Integrals – Chapter 7 | AAI ATC Exam Prep

📖 Chapter Overview

Chapter 7 – Integrals is one of the most important topics for AAI ATC written examination. It covers anti-derivatives, integration techniques, definite integrals, and their applications. Questions from this chapter frequently appear in Section B of the AAI ATC paper.

✦ Anti-derivatives

✦ Standard Integrals

✦ Integration by Substitution

✦ Integration by Parts

✦ Partial Fractions

✦ Definite Integrals

✦ Properties of Integrals

✦ Trig Identities

7.1

Introduction to Integrals

Integration as the inverse of differentiation · Anti-derivatives · Notation

🔄 What is Integration?

Integration is the inverse process of differentiation. While differentiation finds the rate of change of a function, integration recovers the original function from its derivative.

If F'(x) = f(x), then F(x) is called the anti-derivative (primitive) of f(x)
The family of all anti-derivatives is {F(x) + C, C ∈ ℝ}, where C is the constant of integration
Two functions with the same derivative differ by a constant only
Integration solves: (a) finding a function from its derivative; (b) finding area under a curve

Standard Notation

∫ f(x) dx = F(x) + C [Indefinite Integral]

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Insert Image Here: Graphical representation of anti-derivative concept
(Add your diagram/screenshot in this space)

Term	Meaning
∫ f(x) dx	Integral of f w.r.t. x
f(x)	Integrand
x	Variable of integration
C	Constant of integration
F(x)	Anti-derivative such that F'(x) = f(x)

📝 Practice MCQs – Section 7.1

Q1. If F(x) = x³/3 + C, then F'(x) equals:

A) 3x²

B) x²

C) x³

D) x/3

✅ Answer: (B) x² | Differentiating x³/3 + C gives (3x²)/3 = x²

Q2. The anti-derivative of cos x is:

A) –sin x + C

B) sin x + C

C) tan x + C

D) cos x + C

✅ Answer: (B) sin x + C | Since d/dx(sin x) = cos x, ∫cos x dx = sin x + C

Q3. If F(0) = 3 and F'(x) = 4x³ – 6, then F(1) = ?

A) 0

B) 1

C) –2

D) 4

✅ Answer: (C) –2 | F(x) = x⁴ – 6x + C. F(0)=C=3, so F(x)=x⁴–6x+3. F(1)=1–6+3=–2

7.2

Standard Integrals & Properties

Power rule · Trigonometric · Exponential · Log · Properties I–V

📋 Standard Formulae Table

These must be memorized for AAI ATC exam. Questions directly test application of these formulae.

Derivative	Integral
d/dx(xⁿ⁺¹/(n+1)) = xⁿ	∫xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ –1
d/dx(sin x) = cos x	∫cos x dx = sin x + C
d/dx(–cos x) = sin x	∫sin x dx = –cos x + C
d/dx(tan x) = sec²x	∫sec²x dx = tan x + C
d/dx(–cot x) = cosec²x	∫cosec²x dx = –cot x + C
d/dx(sec x) = sec x tan x	∫sec x tan x dx = sec x + C
d/dx(eˣ) = eˣ	∫eˣ dx = eˣ + C
d/dx(log\|x\|) = 1/x	∫(1/x) dx = log\|x\| + C
d/dx(sin⁻¹x) = 1/√(1–x²)	∫dx/√(1–x²) = sin⁻¹x + C
d/dx(tan⁻¹x) = 1/(1+x²)	∫dx/(1+x²) = tan⁻¹x + C

Key Properties

(III) ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
(IV) ∫k·f(x) dx = k·∫f(x) dx, for k ∈ ℝ

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Insert Image Here: Standard integrals summary chart or your handwritten notes
(Add your study material image here)

📝 Practice MCQs – Section 7.2

Q1. Evaluate: ∫(2x² + eˣ) dx

A) 4x + eˣ + C

B) 2x³/3 + eˣ + C

C) x³/3 + eˣ + C

D) 2x³ + eˣ + C

✅ Answer: (B) | ∫2x² dx = 2·x³/3, and ∫eˣ dx = eˣ. So answer = 2x³/3 + eˣ + C

Q2. The value of ∫(x³ – 1)/x² dx is:

