Class 12 Notes

Class 12 Mathematics Notes Chapter 1 (Integrals) – Mathematics Part-II Book

Philoid

Philoid

— 8 min read

Mathematics Part-II
Alright class, let's get straight into one of the most crucial chapters for your Class 12 board exams and especially for various government exams – Integrals. This chapter, Chapter 7 in your NCERT Part II book, forms the backbone of calculus alongside differentiation and has wide applications. Mastering it is essential.

We'll break down the concepts systematically.

Chapter 7: Integrals - Detailed Notes for Government Exam Preparation

1. Introduction: What is Integration?

  • Antiderivative: Integration is fundamentally the reverse process of differentiation. If the derivative of F(x) is f(x) (i.e., F'(x) = f(x)), then the integral (or antiderivative or primitive) of f(x) with respect to x is F(x).
  • Notation: We write ∫ f(x) dx = F(x) + C.
    • ∫ is the integral sign.
    • f(x) is the integrand (the function being integrated).
    • dx indicates that the integration is performed with respect to the variable x.
    • F(x) is the antiderivative.
    • C is the Constant of Integration: Since the derivative of any constant is zero, when we reverse the process, we must add an arbitrary constant 'C' to represent all possible antiderivatives. This is crucial for indefinite integrals.
  • Geometrical Interpretation: Geometrically, the indefinite integral represents a family of curves parallel to each other, whose tangents at points with the same x-coordinate have the same slope f(x).
  • Types:
    • Indefinite Integrals: Find the general antiderivative (includes '+ C').
    • Definite Integrals: Calculate the value of an integral between specific limits (represents area, net change, etc.).

2. Indefinite Integrals

(a) Standard Formulas (Essential - Memorize!)

These are derived directly from differentiation formulas:

  1. ∫ xⁿ dx = (xⁿ⁺¹ / (n+1)) + C, (n ≠ -1)
  2. ∫ 1 dx = ∫ x⁰ dx = x + C
  3. ∫ (1/x) dx = ln|x| + C
  4. ∫ eˣ dx = eˣ + C
  5. ∫ aˣ dx = (aˣ / ln a) + C, (a > 0, a ≠ 1)
  6. ∫ sin x dx = -cos x + C
  7. ∫ cos x dx = sin x + C
  8. ∫ sec² x dx = tan x + C
  9. ∫ cosec² x dx = -cot x + C
  10. ∫ sec x tan x dx = sec x + C
  11. ∫ cosec x cot x dx = -cosec x + C
  12. ∫ (1 / √(1 - x²)) dx = sin⁻¹ x + C = -cos⁻¹ x + C'
  13. ∫ (1 / (1 + x²)) dx = tan⁻¹ x + C = -cot⁻¹ x + C'
  14. ∫ (1 / (x √(x² - 1))) dx = sec⁻¹ x + C = -cosec⁻¹ x + C'
  15. ∫ tan x dx = ln|sec x| + C = -ln|cos x| + C
  16. ∫ cot x dx = ln|sin x| + C = -ln|cosec x| + C
  17. ∫ sec x dx = ln|sec x + tan x| + C = ln|tan(π/4 + x/2)| + C
  18. ∫ cosec x dx = ln|cosec x - cot x| + C = ln|tan(x/2)| + C

(b) Properties of Indefinite Integrals

  1. d/dx (∫ f(x) dx) = f(x)
  2. ∫ f'(x) dx = f(x) + C
  3. ∫ [k * f(x)] dx = k * ∫ f(x) dx (where k is a constant)
  4. ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx (Linearity Property)

(c) Methods of Integration

When the integrand is not a standard form, we use these techniques:

  1. Integration by Substitution:

    • Idea: Transform the integral into a standard form by substituting a part of the integrand with a new variable (say, t). Choose the substitution such that its derivative is also present (possibly with a constant factor) in the integrand.
    • Process: If ∫ f(g(x)) * g'(x) dx, substitute t = g(x), then dt = g'(x) dx. The integral becomes ∫ f(t) dt, which might be easier to solve.
    • Common Forms:
      • ∫ f'(x) / f(x) dx = ln|f(x)| + C (Substitute t = f(x))
      • ∫ f'(x) * [f(x)]ⁿ dx = ([f(x)]ⁿ⁺¹ / (n+1)) + C (Substitute t = f(x))
      • Integrals involving tan x, cot x, sec x, cosec x are often solved using substitution.

