Determine The Following Indefinite Integral. Check Your Work By Differentiation

Unlocking the secrets of indefinite integrals doesn't have to feel like navigating a labyrinth. This guide provides a comprehensive walkthrough, equipping you with the knowledge and skills to confidently tackle these fundamental calculus problems. We'll explore not only the techniques for finding indefinite integrals but also how to verify your solutions using the power of differentiation.

Understanding Indefinite Integrals

At its core, an indefinite integral represents the family of all functions that have a given derivative. In simpler terms, if you know the rate of change of a function, the indefinite integral helps you find the original function (up to a constant). This "constant of integration," often denoted as C, is crucial because the derivative of a constant is always zero.

The Notation

The indefinite integral is represented by the following notation:

∫ f(x) dx = F(x) + C

Where:

• ∫ is the integral symbol
• f(x) is the integrand (the function you are integrating)
• dx indicates that the integration is with respect to the variable x
• F(x) is the antiderivative of f(x)
• C is the constant of integration

Basic Integration Rules

Mastering these basic rules is essential for solving indefinite integrals:

• Power Rule: ∫ xⁿ dx = (xⁿ⁺¹) / (n+1) + C (where n ≠ -1)
• Constant Multiple Rule: ∫ kf(x) dx = k ∫ f(x) dx (where k is a constant)
• Sum/Difference Rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
• Integral of a Constant: ∫ k dx = kx + C
• Integral of 1/x: ∫ (1/x) dx = ln|x| + C
• Integral of e^x: ∫ e^x dx = e^x + C
• Integral of a^x: ∫ a^x dx = (a^x / ln(a)) + C
• Integrals of Trigonometric Functions:
  • ∫ sin(x) dx = -cos(x) + C
  • ∫ cos(x) dx = sin(x) + C
  • ∫ sec²(x) dx = tan(x) + C
  • ∫ csc²(x) dx = -cot(x) + C
  • ∫ sec(x)tan(x) dx = sec(x) + C
  • ∫ csc(x)cot(x) dx = -csc(x) + C

Step-by-Step Guide to Determining Indefinite Integrals

Let's break down the process of finding indefinite integrals with examples and verification through differentiation.

Example 1: ∫ (3x² + 4x - 5) dx

1. Apply the Sum/Difference Rule:

∫ (3x² + 4x - 5) dx = ∫ 3x² dx + ∫ 4x dx - ∫ 5 dx

2. Apply the Constant Multiple Rule:

= 3 ∫ x² dx + 4 ∫ x dx - 5 ∫ dx

3. Apply the Power Rule and Integral of a Constant:

= 3 * (x³ / 3) + 4 * (x² / 2) - 5x + C

4. Simplify:

= x³ + 2x² - 5x + C

Therefore, ∫ (3x² + 4x - 5) dx = x³ + 2x² - 5x + C

Verification by Differentiation:

To verify, we differentiate the result:

d/dx (x³ + 2x² - 5x + C) = 3x² + 4x - 5 + 0 = 3x² + 4x - 5

Since the derivative of our result matches the original integrand, our integration is correct.

Example 2: ∫ (e^x + sin(x)) dx

1. Apply the Sum Rule:

∫ (e^x + sin(x)) dx = ∫ e^x dx + ∫ sin(x) dx

2. Apply the Integration Rules for e^x and sin(x):

= e^x - cos(x) + C

Therefore, ∫ (e^x + sin(x)) dx = e^x - cos(x) + C

Verification by Differentiation:

d/dx (e^x - cos(x) + C) = e^x + sin(x) + 0 = e^x + sin(x)

The derivative matches the integrand, confirming our solution.

Example 3: ∫ (5/x) dx

1. Apply the Constant Multiple Rule:

∫ (5/x) dx = 5 ∫ (1/x) dx

2. Apply the Integration Rule for 1/x:

= 5 ln|x| + C

Therefore, ∫ (5/x) dx = 5 ln|x| + C

Verification by Differentiation:

d/dx (5 ln|x| + C) = 5 * (1/x) + 0 = 5/x

The derivative matches the integrand, confirming our solution.