A) x²/2 – x + C

B) x²/2 + 1/x + C

C) x³/3 – 1/x + C

D) x²/2 – 1/x + C

✅ Answer: (B) | (x³–1)/x² = x – x⁻². So ∫(x – x⁻²) dx = x²/2 – x⁻¹/(–1) = x²/2 + 1/x + C

Q3. ∫sec x (sec x + tan x) dx equals:

A) –cot x + sec x + C

B) tan x – sec x + C

C) tan x + sec x + C

D) tan²x + C

✅ Answer: (C) | sec x(sec x + tan x) = sec²x + sec x·tan x. ∫sec²x dx + ∫sec x·tan x dx = tan x + sec x + C

7.3

Methods of Integration

Integration by Substitution · Trigonometric Identities

🔁 Integration by Substitution

Change the variable x to t via x = g(t). Then dx = g'(t) dt, transforming ∫f(x) dx → ∫f(g(t))·g'(t) dt.

Key Rule

If x = g(t), then ∫f(x) dx = ∫f(g(t))·g'(t) dt

∫tan x dx = log|sec x| + C (Put cos x = t)
∫cot x dx = log|sin x| + C (Put sin x = t)
∫sec x dx = log|sec x + tan x| + C
∫cosec x dx = log|cosec x – cot x| + C

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Insert Image Here: Worked examples of substitution method with step-by-step working
(Add your solved examples here)

🔺 Integration Using Trig Identities

cos²x = (1 + cos2x)/2 → ∫cos²x dx = x/2 + sin2x/4 + C
sin²x = (1 – cos2x)/2 → ∫sin²x dx = x/2 – sin2x/4 + C
sin 3x = 3sin x – 4sin³x → use to integrate sin³x
sin x cos y = ½[sin(x+y) + sin(x–y)]

📝 Practice MCQs – Section 7.3

Q1. ∫2x sin(x² + 1) dx equals:

A) cos(x² + 1) + C

B) –cos(x² + 1) + C

C) sin(x² + 1) + C

D) –2cos(x² + 1) + C

✅ Answer: (B) | Put t = x² + 1, dt = 2x dx. Then ∫sin t dt = –cos t + C = –cos(x² + 1) + C

Q2. ∫cos²x dx equals:

A) x + sin2x/2 + C

B) x/2 + sin2x/4 + C

C) sin2x/4 + C

D) x/2 – sin2x/4 + C

✅ Answer: (B) | cos²x = (1+cos2x)/2. So ∫cos²x dx = ½∫(1+cos2x) dx = x/2 + sin2x/4 + C

Q3. The value of ∫sin³x dx is:

A) –cos x + cos³x/3 + C

B) –3cosx/4 + cos3x/12 + C

C) 3cosx/4 – cos3x/12 + C

D) –cos x + C

✅ Answer: (B) | sin³x = (3sinx – sin3x)/4. ∫sin³x dx = ¾∫sinx dx – ¼∫sin3x dx = –3cosx/4 + cos3x/12 + C

7.4

Integrals of Some Particular Functions

Six special standard formulae · Completing the square method

📐 Six Important Formulae

#	Integral	Result
1	∫dx/(x²–a²)	(1/2a) log\|(x–a)/(x+a)\| + C
2	∫dx/(a²–x²)	(1/2a) log\|(a+x)/(a–x)\| + C
3	∫dx/(x²+a²)	(1/a) tan⁻¹(x/a) + C
4	∫dx/√(x²–a²)	log\|x + √(x²–a²)\| + C
5	∫dx/√(a²–x²)	sin⁻¹(x/a) + C
6	∫dx/√(x²+a²)	log\|x + √(x²+a²)\| + C

Completing the Square Trick

ax² + bx + c = a[(x + b/2a)² + (c/a – b²/4a²)]
Then substitute t = x + b/2a to reduce to standard form

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Insert Image Here: Worked example of completing the square to evaluate integral
(Add your solved example of type 7 or 8 here)

📝 Practice MCQs – Section 7.4

Q1. ∫dx/(x² – 16) equals:

