Integration Techniques: Substitution and Powers

Integration Techniques

Substitution Method in Integration

The substitution method is a fundamental technique in calculus for evaluating integrals, especially when the integrand contains composite functions or expressions that can be simplified by changing variables. This method is often referred to as "u-substitution."

• Definition: Substitution involves replacing a complicated expression with a new variable, typically u, to simplify the integral.

• Procedure:

  1. Identify a substitution u = g(x) that simplifies the integrand.

  2. Compute du = g'(x) dx.

  3. Rewrite the integral in terms of u and du.

  4. Integrate with respect to u.

  5. Substitute back to the original variable if necessary.

• Definite Integrals: When integrating definite integrals, adjust the bounds according to the substitution.

Examples of Substitution

Example 1: Definite Integral with Substitution

Evaluate $\int_{0}^{1} \left[10(2x-1)^4 + 9(2x-1)^2 - 8\right] dx$ using substitution.

• Substitution: Let $u = 2x - 1$, so $du = 2dx \Rightarrow dx = \frac{du}{2}$.

• Bounds: When $x = 0$, $u = -1$; when $x = 1$, $u = 1$.

• Integral: $\int_{0}^{1} \left[10(2x-1)^4 + 9(2x-1)^2 - 8\right] dx = \frac{1}{2} \int_{-1}^{1} \left[10u^4 + 9u^2 - 8\right] du$

• Antiderivative: $\frac{1}{2} \left[2u^5 + 3u^3 - 8u\right] \Big|_{-1}^{1}$

• Result: $-3$

Example 2: Substitution with Powers

Evaluate $\int (3x^2 + 4)(x^3 + 4x - 8)^{14} dx$.

• Substitution: Let $u = x^3 + 4x - 8$, $du = (3x^2 + 4)dx$.

• Integral: $\int u^{14} du = \frac{1}{15} u^{15} + C$

• Result: $\frac{1}{15}(x^3 + 4x - 8)^{15} + C$

Example 3: Substitution with Rational Functions

Evaluate $\int \frac{2x-1}{3x^2 - 3x + 1} dx$.

• Substitution: Let $u = 3x^2 - 3x + 1$, $du = (6x - 3)dx = 3(2x - 1)dx \Rightarrow (2x - 1)dx = \frac{du}{3}$.

• Integral: $\int \frac{du}{3u} = \frac{1}{3} \int \frac{du}{u} = \frac{1}{3} \ln|u| + C$

• Result: $\frac{1}{3} \ln|3x^2 - 3x + 1| + C$

Example 4: Substitution with Linear Expressions and Powers

Evaluate $\int \frac{x+3}{(x-2)^{16}} dx$.

• Substitution: Let $u = x - 2$, $du = dx$, $x = u + 2$.

• Integral: $\int \frac{u + 5}{u^{16}} du = \int u^{-15} du + 5 \int u^{-16} du$

• Antiderivative: $\frac{u^{-14}}{-14} + 5 \frac{u^{-15}}{-15} + C$

• Result: $-\frac{1}{14}(x-2)^{-14} - \frac{1}{3}(x-2)^{-15} + C$

Example 5: Substitution with Square Roots

Evaluate $\int (x^2 + 1) \sqrt{x-3} dx$.

• Substitution: Let $u = x - 3$, $du = dx$, $x = u + 3$.

• Expand: $(x^2 + 1) = (u + 3)^2 + 1 = u^2 + 6u + 9 + 1 = u^2 + 6u + 10$

• Integral: $\int (u^2 + 6u + 10) u^{1/2} du = \int (u^{5/2} + 6u^{3/2} + 10u^{1/2}) du$

• Antiderivative: $\frac{u^{7/2}}{7/2} + 6 \frac{u^{5/2}}{5/2} + 10 \frac{u^{3/2}}{3/2} + C$

• Result: $\frac{2}{7}(x-3)^{7/2} + \frac{12}{5}(x-3)^{5/2} + \frac{20}{3}(x-3)^{3/2} + C$

Summary Table: Substitution Integrals

Integral	Substitution	Result
$\int_{0}^{1} [10(2x-1)^4 + 9(2x-1)^2 - 8] dx$	$u = 2x-1$	$-3$
$\int (3x^2 + 4)(x^3 + 4x - 8)^{14} dx$	$u = x^3 + 4x - 8$	$\frac{1}{15}(x^3 + 4x - 8)^{15} + C$
$\int \frac{2x-1}{3x^2 - 3x + 1} dx$	$u = 3x^2 - 3x + 1$	$\frac{1}{3} \ln|3x^2 - 3x + 1| + C$
$\int \frac{x+3}{(x-2)^{16}} dx$	$u = x - 2$	$-\frac{1}{14}(x-2)^{-14} - \frac{1}{3}(x-2)^{-15} + C$
$\int (x^2 + 1) \sqrt{x-3} dx$	$u = x - 3$	$\frac{2}{7}(x-3)^{7/2} + \frac{12}{5}(x-3)^{5/2} + \frac{20}{3}(x-3)^{3/2} + C$

Additional info: All examples above illustrate the use of substitution in integration, a key topic in Calculus Chapter 8 (Integration Techniques). The notes expand on the original brief solutions to provide full academic context and step-by-step explanations.