  2. Integration using Trigonometric Identities:

    • Idea: Simplify the integrand using trigonometric identities before integrating.
    • Examples: Use sin²x = (1-cos2x)/2, cos²x = (1+cos2x)/2, sin3x, cos3x formulas, product-to-sum formulas (2sinAcosB, etc.), etc.

  3. Integration by Parts:

    • Idea: Used to integrate the product of two functions. Based on the product rule of differentiation.
    • Formula: ∫ u v dx = u ∫ v dx - ∫ [ (d/dx u) * (∫ v dx) ] dx
    • Choosing 'u' (First function) and 'v' (Second function): Use the ILATE or LIATE rule for prioritizing the first function (u):
      • Inverse Trigonometric (sin⁻¹x, cos⁻¹x, etc.)
      • Logarithmic (ln x, log x)
      • Algebraic (x², x+1, etc.)
      • Trigonometric (sin x, cos x, etc.)
      • Exponential (eˣ, aˣ)
        The function that comes first in this order is usually chosen as 'u'.
    • Example: ∫ x sin x dx (Here, u=x (Algebraic), v=sin x (Trigonometric))

  4. Integration by Partial Fractions:

    • Idea: Used to integrate rational functions P(x)/Q(x), where degree(P(x)) < degree(Q(x)). If not, perform long division first. Decompose the rational function into simpler fractions.
    • Process: Factorize the denominator Q(x). Based on the factors, decompose P(x)/Q(x):
      • Distinct Linear Factors: (x-a)(x-b) -> A/(x-a) + B/(x-b)
      • Repeated Linear Factors: (x-a)² -> A/(x-a) + B/(x-a)²
      • Non-factorizable Quadratic Factors: (ax²+bx+c) -> (Ax+B)/(ax²+bx+c)
      • Repeated Quadratic Factors: (ax²+bx+c)² -> (Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)²
    • Find the constants (A, B, C, etc.) by comparing coefficients or substituting values of x. Then integrate the simpler fractions.

(d) Some Special Integral Forms (Important for direct application)

  1. ∫ dx / (x² - a²) = (1 / 2a) ln |(x-a)/(x+a)| + C
  2. ∫ dx / (a² - x²) = (1 / 2a) ln |(a+x)/(a-x)| + C
  3. ∫ dx / (x² + a²) = (1 / a) tan⁻¹(x/a) + C
  4. ∫ dx / √(x² - a²) = ln |x + √(x² - a²)| + C
  5. ∫ dx / √(a² - x²) = sin⁻¹(x/a) + C
  6. ∫ dx / √(x² + a²) = ln |x + √(x² + a²)| + C

(e) Integrals of the type ∫ (px+q) / (ax²+bx+c) dx, ∫ (px+q) / √(ax²+bx+c) dx

  • Method: Express px+q = A * d/dx(ax²+bx+c) + B = A(2ax+b) + B.
  • Find A and B by comparing coefficients.
  • Split the integral into two parts: One solvable by substituting the quadratic term (or its derivative is in the numerator), and the other reducing to one of the special integral forms after completing the square in the denominator.