A) (1/8) log|(x–4)/(x+4)| + C

B) (1/8) log|(x–4)/(x+4)| + C ✓ (a=4)

C) (1/4) tan⁻¹(x/4) + C

D) log|x–4| + C

✅ Answer: (B) | Using formula 1 with a=4: (1/2·4)log|(x–4)/(x+4)| = (1/8)log|(x–4)/(x+4)| + C

Q2. ∫dx/√(2x – x²) equals: [Hint: 2x–x² = 1–(x–1)²]

A) cos⁻¹(x–1) + C

B) sin⁻¹(x–1) + C

C) tan⁻¹(x–1) + C

D) log|x–1| + C

✅ Answer: (B) | 2x–x² = 1–(x–1)². Put t = x–1. ∫dt/√(1–t²) = sin⁻¹t + C = sin⁻¹(x–1) + C

Q3. ∫dx/(x² – 6x + 13) equals:

A) (1/4) tan⁻¹(x–3) + C

B) (1/2) tan⁻¹((x–3)/2) + C

C) log|x² – 6x + 13| + C

D) (1/2) sin⁻¹(x–3) + C

✅ Answer: (B) | x²–6x+13 = (x–3)²+4. Put t=x–3, a=2. ∫dt/(t²+4) = (1/2)tan⁻¹(t/2) + C = (1/2)tan⁻¹((x–3)/2) + C

7.5

Integration by Partial Fractions

Rational functions · Decomposition types · Long division

🧮 When to Use Partial Fractions

Use when integrating a rational function P(x)/Q(x). If degree of P ≥ degree of Q, first do polynomial long division.

Form of Rational Function	Partial Fraction Form
(px+q)/[(x–a)(x–b)], a≠b	A/(x–a) + B/(x–b)
(px+q)/(x–a)²	A/(x–a) + B/(x–a)²
(px²+qx+r)/[(x–a)(x–b)(x–c)]	A/(x–a)+B/(x–b)+C/(x–c)
(px²+qx+r)/[(x–a)(x²+bx+c)]	A/(x–a) + (Bx+C)/(x²+bx+c)

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Insert Image Here: Worked example of partial fraction decomposition for AAI ATC level
(Add your solved problem here)

📝 Practice MCQs – Section 7.5

Q1. ∫dx/[(x+1)(x+2)] = log|x+1| – log|x+2| + C. The value of A in the partial fraction A/(x+1) + B/(x+2) is:

A) 1

B) –1

C) 2

D) 1/2

✅ Answer: (A) 1 | 1 = A(x+2)+B(x+1). Put x=–1: 1=A(1), so A=1. Put x=–2: 1=B(–1), so B=–1

Q2. For integrating x/(x–1)(x–2)(x–3), the partial fraction has how many constants to find?

A) 1

B) 2

C) 3

D) 4

✅ Answer: (C) 3 | Three distinct linear factors → A/(x–1)+B/(x–2)+C/(x–3) → 3 constants A, B, C

Q3. ∫x dx/[(x–1)(x–2)(x–3)]: Value of A + B + C = ? (where A/(x–1)+B/(x–2)+C/(x–3) is the decomposition)

A) 0

B) 1

C) –1

D) 3

✅ Answer: (B) 1 | A=1/(−1·−2)=½, B=2/(1·−1)=−2, C=3/(2·1)=3/2. A+B+C = ½–2+3/2 = 1

7.6

Integration by Parts

ILATE rule · eˣ[f(x)+f'(x)] formula · Special integrals

🤝 Integration by Parts Formula

Main Formula

∫u·v dx = u·∫v dx – ∫[u' · ∫v dx] dx
(First function × Integral of second) – ∫(derivative of first × integral of second) dx

ILATE Rule for choosing first function: Inverse trig → Logarithmic → Algebraic → Trigonometric → Exponential
∫log x dx = x log x – x + C
∫xeˣ dx = xeˣ – eˣ + C
∫x cos x dx = x sin x + cos x + C
Special: ∫eˣ[f(x)+f'(x)] dx = eˣ f(x) + C

Special Integrals (7.6.2)

∫√(x²–a²) dx = (x/2)√(x²–a²) – (a²/2)log|x+√(x²–a²)| + C
∫√(a²–x²) dx = (x/2)√(a²–x²) + (a²/2)sin⁻¹(x/a) + C
∫√(x²+a²) dx = (x/2)√(x²+a²) + (a²/2)log|x+√(x²+a²)| + C

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Insert Image Here: ILATE rule diagram + worked example of ∫x·eˣ dx step-by-step
(Add your teaching diagram here)

📝 Practice MCQs – Section 7.6

Q1. ∫x·eˣ dx = ?

