Problems based on substitution - algebraic substitution

Algebraic substitution is a technique used in calculus, particularly in integration, to simplify an expression or to transform it into a form that is easier to integrate. This method involves replacing a part of the integral with a new variable, which simplifies the integral into a more manageable form.

Why use algebraic substitution?

Algebraic substitution is used to:

• Simplify complex expressions
• Make the integrand resemble standard forms
• Reduce the integral to a form that can be integrated using basic integration rules

How to perform algebraic substitution

To perform algebraic substitution, follow these steps:

1. Identify a part of the integrand that can be substituted with a new variable to simplify the integral.
2. Choose a substitution that will make the integral easier to evaluate.
3. Replace the identified part with the new variable in the integral.
4. Express the differential of the original variable in terms of the new variable.
5. Perform the integration with respect to the new variable.
6. Substitute back the original variable to get the final result.

Table of Differences and Important Points

Aspect	Without Substitution	With Substitution
Complexity	May be complex and difficult to integrate	Simplified, easier to integrate
Integrals	May not resemble standard forms	Often resembles standard forms
Integration Technique	May require advanced techniques	Basic integration rules can be applied
Time and Effort	Can be time-consuming and require more effort	Usually less time-consuming and requires less effort

Formulas

When performing algebraic substitution, the following relationship is used to express the differential in terms of the new variable:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Where:

• ( y ) is the function to be integrated
• ( x ) is the original variable
• ( u ) is the new variable after substitution

The differential ( dx ) can be expressed as:

$$dx = \frac{du}{\frac{dy}{du}}$$

Examples

Example 1: Basic Substitution

Consider the integral:

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

Here, we can use the substitution ( u = 1 + x^2 ). Then, ( du = 2x \, dx ), or ( \frac{1}{2} du = x \, dx ).

The integral becomes:

$$\int \frac{\sqrt{u}}{2} \, du$$

Which can be integrated to get:

$$\frac{1}{3} u^{3/2} + C$$

Substituting back ( u = 1 + x^2 ), we get:

$$\frac{1}{3} (1 + x^2)^{3/2} + C$$

Example 2: Trigonometric Substitution

Consider the integral:

$$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx$$

We can use the substitution ( x = a \sin \theta ). Then, ( dx = a \cos \theta \, d\theta ).

The integral becomes:

$$\int \frac{1}{\sqrt{a^2 - a^2 \sin^2 \theta}} \cdot a \cos \theta \, d\theta$$

Simplifying, we get:

$$\int \frac{1}{\sqrt{a^2(1 - \sin^2 \theta)}} \cdot a \cos \theta \, d\theta = \int \frac{1}{a \cos \theta} \cdot a \cos \theta \, d\theta = \int d\theta$$

Which integrates to:

$$\theta + C$$

Substituting back ( \theta = \arcsin(\frac{x}{a}) ), we get:

$$\arcsin(\frac{x}{a}) + C$$

Example 3: Partial Fraction Decomposition

Consider the integral:

$$\int \frac{1}{x^2 - 1} \, dx$$

We can decompose the fraction as:

$$\frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$

Solving for ( A ) and ( B ), we get ( A = \frac{1}{2} ) and ( B = -\frac{1}{2} ).

The integral becomes:

$$\int \left( \frac{1}{2(x - 1)} - \frac{1}{2(x + 1)} \right) dx$$

Which can be integrated to get:

$$\frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C$$

Algebraic substitution is a powerful tool in integration, and understanding how to apply it can greatly simplify many complex integrals. Practice with various types of integrals is essential to mastering this technique.