Advanced Techniques for Indefinite Integration

While the basic rules cover many cases, some integrals require more advanced techniques:

1. u-Substitution (Substitution Rule)

u-substitution is a powerful technique for simplifying integrals by replacing a part of the integrand with a new variable, u. The goal is to transform the integral into a form that is easier to integrate using the basic rules.

Steps for u-Substitution:

1. Choose a suitable u: Look for a function within the integrand whose derivative is also present (up to a constant multiple).
2. Find du/dx: Calculate the derivative of u with respect to x.
3. Solve for dx: Express dx in terms of du.
4. Substitute: Replace the original expression in terms of x with the new expression in terms of u.
5. Integrate with respect to u: Evaluate the integral with respect to u.
6. Substitute back: Replace u with its original expression in terms of x.
7. Add the constant of integration, C.

Example: ∫ 2x * cos(x²) dx

1. Choose u: Let u = x²
2. Find du/dx: du/dx = 2x
3. Solve for dx: dx = du / (2x)
4. Substitute: ∫ 2x * cos(x²) dx = ∫ 2x * cos(u) * (du / 2x) = ∫ cos(u) du
5. Integrate with respect to u: ∫ cos(u) du = sin(u) + C
6. Substitute back: sin(u) + C = sin(x²) + C

Therefore, ∫ 2x * cos(x²) dx = sin(x²) + C

Verification by Differentiation:

d/dx (sin(x²) + C) = cos(x²) * 2x + 0 = 2x * cos(x²)

The derivative matches the integrand, confirming our solution.

2. Integration by Parts

Integration by parts is used to integrate the product of two functions. It is based on the product rule for differentiation.

The Formula:

∫ u dv = uv - ∫ v du

Where:

• u and v are functions of x.
• du is the derivative of u with respect to x.
• dv is the derivative of v with respect to x.

Steps for Integration by Parts:

1. Choose u and dv: Select which part of the integrand will be u and which will be dv. A helpful guideline is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which suggests the order of preference for choosing u.
2. Find du and v: Calculate the derivative of u (du) and the integral of dv (v).
3. Apply the formula: Substitute u, v, du, and dv into the integration by parts formula.
4. Evaluate the new integral: ∫ v du. This integral should be simpler than the original. If not, you may need to adjust your choice of u and dv, or apply integration by parts again.
5. Add the constant of integration, C.

Example: ∫ x * e^x dx

1. Choose u and dv: Let u = x (Algebraic) and dv = e^x dx (Exponential). According to LIATE, Algebraic comes before Exponential, so this is a suitable choice.
2. Find du and v: du = dx and v = ∫ e^x dx = e^x
3. Apply the formula: ∫ x * e^x dx = x * e^x - ∫ e^x dx
4. Evaluate the new integral: ∫ e^x dx = e^x + C
5. Substitute and Simplify: x * e^x - e^x + C = e^x(x - 1) + C

Therefore, ∫ x * e^x dx = e^x(x - 1) + C

Verification by Differentiation:

d/dx [e^x(x - 1) + C] = e^x(x - 1) + e^x(1) + 0 = xe^x - e^x + e^x = xe^x

The derivative matches the integrand, confirming our solution.

3. Trigonometric Substitution

Trigonometric substitution is used to integrate expressions containing square roots of the form √(a² - x²), √(a² + x²), or √(x² - a²). It involves substituting x with a trigonometric function to eliminate the square root.

Common Substitutions:

• For √(a² - x²), let x = a sin(θ)
• For √(a² + x²), let x = a tan(θ)
• For √(x² - a²), let x = a sec(θ)

Steps for Trigonometric Substitution:

1. Choose the appropriate substitution: Based on the form of the square root in the integrand.
2. Find dx: Calculate the derivative of x with respect to θ.
3. Substitute: Replace the original expression in terms of x with the new expression in terms of θ.
4. Simplify: Use trigonometric identities to simplify the integral.
5. Integrate with respect to θ: Evaluate the integral with respect to θ.
6. Substitute back: Replace θ with its original expression in terms of x. You may need to use a right triangle to determine the relationships between the trigonometric functions and x.
7. Add the constant of integration, C.

