoiding Bad Choices)

Choosing $u$ poorly leads to an integral still containing the original variable.

Example 2.2.1: $\displaystyle\int (x + 8)^{1/7},dx$

  • Correct: Let $u = x + 8$, therefore $du = dx$. Then $\displaystyle\int u^{1/7},du$
  • Incorrect: Let $u = x$, which just gives $\displaystyle\int (u + 8)^{1/7},du$ — no simplification.

Example 2.2.2: $\displaystyle\int \frac{x + 1}{x^2 + 2x + 3},dx$

  • Correct: Let $u = x^2 + 2x + 3$, then $du = (2x + 2),dx = 2(x + 1),dx$
  • Incorrect: Let $u = x + 1$, which leaves $x$ in the denominator.

5.4 Trigonometric Function Substitution

Example 2.3.1: Find $\displaystyle\int x\cos(3x^2),dx$

  • Let $u = 3x^2$, then $du = 6x,dx$ so $\frac{du}{6} = x,dx$
  • $\displaystyle\frac{1}{6}\int \cos u,du = \frac{1}{6}\sin u + C = \frac{1}{6}\sin(3x^2) + C$

Example 2.3.2: Find $\displaystyle\int (\sin x + 1)^7 \cos x,dx$

  • Let $u = \sin x + 1$, then $du = \cos x,dx$
  • $\displaystyle\int u^7,du = \frac{u^8}{8} + C = \frac{1}{8}(\sin x + 1)^8 + C$

5.5 Exponential Function Substitution

Example 2.4.1: Find $\displaystyle\int e^{3x},dx$

  • Let $u = 3x$, then $\frac{du}{3} = dx$
  • $\displaystyle\frac{1}{3}\int e^u,du = \frac{1}{3}e^{3x} + C$

Example 2.4.2: Find $\displaystyle\int 5^{x^2} x,dx$

  • Let $u = x^2$, then $\frac{du}{2} = x,dx$
  • $\displaystyle\frac{1}{2}\int 5^u,du = \frac{5^u}{2\ln 5} + C = \frac{5^{x^2}}{2\ln 5} + C$

Example 2.4.3: Find $\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}},dx$

  • Let $u = \sqrt{x}$, then $2,du = \frac{1}{\sqrt{x}},dx$
  • $\displaystyle 2\int e^u,du = 2e^{\sqrt{x}} + C$

5.6 Resulting Logarithmic Function

When substitution produces $\displaystyle\int \frac{du}{u}$, the result is $\ln|u| + C$.

Example 2.1.5: Find $\displaystyle\int \frac{x^2 + 1}{x^3 + 3x},dx$

  • Let $u = x^3 + 3x$, then $du = (3x^2 + 3),dx = 3(x^2 + 1),dx$
  • $\displaystyle\frac{1}{3}\int \frac{du}{u} = \frac{1}{3}\ln|x^3 + 3x| + C$

Example 2.1.9: Find $\displaystyle\int \frac{e^{3t}}{e^{3t} + 2},dt$

  • Let $u = e^{3t} + 2$, then $\frac{du}{3e^{3t}} = dt$
  • $\displaystyle\frac{1}{3}\int \frac{du}{u} = \frac{1}{3}\ln(e^{3t} + 2) + C$

Example 2.5.3: Find $\displaystyle\int \frac{24x + 9}{8x^2 + 6x + 2},dx$

  • Factor numerator: $3(8x + 3)$
  • Let $u = 8x^2 + 6x + 2$, then $du = (16x + 6),dx = 2(8x + 3),dx$
  • $\displaystyle\frac{3}{2}\int \frac{du}{u} = \frac{3}{2}\ln|8x^2 + 6x + 2| + C$

5.7 Resulting Inverse Trigonometric Function

Two fundamental integration formulas arise frequently: $$\int \frac{1}{\sqrt{1 - x^2}},dx = \sin^{-1} x + C$$ $$\int \frac{1}{x^2 + 1},dx = \tan^{-1} x + C$$

Generalized forms via substitution: $$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C \quad (a > 0)$$ $$\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C \quad (a \neq 0)$$

Example 2.6.1: Find $\displaystyle\int \frac{dx}{\sqrt{9 - x^2}}$

  • Rewrite: $\displaystyle\int \frac{dx}{3\sqrt{1 - (x/3)^2}}$
  • Let $u = x/3$, then $du = \frac{1}{3},dx$
  • $\displaystyle\int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}\frac{x}{3} + C$

Example 2.6.3: Find $\displaystyle\int \frac{x}{x^4 + 9},dx$

  • Let $u = x^2$, then $du = 2x,dx$
  • $\displaystyle\frac{1}{2}\int \frac{du}{u^2 + 9} = \frac{1}{2} \cdot \frac{1}{3}\tan^{-1}\frac{u}{3} + C = \frac{1}{6}\tan^{-1}\frac{x^2}{3} + C$

6. Summary of Key Patterns

Pattern Choice of $u$ Notes
Power/Root Quantity inside power or under root Often requires constant adjustment
Rational Quantity in denominator Check if numerator is derivative of denominator
Exponential Exponent on $e$ or base $a$ Remember $\int a^x,dx = \frac{a^x}{\ln a} + C$
Logarithmic Quantity in denominator Result is $\ln
Inverse Trig Forms matching $\sqrt{a^2 - x^2}$ or $a^2 + x^2$ May need substitution to match standard form

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