Class 12 Mathematics Notes Chapter 7 (Integrals) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 7: Integrals from the NCERT Exemplar. This is a cornerstone chapter for calculus and frequently tested in various government examinations. Pay close attention as we break down the key concepts and techniques.

Chapter 7: Integrals - Detailed Notes (NCERT Exemplar Perspective)

1. Introduction: Integration as an Antiderivative

• Concept: Integration is the inverse process of differentiation. If the derivative of F(x) is f(x) (i.e., d/dx [F(x)] = f(x)), then the integral of f(x) with respect to x is F(x) + C.
  • ∫ f(x) dx = F(x) + C
• Terminology:
  • ∫: Integral sign
  • f(x): Integrand (the function being integrated)
  • dx: Indicates integration with respect to the variable x
  • F(x): Antiderivative or primitive of f(x)
  • C: Constant of Integration (accounts for the fact that the derivative of any constant is zero). This is crucial for indefinite integrals.

2. Standard Elementary Integrals (Based on Derivatives)

You must memorize these fundamental integration formulas:

• ∫ xⁿ dx = (xⁿ⁺¹ / (n+1)) + C, (n ≠ -1)
• ∫ 1 dx = ∫ dx = x + C
• ∫ (1/x) dx = ln|x| + C
• ∫ eˣ dx = eˣ + C
• ∫ aˣ dx = (aˣ / ln a) + C, (a > 0, a ≠ 1)
• ∫ sin x dx = -cos x + C
• ∫ cos x dx = sin x + C
• ∫ sec² x dx = tan x + C
• ∫ cosec² x dx = -cot x + C
• ∫ sec x tan x dx = sec x + C
• ∫ cosec x cot x dx = -cosec x + C
• ∫ (1 / √(1 - x²)) dx = sin⁻¹ x + C = -cos⁻¹ x + C'
• ∫ (1 / (1 + x²)) dx = tan⁻¹ x + C = -cot⁻¹ x + C'
• ∫ (1 / (x √(x² - 1))) dx = sec⁻¹ x + C = -cosec⁻¹ x + C'

3. Properties of Indefinite Integrals

• Linearity:
  • ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
  • ∫ k * f(x) dx = k * ∫ f(x) dx (where k is a constant)
• d/dx [∫ f(x) dx] = f(x) (Derivative of an integral is the original function)
• ∫ [d/dx F(x)] dx = F(x) + C (Integral of a derivative gives the function plus a constant)

4. Methods of Integration

(a) Integration by Substitution:

• Purpose: To transform the integral into a standard form by substituting a part of the integrand with a new variable.
• Technique:
  1. Choose a suitable substitution, often u = g(x).
  2. Find du/dx = g'(x) => du = g'(x) dx.
  3. Substitute u and du into the integral. The integral should now be entirely in terms of u.
  4. Integrate with respect to u.
  5. Replace u back with g(x) to get the final answer in terms of x.
• Common Forms: ∫ f(g(x)) * g'(x) dx; substitute u = g(x).
• Important Substitutions:
  • For √(a² - x²), substitute x = a sin θ or x = a cos θ
  • For √(a² + x²), substitute x = a tan θ or x = a cot θ
  • For √(x² - a²), substitute x = a sec θ or x = a cosec θ

(b) Integration using Trigonometric Identities:

• Purpose: To simplify trigonometric integrands using identities before integrating.
• Key Identities: Double angle formulas (sin 2x, cos 2x), product-to-sum formulas (2 sin A cos B, etc.), power reduction formulas (sin²x, cos²x).
  • Example: ∫ sin²x dx = ∫ (1 - cos 2x)/2 dx

(c) Integration by Parts:

• Purpose: To integrate the product of two functions.
• Formula: ∫ u v dx = u ∫ v dx - ∫ [ (du/dx) * (∫ v dx) ] dx
  • u: First function
  • v: Second function
• Choosing u and v (ILATE Rule): Choose the first function (u) based on this preference order:
  • Inverse Trigonometric
  • Logarithmic
  • Algebraic (polynomials)
  • Trigonometric
  • Exponential
    The function appearing earlier in ILATE is typically chosen as 'u'.
• Note: Sometimes repeated application of integration by parts is needed.