(f) Integrals of the type ∫ √(ax²+bx+c) dx, ∫ (px+q)√(ax²+bx+c) dx

  • These often involve completing the square and using standard results like:
    • ∫ √(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
    • ∫ √(x² + a²) dx = (x/2)√(x² + a²) + (a²/2)ln|x + √(x² + a²)| + C

3. Definite Integrals

(a) Definition

  • If ∫ f(x) dx = F(x) + C, then the definite integral of f(x) from x=a to x=b is:
    ∫[a to b] f(x) dx = [F(x)]<0xE1><0xB5><0xA6> = F(b) - F(a)
  • 'a' is the lower limit, 'b' is the upper limit.
  • The constant 'C' cancels out.
  • Represents the algebraic sum of areas under the curve y=f(x) from x=a to x=b. Areas above the x-axis are positive, below are negative.

(b) Fundamental Theorem of Calculus

  • First Fundamental Theorem: If f is continuous on [a, b], then the function g(x) = ∫[a to x] f(t) dt is continuous on [a, b] and differentiable on (a, b), and g'(x) = f(x).
  • Second Fundamental Theorem: If f is continuous on [a, b] and F is an antiderivative of f, then ∫[a to b] f(x) dx = F(b) - F(a). (This is the one used for evaluation).

(c) Properties of Definite Integrals (Very Important for Simplification)

Let f and g be continuous functions:

  • P₀: ∫[a to b] f(x) dx = ∫[a to b] f(t) dt (Changing variable doesn't change value)
  • P₁: ∫[a to b] f(x) dx = - ∫[b to a] f(x) dx (Interchanging limits negates value)
  • P₂: ∫[a to b] f(x) dx = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx (Splitting the interval)
  • P₃: ∫[a to b] f(x) dx = ∫[a to b] f(a+b-x) dx (Important for many problems)
  • P₄: ∫[0 to a] f(x) dx = ∫[0 to a] f(a-x) dx (A special case of P₃, very frequently used)
  • P₅: ∫[0 to 2a] f(x) dx = ∫[0 to a] f(x) dx + ∫[0 to a] f(2a-x) dx
  • P₆: ∫[0 to 2a] f(x) dx = 2 ∫[0 to a] f(x) dx, if f(2a-x) = f(x)
    = 0, if f(2a-x) = -f(x)
  • P₇: (Even and Odd Functions)
    • ∫[-a to a] f(x) dx = 2 ∫[0 to a] f(x) dx, if f(-x) = f(x) (f is an even function)
    • ∫[-a to a] f(x) dx = 0, if f(-x) = -f(x) (f is an odd function)

(d) Integration as the Limit of a Sum

  • This is the definition of definite integral based on Riemann sums.
  • ∫[a to 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 in MCQs for government exams compared to direct evaluation or using properties, but the concept is fundamental.

Exam Focus Points:

  1. Master Standard Formulas: Speed and accuracy start here.
  2. Identify the Correct Method: Quickly recognize whether substitution, by parts, partial fractions, or a special form is needed.
  3. Definite Integral Properties: These are key to solving many competitive exam problems quickly, especially P₃, P₄, P₇.
  4. Accuracy: Pay attention to signs, constants, and limits of integration.
  5. Practice: Work through a variety of problems from NCERT, NCERT Exemplar, and previous year question papers of relevant exams.

Multiple Choice Questions (MCQs)

  1. The value of ∫ x⁵ dx is:
    (A) 5x⁴ + C
    (B) x⁶ + C
    (C) x⁶ / 6 + C
    (D) x⁵ / 5 + C

  2. ∫ sec²(3x+5) dx equals:
    (A) tan(3x+5) + C
    (B) (1/3) tan(3x+5) + C
    (C) -cot(3x+5) + C
    (D) (-1/3) cot(3x+5) + C

  3. ∫ (2x / (1+x²)) dx is:
    (A) tan⁻¹(x) + C
    (B) 2 tan⁻¹(x) + C
    (C) ln(1+x²) + C
    (D) 2 ln(1+x²) + C

  4. Using integration by parts, ∫ x eˣ dx equals:
    (A) eˣ (x - 1) + C
    (B) eˣ (x + 1) + C
    (C) x eˣ + C
    (D) eˣ + C

