Calculus 1: The Substitution Rule and Symmetry in Rule of Substitution

Introduction to the Substitution Rule

The Substitution Rule, frequently referred to as "u-sub," is a fundamental technique in integral calculus used to evaluate antiderivatives of composite functions. Many functions do not have antiderivatives that can be expressed straightforwardly in terms of familiar functions. The substitution rule essentially "undoes" the chain rule used in differentiation.

Conceptual Basis of the Substitution Rule

Let $F(x)$ be an antiderivative of $f(x)$. If an integral can be written in the form $\int f(g(x))g'(x)\,dx$, then, since $F' = f$, we can rephrase the integral as:

$\int F'(g(x))g'(x)\,dx$

According to the chain rule, this translates to:

$F(g(x)) + C$

Formal Definition

Let $u = g(x)$, where $g'(x)$ is continuous over a specific interval. Let $f(x)$ be continuous over the corresponding range of $g(x)$. In this case:

$\frac{du}{dx} = g'(x) \text{ or } du = g'(x)\,dx$

Substituting these into the original integral yields the Substitution Rule formula:

$\int f(g(x))g'(x)\,dx = \int f(u)\,du$

Indefinite Integral Examples

Example 1: Basic Substitution

Evaluate $\int 2x(x^2 + 1)^4\,dx$

1. Identify the substitution: Let $u = x^2 + 1$.
2. Compute the differential: $\frac{du}{dx} = 2x$, which gives $du = 2x\,dx$.
3. Substitute into the integral: $\int u^4\,du$.
4. Integrate with respect to $u$: $\frac{u^5}{5} + C$.
5. Back-substitute to express the result in terms of $x$: $\frac{(x^2 + 1)^5}{5} + C$.

Example 2: Handling Constants

Evaluate $\int x e^{x^2 + 1}\,dx$

1. Let $u = x^2 + 1$.
2. Compute the differential: $\frac{du}{dx} = 2x \rightarrow du = 2x\,dx \rightarrow \frac{1}{2}\,du = x\,dx$.
3. Substitute: $\frac{1}{2} \int e^u\,du$.
4. Integrate: $\frac{1}{2} e^u + C$.
5. Back-substitute: $\frac{1}{2} e^{x^2 + 1} + C$.

Example 3: Trigonometric Composite Functions

Evaluate $\int x^3 \sin(x^4 + 7)\,dx$

1. Let $u = x^4 + 7$.
2. Compute the differential: $\frac{du}{dx} = 4x^3 \rightarrow du = 4x^3\,dx \rightarrow \frac{1}{4}\,du = x^3\,dx$.
3. Substitute: $\frac{1}{4} \int \sin(u)\,du$.
4. Integrate: $\frac{1}{4}(-\cos(u)) + C = -\frac{1}{4}\cos(u) + C$.
5. Back-substitute: $-\frac{1}{4}\cos(x^4 + 7) + C$.

Example 4: Integrating Tangent Functions

Evaluate $\int \tan(x)\,dx$

1. Rewrite using trigonometric identities: $\int \frac{\sin(x)}{\cos(x)}\,dx$.
2. Let $u = \cos(x)$.
3. Compute the differential: $\frac{du}{dx} = -\sin(x) \rightarrow -du = \sin(x)\,dx$.
4. Substitute: $\int -\frac{1}{u}\,du = -\int \frac{1}{u}\,du$.
5. Integrate: $-\ln|u| + C$.
6. Back-substitute: $-\ln|\cos(x)| + C$.

Example 5: Substitution via Algebraic Manipulation

Evaluate $\int \sqrt{1 + \sqrt{x}}\,dx$

1. Identify the substitution: Let $u = 1 + \sqrt{x} = 1 + x^{1/2}$.
2. Compute the differential: $\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \rightarrow 2\sqrt{x}\,du = dx$.
3. Express $dx$ solely in terms of $u$: Since $u = 1 + \sqrt{x}$, then $\sqrt{x} = u - 1$. Therefore, $dx = 2(u - 1)\,du$.
4. Substitute: $\int \sqrt{u} \cdot 2(u - 1)\,du = 2 \int u^{1/2}(u - 1)\,du = 2 \int (u^{3/2} - u^{1/2})\,du$.
5. Integrate: $2(\frac{u^{5/2}}{5/2} - \frac{u^{3/2}}{3/2}) + C = 2(\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}) + C$.
6. Back-substitute: $\frac{4}{5}(1 + \sqrt{x})^{5/2} - \frac{4}{3}(1 + \sqrt{x})^{3/2} + C$.