(d) Integration by Partial Fractions:

• Purpose: To integrate rational functions P(x)/Q(x) where degree(P(x)) < degree(Q(x)) (proper rational function). If not proper, perform polynomial long division first.
• Technique: Decompose the rational function into simpler fractions based on the factors of the denominator Q(x).
  • Case 1: Distinct Linear Factors: Q(x) = (x-a)(x-b)... => P(x)/Q(x) = A/(x-a) + B/(x-b) + ...
  • Case 2: Repeated Linear Factors: Q(x) = (x-a)ᵏ... => P(x)/Q(x) = A₁/(x-a) + A₂/(x-a)² + ... + Aₖ/(x-a)ᵏ + ...
  • Case 3: Irreducible Quadratic Factors: Q(x) = (ax²+bx+c)... (where b²-4ac < 0) => P(x)/Q(x) = (Ax+B)/(ax²+bx+c) + ...
  • Case 4: Repeated Irreducible Quadratic Factors: Similar structure to Case 2, but with linear numerators (Ax+B) over powers of the quadratic.
• Finding Constants (A, B, C...): Equate numerators after taking LCM and either compare coefficients of like powers of x or substitute strategic values of x (roots of factors).

(e) Integrals of Some Particular Functions (Standard Forms):

• ∫ dx / (x² - a²) = (1/2a) ln |(x-a)/(x+a)| + C

• ∫ dx / (a² - x²) = (1/2a) ln |(a+x)/(a-x)| + C

• ∫ dx / (x² + a²) = (1/a) tan⁻¹(x/a) + C

• ∫ dx / √(x² - a²) = ln |x + √(x² - a²)| + C

• ∫ dx / √(a² - x²) = sin⁻¹(x/a) + C

• ∫ dx / √(x² + a²) = ln |x + √(x² + a²)| + C

• ∫ √(x² - a²) dx = (x/2)√(x² - a²) - (a²/2) ln |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) ln |x + √(x² + a²)| + C

• Integrals of the form: ∫ (px+q) / (ax²+bx+c) dx, ∫ (px+q) / √(ax²+bx+c) dx

  • Technique: Express px+q = A * d/dx(ax²+bx+c) + B. Find A and B. Split the integral into two parts, one solvable by substitution (using the derivative) and the other reducing to the standard forms above (after completing the square in the denominator if necessary).

• Integrals of the form: ∫ dx / (ax²+bx+c), ∫ dx / √(ax²+bx+c)

  • Technique: Complete the square in the quadratic expression in the denominator to reduce it to one of the standard forms involving x² ± a² or a² - x².

5. Definite Integrals

• Concept: Represents the algebraic sum of areas under the curve y = f(x) from x = a to x = b.
• Notation: ∫[a, b] f(x) dx
  • a: Lower limit of integration
  • b: Upper limit of integration
• Fundamental Theorem of Calculus (Part 2 - Evaluation):
  If F(x) is an antiderivative of f(x) (i.e., ∫ f(x) dx = F(x)), then:
  ∫[a, b] f(x) dx = F(b) - F(a)
  (No constant 'C' needed for definite integrals as it cancels out).
• Definite Integral as the Limit of a Sum (Conceptual Basis):
  ∫[a, b] f(x) dx = lim (h→0) h [ f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h) ]
  where h = (b-a)/n and n → ∞. (Less common for direct evaluation in exams but important conceptually).

6. Properties of Definite Integrals (Extremely Important)

Let f and g be continuous functions.

• P₀: ∫[a, b] f(x) dx = ∫[a, b] f(t) dt (Change of variable doesn't change value)
• P₁: ∫[a, b] f(x) dx = - ∫[b, a] f(x) dx (Interchanging limits changes sign)
• P₂: ∫[a, a] f(x) dx = 0 (Integral over zero width is zero)
• P₃: ∫[a, b] f(x) dx = ∫[a, c] f(x) dx + ∫[c, b] f(x) dx (Splitting the interval)
• P₄: ∫[a, b] f(x) dx = ∫[a, b] f(a+b-x) dx ("King's Property" - Very useful)
• P₅: ∫[0, a] f(x) dx = ∫[0, a] f(a-x) dx (Special case of P₄)
• P₆: ∫[0, 2a] f(x) dx = ∫[0, a] f(x) dx + ∫[0, a] f(2a-x) dx
  • Corollary 1: If f(2a-x) = f(x), then ∫[0, 2a] f(x) dx = 2 * ∫[0, a] f(x) dx
  • Corollary 2: If f(2a-x) = -f(x), then ∫[0, 2a] f(x) dx = 0
• P₇: (Even and Odd Functions)
  • ∫[-a, a] f(x) dx = 2 * ∫[0, a] f(x) dx, if f is an even function (f(-x) = f(x))
  • ∫[-a, a] f(x) dx = 0, if f is an odd function (f(-x) = -f(x))

Key Takeaways for Exams:

1. Master the standard integral formulas.
2. Be proficient in all methods: Substitution, By Parts, Partial Fractions. Recognize when to use which.
3. Memorize and understand how to apply the properties of definite integrals, especially P₄, P₅, P₆, and P₇, as they often simplify complex problems significantly.
4. Practice completing the square for integrals involving quadratics.
5. Pay attention to the constant of integration 'C' for indefinite integrals.

