How to Solve: Indefinite Integration (Substitution, Integration by Parts, Partial Fractions, Trigonometric Integrals) – IIT JEE Guide

How to Solve: Indefinite Integration (Substitution, Integration by Parts, Partial Fractions, Trigonometric Integrals) – IIT JEE Guide

Introduction

Mastering indefinite integration unlocks 10-15% of your IIT JEE score—directly in 10-12 marks in JEE Main and 15-20 marks in JEE Advanced. Whether it’s calculating work done in physics or optimizing engineering designs, integration is the backbone of competitive exams and real-world problem-solving.

What You Need To Know First

1. Basic differentiation rules (power rule, product rule, chain rule).
2. Algebraic manipulation (factoring, polynomial division, trigonometric identities).
3. Trigonometric identities (Pythagorean, double-angle, half-angle).

Key Vocabulary

Formulas To Know

1. Substitution Method

Formula: ∫f(g(x)) g'(x) dx = ∫f(u) du, where u = g(x)

Variables: - u = substitution variable - du/dx = g'(x) → du = g'(x) dx

MEMORISE THIS

2. Integration by Parts

Formula: ∫u dv = uv - ∫v du

Variables: - u = differentiable function (choose using LIATE: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential) - dv = integrable part

MEMORISE THIS

3. Partial Fractions (for Rational Functions)

Case 1: Distinct Linear Factors Formula: 1/(x-a)(x-b) = A/(x-a) + B/(x-b)

Case 2: Repeated Linear Factors Formula: 1/(x-a)² = A/(x-a) + B/(x-a)²

Case 3: Irreducible Quadratic Factors Formula: 1/(x² + bx + c) = (Ax + B)/(x² + bx + c)

MEMORISE THESE FORMS

4. Trigonometric Integrals

Basic Integrals (Given on Exam Sheet, but memorize for speed): - ∫sin x dx = -cos x + C - ∫cos x dx = sin x + C - ∫sec²x dx = tan x + C - ∫csc²x dx = -cot x + C - ∫sec x tan x dx = sec x + C - ∫csc x cot x dx = -csc x + C

Power-Reducing Identities (MEMORISE THIS): - sin²x = (1 - cos2x)/2 - cos²x = (1 + cos2x)/2 - sin x cos x = sin2x/2

Step-by-Step Method

1. Substitution Method

When to use: When the integrand is a composite function (f(g(x))) and its derivative (g'(x)) is present.

Steps: 1. Identify u: Choose the inner function (g(x)) as u. 2. Find du: Differentiate u to get du = g'(x) dx. 3. Rewrite integral: Replace g(x) with u and g'(x) dx with du. 4. Integrate: Solve the new integral in terms of u. 5. Back-substitute: Replace u with g(x) and add +C.

Example: ∫2x e^(x²) dx

1. Let u = x² (inner function).
2. du = 2x dx (derivative of u).
3. Rewrite: ∫e^u du.
4. Integrate: e^u + C.
5. Back-substitute: e^(x²) + C.

2. Integration by Parts

When to use: When the integrand is a product of two functions (e.g., x eˣ, x sin x, ln x).

Steps: 1. Choose u and dv: Use LIATE (Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential). 2. Differentiate u: Find du. 3. Integrate dv: Find v. 4. Apply formula: ∫u dv = uv - ∫v du. 5. Simplify: Solve the remaining integral.

Example: ∫x eˣ dx

1. Let u = x (Algebraic), dv = eˣ dx.
2. du = dx.
3. v = eˣ.
4. Apply formula: x eˣ - ∫eˣ dx.
5. Simplify: x eˣ - eˣ + C = eˣ(x - 1) + C.

3. Partial Fractions

When to use: When the integrand is a rational function (polynomial/polynomial) with a denominator that can be factored.

Steps: 1. Factor denominator: Break into linear/quadratic factors. 2. Set up partial fractions: Write as A/(x-a) + B/(x-b) + ... (adjust for repeated/quadratic factors). 3. Solve for A, B, C: Multiply both sides by denominator and equate coefficients. 4. Rewrite integral: Split into simpler fractions. 5. Integrate each term: Use basic integral rules.