Example: ∫ √(4 - x²) dx

1. Choose the appropriate substitution: Since we have √(4 - x²), let x = 2 sin(θ). Here, a = 2.
2. Find dx: dx = 2 cos(θ) dθ
3. Substitute: ∫ √(4 - x²) dx = ∫ √(4 - (2 sin(θ))²) * 2 cos(θ) dθ = ∫ √(4 - 4 sin²(θ)) * 2 cos(θ) dθ
4. Simplify: = ∫ √(4(1 - sin²(θ))) * 2 cos(θ) dθ = ∫ 2 cos(θ) * 2 cos(θ) dθ = 4 ∫ cos²(θ) dθ
5. Integrate with respect to θ: Use the identity cos²(θ) = (1 + cos(2θ))/2. 4 ∫ cos²(θ) dθ = 4 ∫ (1 + cos(2θ))/2 dθ = 2 ∫ (1 + cos(2θ)) dθ = 2(θ + (1/2)sin(2θ)) + C = 2θ + sin(2θ) + C
6. Substitute back: We know x = 2 sin(θ), so sin(θ) = x/2, and θ = arcsin(x/2). Also, sin(2θ) = 2 sin(θ) cos(θ). Since sin(θ) = x/2, we can draw a right triangle where the opposite side is x, the hypotenuse is 2, and the adjacent side is √(4 - x²). Therefore, cos(θ) = √(4 - x²)/2. Substituting back: 2θ + sin(2θ) + C = 2 arcsin(x/2) + 2(x/2)(√(4 - x²)/2) + C = 2 arcsin(x/2) + (x√(4 - x²))/2 + C

Therefore, ∫ √(4 - x²) dx = 2 arcsin(x/2) + (x√(4 - x²))/2 + C

Verification by Differentiation:

d/dx [2 arcsin(x/2) + (x√(4 - x²))/2 + C] = 2 * (1/√(4 - x²)) * (1/2) + (1/2)[√(4 - x²) + x * (1/2)(4 - x²)^-1/2 * (-2x)] + 0 = 1/√(4 - x²) + (√(4 - x²)/2) - (x² / (2√(4 - x²))) = (2 + 4 - x² - x²) / (2√(4 - x²)) = (4-x²)/√(4-x²) = √(4-x²).

The derivative matches the integrand, confirming our solution.

Common Mistakes to Avoid

• Forgetting the Constant of Integration (C): Always include C in indefinite integrals.
• Incorrectly Applying the Power Rule: Ensure that n ≠ -1 when using the power rule.
• Choosing the Wrong u in u-Substitution: Practice choosing appropriate u values to simplify the integral.
• Errors in Trigonometric Identities: Double-check trigonometric identities used in trigonometric substitution.
• Not Verifying by Differentiation: Always verify your result by differentiating the antiderivative to ensure it matches the integrand.

Tips for Success

• Practice Regularly: Consistent practice is key to mastering indefinite integrals.
• Master Basic Integration Rules: Know the basic rules inside and out.
• Recognize Patterns: Identify common patterns in integrands to choose the appropriate integration technique.
• Use Online Resources: Utilize online calculators and solvers to check your work.
• Break Down Complex Integrals: Simplify complex integrals into smaller, manageable parts.
• Understand the Theory: Knowing the underlying concepts helps in problem-solving.

Conclusion

Determining indefinite integrals and verifying the results by differentiation is a fundamental skill in calculus. By understanding the basic rules, mastering advanced techniques like u-substitution, integration by parts, and trigonometric substitution, and consistently practicing, you can confidently tackle a wide range of integration problems. Always remember to verify your solutions by differentiation to ensure accuracy. With dedication and perseverance, you can unlock the power of indefinite integrals and enhance your calculus proficiency.