Multiple Choice Questions (MCQs)

Here are 10 MCQs covering various concepts from the chapter:

1. The anti-derivative of (√x + 1/√x) is:
   (A) (1/2)x¹ᐟ² + 2x⁻¹ᐟ² + C
   (B) (2/3)x³/² + 2x¹ᐟ² + C
   (C) (3/2)x³/² + (1/2)x¹ᐟ² + C
   (D) (2/3)x³/² + (1/2)x⁻¹ᐟ² + C

2. ∫ eˣ (sec x + sec x tan x) dx equals:
   (A) eˣ tan x + C
   (B) eˣ sec x + C
   (C) eˣ log|sec x| + C
   (D) eˣ log|tan x| + C
   (Hint: Use the form ∫ eˣ [f(x) + f'(x)] dx = eˣ f(x) + C)

3. ∫ dx / (x² + 2x + 2) equals:
   (A) tan⁻¹(x+1) + C
   (B) (1/2) tan⁻¹(x+1) + C
   (C) log|x² + 2x + 2| + C
   (D) -1 / (x+1) + C
   (Hint: Complete the square in the denominator)

4. The value of ∫[0, π/2] (sin x / (sin x + cos x)) dx is:
   (A) π/4
   (B) π/2
   (C) 0
   (D) π
   (Hint: Use property P₅: ∫[0, a] f(x) dx = ∫[0, a] f(a-x) dx)

5. ∫ x sec²x dx equals:
   (A) x tan x + log|cos x| + C
   (B) x tan x - log|sec x| + C
   (C) x sec x - log|sec x + tan x| + C
   (D) x tan x + log|sec x| + C
   (Hint: Use Integration by Parts - ILATE)

6. If ∫ f(x) dx = g(x) + C, then ∫ f(ax+b) dx equals:
   (A) g(ax+b) + C
   (B) a * g(ax+b) + C
   (C) (1/a) * g(ax+b) + C
   (D) g(ax+b) / b + C

7. The value of ∫[-1, 1] sin⁵x cos⁴x dx is:
   (A) 1
   (B) 2 * ∫[0, 1] sin⁵x cos⁴x dx
   (C) 0
   (D) π/2
   (Hint: Check if the integrand is an odd or even function)

8. ∫ dx / (x(x²+1)) equals:
   (A) log|x| - (1/2)log(x²+1) + C
   (B) log|x| + (1/2)log(x²+1) + C
   (C) -log|x| + (1/2)log(x²+1) + C
   (D) (1/2)log|x| + log(x²+1) + C
   (Hint: Use Partial Fractions)

9. The value of ∫[0, 1] tan⁻¹((2x-1)/(1+x-x²)) dx is:
   (A) 1
   (B) 0
   (C) -1
   (D) π/4
   (Hint: Try simplifying the argument using tan⁻¹x - tan⁻¹y)

10. ∫ (cos 2x - cos 2α) / (cos x - cos α) dx equals:
    (A) 2(sin x + x cos α) + C
    (B) 2(sin x - x cos α) + C
    (C) 2(cos x + x sin α) + C
    (D) 2(cos x - x sin α) + C
    (Hint: Use cos 2θ = 2cos²θ - 1)

Answer Key for MCQs:

1. (B)
2. (B)
3. (A)
4. (A)
5. (D) (Correction: It should be x tan x - ln|sec x| + C or x tan x + ln|cos x| + C. Option D is the closest standard form if we interpret log as ln) Let's re-evaluate Q5. ∫ x sec²x dx. u=x, dv=sec²x dx => du=dx, v=tan x. ∫ = x tan x - ∫ tan x dx = x tan x - (-ln|cos x|) + C = x tan x + ln|cos x| + C. OR x tan x - ln|sec x| + C. Option (D) has +log|sec x|, Option (A) has +log|cos x|. Let's assume log means natural log (ln). So (A) is correct. Let's stick with (A).
6. (C)
7. (C) (Integrand is odd)
8. (A)
9. (B) (The integral simplifies to ∫[0, 1] (tan⁻¹x - tan⁻¹(x-1)) dx. Using property P4 or direct integration leads to 0).
10. (A) (Simplify numerator: (2cos²x - 1) - (2cos²α - 1) = 2(cos²x - cos²α) = 2(cos x - cos α)(cos x + cos α). The integral becomes ∫ 2(cos x + cos α) dx = 2(sin x + x cos α) + C)

Remember to practice a wide variety of problems from the NCERT book and the Exemplar to gain mastery over these concepts. Good luck with your preparation!