Substitution Rule with Definite Integrals

There are two primary methods for handling definite integrals when using substitution:

1. Method 1: Find the indefinite integral first (switching back to the original variable $x$), and then apply the Fundamental Theorem of Calculus (FTOC) Part 2 using the original limits of integration.
2. Method 2: Change the original limits of integration ($x$-limits) into new limits corresponding to the substitution variable ($u$-limits) and perform the entire integration in terms of $u$ alone.

Theoretical Formula for Method 2:$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$

Note: It is crucial to be careful when changing the bounds to reflect the substitution. Avoid mixed notation (having both $x$ and $u$ in the same integral) as it is mathematically inconsistent.

Definite Integral: Method 1 (Indefinite integral first)

Evaluate $\int_0^{\pi/4} \frac{\tan^3(x)}{\cos^2(x)}\,dx$

1. Let $u = \tan(x)$.
2. Compute the differential: $du = \sec^2(x)\,dx = \frac{1}{\cos^2(x)}\,dx$.
3. Find the indefinite integral in $u$ (temporarily ignoring bounds): $\int u^3\,du = \frac{u^4}{4}$.
4. Switch back to $x$ to use original bounds:    $[\frac{\tan^4(x)}{4}]_0^{\pi/4} = \frac{\tan^4(\pi/4)}{4} - \frac{\tan^4(0)}{4}$
5. Evaluate: $\frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$.

Definite Integral: Method 2 (Changing bounds)

Evaluate $\int_0^2 \frac{x}{(1 + x^2)^2}\,dx$

1. Let $u = 1 + x^2$.
2. Compute the differential: $du = 2x\,dx \rightarrow \frac{1}{2}\,du = x\,dx$.
3. Change the limits:
   • If $x = 0$, then $u = 1 + 0^2 = 1$.
   • If $x = 2$, then $u = 1 + 2^2 = 5$.
4. Set up the new integral in $u$: $\frac{1}{2} \int_1^5 u^{-2}\,du$.
5. Integrate and evaluate directly:    $\frac{1}{2} [-\frac{1}{u}]_1^5 = -\frac{1}{2} (\frac{1}{5} - \frac{1}{1}) = -\frac{1}{2} (\frac{1}{5} - 1) = -\frac{1}{2} (-\frac{4}{5}) = \frac{2}{5}$.

Symmetry in Integration

Symmetry provides a significant shortcut for calculating integrals that are centered at the origin (from $-a$ to $a$).

Even Functions

• Definition: $f(-x) = f(x)$. The graph is symmetric with respect to the y-axis (e.g., $y = \cos(x)$).
• Property: If $(x, y)$ is a point on the function, then $(-x, y)$ is also on the function.
• Integration Rule:$\int_{-a}^{a} f(x)\,dx = 2 \int_0^a f(x)\,dx$

Odd Functions

• Definition: $f(-x) = -f(x)$. The graph is symmetric with respect to the origin (e.g., $y = \sin(x)$).
• Property: If $(x, y)$ is a point on the function, then $(-x, -y)$ is also on the function.
• Integration Rule:$\int_{-a}^{a} f(x)\,dx = 0$

Example: Applying Symmetry

Evaluate $\int_{-2}^2 (x^9 - 3x^5 + 3x^2 + 10)\,dx$

1. Observe the function. The terms $x^9$ and $-3x^5$ are odd functions, while the terms $3x^2$ and $10$ are even functions.
2. Apply symmetry rules: The integral of the odd terms from $-2$ to $2$ is zero.
3. The integral becomes: $\int_{-2}^2 (3x^2 + 10)\,dx = 2 \int_0^2 (3x^2 + 10)\,dx$.
4. Integrate: $2 [x^3 + 10x]_0^2$.
5. Evaluate: $2 [(2^3 + 10(2)) - (0^3 + 10(0))] = 2 [8 + 20] = 2(28) = 56$.