  5. ∫ dx / (x² - 9) equals:
    (A) (1/6) ln |(x-3)/(x+3)| + C
    (B) (1/6) ln |(x+3)/(x-3)| + C
    (C) (1/3) tan⁻¹(x/3) + C
    (D) ln |x² - 9| + C

  6. The value of the definite integral ∫[0 to π/2] cos x dx is:
    (A) 0
    (B) 1
    (C) -1
    (D) π/2

  7. The value of ∫[-1 to 1] x³ dx is:
    (A) 1/4
    (B) 1/2
    (C) 0
    (D) 2

  8. Using properties of definite integrals, ∫[0 to π/2] (sin x / (sin x + cos x)) dx equals:
    (A) π/4
    (B) π/2
    (C) 0
    (D) 1

  9. ∫ dx / (4 + 9x²) equals:
    (A) (1/2) tan⁻¹(3x/2) + C
    (B) (1/6) tan⁻¹(3x/2) + C
    (C) (1/9) tan⁻¹(x/3) + C
    (D) (1/3) tan⁻¹(2x/3) + C

  10. The value of ∫[0 to 1] (tan⁻¹x / (1+x²)) dx is:
    (A) π²/8
    (B) π²/32
    (C) π/4
    (D) 1

Answer Key for MCQs:

  1. (C)
  2. (B) [Use substitution t = 3x+5, dt = 3dx]
  3. (C) [Use substitution t = 1+x², dt = 2x dx]
  4. (A) [Use ILATE: u=x, v=eˣ. ∫xeˣ dx = x∫eˣdx - ∫(d/dx(x) * ∫eˣdx)dx = xeˣ - ∫1*eˣdx = xeˣ - eˣ + C]
  5. (A) [Use ∫ dx / (x² - a²) = (1 / 2a) ln |(x-a)/(x+a)| + C with a=3]
  6. (B) [∫cos x dx = sin x. [sin x]<0xE1><0xB5><0xA6>⁽⁰﹐ 𝜋/2⁾ = sin(π/2) - sin(0) = 1 - 0 = 1]
  7. (C) [f(x) = x³ is an odd function (f(-x) = (-x)³ = -x³ = -f(x)). Using property P₇, ∫[-a to a] f(x) dx = 0 if f is odd.]
  8. (A) [Let I = ∫[0 to π/2] (sin x / (sin x + cos x)) dx. Using P₄, I = ∫[0 to π/2] (sin(π/2-x) / (sin(π/2-x) + cos(π/2-x))) dx = ∫[0 to π/2] (cos x / (cos x + sin x)) dx. Adding the two equations for I: 2I = ∫[0 to π/2] ((sin x + cos x) / (sin x + cos x)) dx = ∫[0 to π/2] 1 dx = [x]<0xE1><0xB5><0xA6>⁽⁰﹐ 𝜋/2⁾ = π/2. So, I = π/4.]
  9. (B) [∫ dx / (4 + 9x²) = ∫ dx / (9 * (4/9 + x²)) = (1/9) ∫ dx / ((2/3)² + x²). Use ∫ dx / (a² + x²) = (1/a) tan⁻¹(x/a) + C with a=2/3. (1/9) * (1/(2/3)) tan⁻¹(x/(2/3)) + C = (1/9) * (3/2) tan⁻¹(3x/2) + C = (1/6) tan⁻¹(3x/2) + C]
  10. (B) [Substitute t = tan⁻¹x, dt = dx / (1+x²). Limits change: when x=0, t=tan⁻¹(0)=0; when x=1, t=tan⁻¹(1)=π/4. Integral becomes ∫[0 to π/4] t dt = [t²/2]<0xE1><0xB5><0xA6>⁽⁰﹐ 𝜋/4⁾ = ((π/4)² / 2) - 0 = (π²/16)/2 = π²/32.]

Study these notes thoroughly, paying special attention to the formulas and properties. Practice consistently, and you'll find Integrals becoming much more manageable for your exams. Good luck!

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