A) xeˣ + C

B) xeˣ – eˣ + C

C) x²eˣ/2 + C

D) eˣ(x+1) + C

✅ Answer: (B) | By parts: u=x, v=eˣ. ∫xeˣ dx = x·eˣ – ∫1·eˣ dx = xeˣ – eˣ + C = eˣ(x–1) + C

Q2. ∫eˣ(sin x + cos x) dx = ?

A) eˣ cos x + C

B) eˣ sin x + C

C) eˣ(sin x – cos x) + C

D) eˣ tan x + C

✅ Answer: (B) | Using ∫eˣ[f(x)+f'(x)] = eˣf(x) + C. Here f(x)=sinx, f'(x)=cosx. So result = eˣ sin x + C

Q3. ∫log x dx = ?

A) x/log x + C

B) x log x – x + C

C) log x – x + C

D) x log x + C

✅ Answer: (B) | By parts: u=log x (first), v=1 (second). ∫log x · 1 dx = log x · x – ∫(1/x)·x dx = x log x – x + C

7.7–7.10

Definite Integrals & Properties

Fundamental Theorems · Evaluation by substitution · Key properties P0–P7

📌 Second Fundamental Theorem

Key Result

∫ₐᵇ f(x) dx = F(b) – F(a), where F'(x) = f(x)

⚙️ Properties of Definite Integrals (AAI ATC Favourites)

Property	Statement
P₁	∫ₐᵇ f(x) dx = –∫ᵦₐ f(x) dx
P₂	∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx
P₃	∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b–x) dx
P₄	∫₀ᵃ f(x) dx = ∫₀ᵃ f(a–x) dx
P₇(i)	∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx, if f is even
P₇(ii)	∫₋ₐᵃ f(x) dx = 0, if f is odd

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Insert Image Here: Area interpretation of definite integral diagram (Fig 7.1 style)
(Add your geometric interpretation diagram here)

📝 Practice MCQs – Definite Integrals (7.7–7.10)

Q1. Evaluate ∫₂³ x² dx:

A) 5

B) 6

C) 19/3

D) 9

✅ Answer: (C) | F(x) = x³/3. F(3)–F(2) = 27/3 – 8/3 = 19/3

Q2. ∫₋₁¹ sin⁵x·cos⁴x dx = ?

A) 2/9

B) 0

C) 1

D) –1

✅ Answer: (B) 0 | f(x) = sin⁵x·cos⁴x. f(–x) = (–sinx)⁵·cos⁴x = –sin⁵x·cos⁴x = –f(x). Odd function → integral over [–1,1] = 0

Q3. ∫₀^(π/2) sin⁴x/(sin⁴x + cos⁴x) dx = ? [Use P4]

A) 0

B) π/2

C) π/4

D) 1

✅ Answer: (C) π/4 | Let I = ∫₀^(π/2) sin⁴x/(sin⁴x+cos⁴x) dx. By P4: I = ∫₀^(π/2) cos⁴x/(cos⁴x+sin⁴x) dx. Adding: 2I = ∫₀^(π/2) 1 dx = π/2. So I = π/4

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Integrals – Complete Lesson

📖 Chapter Overview

Introduction to Integrals

🔄 What is Integration?

Standard Integrals & Properties

📋 Standard Formulae Table

Methods of Integration

🔁 Integration by Substitution

🔺 Integration Using Trig Identities

Integrals of Some Particular Functions

📐 Six Important Formulae

Integration by Partial Fractions

🧮 When to Use Partial Fractions

Integration by Parts

🤝 Integration by Parts Formula

Definite Integrals & Properties

📌 Second Fundamental Theorem

⚙️ Properties of Definite Integrals (AAI ATC Favourites)