Example: ∫1/(x² - 1) dx

1. Factor denominator: x² - 1 = (x-1)(x+1).
2. Set up: 1/(x-1)(x+1) = A/(x-1) + B/(x+1).
3. Solve: 1 = A(x+1) + B(x-1).
4. Let x = 1 → A = 1/2.
5. Let x = -1 → B = -1/2.
6. Rewrite: (1/2)∫1/(x-1) dx - (1/2)∫1/(x+1) dx.
7. Integrate: (1/2)ln|x-1| - (1/2)ln|x+1| + C = (1/2)ln|(x-1)/(x+1)| + C.

4. Trigonometric Integrals

When to use: When the integrand is a product of trigonometric functions (e.g., sin²x, sin x cos x, sec x tan x).

Steps: 1. Simplify using identities: Convert powers to single angles (e.g., sin²x → (1 - cos2x)/2). 2. Substitute if needed: For odd powers, save one function and convert the rest. 3. Integrate: Use basic trigonometric integrals. 4. Back-substitute: If substitution was used, replace the variable.

Example: ∫sin³x dx

1. Rewrite: sin³x = sin²x sin x = (1 - cos²x) sin x.
2. Let u = cos x → du = -sin x dx.
3. Rewrite: ∫(1 - u²)(-du) = ∫(u² - 1) du.
4. Integrate: (u³/3) - u + C.
5. Back-substitute: (cos³x)/3 - cos x + C.

Worked Examples

Example 1 – Basic (Substitution)

Problem: ∫(3x² + 2x) e^(x³ + x²) dx

Solution: 1. Let u = x³ + x² (inner function). 2. du = (3x² + 2x) dx (derivative of u). 3. Rewrite: ∫e^u du. 4. Integrate: e^u + C. 5. Back-substitute: e^(x³ + x²) + C.

What we did and why: We recognized the integrand as e^(composite function) × derivative of composite function, so substitution simplifies it to ∫e^u du.

Example 2 – Medium (Integration by Parts)

Problem: ∫x ln x dx

Solution: 1. Let u = ln x (Logarithmic), dv = x dx. 2. du = (1/x) dx, v = x²/2. 3. Apply formula: (ln x)(x²/2) - ∫(x²/2)(1/x) dx. 4. Simplify: (x²/2) ln x - (1/2)∫x dx. 5. Integrate: (x²/2) ln x - (x²/4) + C.

What we did and why: We used LIATE to choose u = ln x (Logarithmic > Algebraic), then applied integration by parts to reduce the integral.

Example 3 – Exam-Style (Partial Fractions + Trigonometric)

Problem: ∫(x² + 1)/(x³ + x) dx

Solution: 1. Factor denominator: x³ + x = x(x² + 1). 2. Set up: (x² + 1)/x(x² + 1) = A/x + (Bx + C)/(x² + 1). 3. Solve: x² + 1 = A(x² + 1) + (Bx + C)x.
- Let x = 0 → A = 1.
- Compare coefficients: B = 0, C = 0. 4. Rewrite: ∫1/x dx + ∫0 dx = ln|x| + C.

What we did and why: We factored the denominator and used partial fractions, but the x² + 1 terms canceled, simplifying the integral to a basic ln|x|.

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute integration cheat sheet for JEE!

1. Substitution? If you see a function and its derivative, let u = inner function and rewrite.
2. Integration by parts? Use LIATE to pick u, then apply uv - ∫v du.
3. Partial fractions? Factor the denominator, set up A/(x-a) + B/(x-b), solve for A and B, then integrate.
4. Trig integrals? Use identities to reduce powers (sin²x → (1 - cos2x)/2) or save one function for substitution.
5. Always add +C! And double-check your algebra—examiners love hiding simple mistakes.

Pro tip: If an integral looks too hard, try substitution first. If it’s a product, think integration by parts. If it’s a fraction, factor the denominator. You’ve got this—now go ace that exam!