Integrals
1. Introduction
Integration is the inverse process of differentiation. It is used to find areas, volumes, and to solve differential equations. This chapter covers both indefinite and definite integrals thoroughly.
2. Indefinite Integrals
2.1 Standard Forms
∫ xⁿ dx = x^{n+1}/(n+1) + C (n ≠ -1) ∫ 1/x dx = ln|x| + C ∫ eˣ dx = eˣ + C ∫ aˣ dx = aˣ/ln a + C ∫ 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 ∫ 1/(1 + x²) dx = tan^{-1} x + C ∫ 1/(x√(x² - 1)) dx = sec^{-1} x + C
2.2 Integration by Substitution
If u = g(x), then ∫ f(g(x)) g'(x) dx = ∫ f(u) du.
'Choose the substitution wisely. Common forms: √(a² - x²) → x = a sin θ, √(a² + x²) → x = a tan θ, √(x² - a²) → x = a sec θ.'
2.3 Integration by Partial Fractions
Express the rational function as a sum of simpler fractions:
Type 1: (px + q)/((x - a)(x - b)) = A/(x - a) + B/(x - b) Type 2: (px + q)/((x - a)²) = A/(x - a) + B/(x - a)² Type 3: (px² + qx + r)/((x - a)(x² + bx + c)) = A/(x - a) + (Bx + C)/(x² + bx + c)
2.4 Integration by Parts
∫ u dv = uv - ∫ v du. Choose u using ILATE rule: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential.
2.5 Special Integrals
∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin^{-1}(x/a) + C ∫ √(a² + x²) dx = (x/2)√(a² + x²) + (a²/2) ln|x + √(a² + x²)| + C ∫ √(x² - a²) dx = (x/2)√(x² - a²) - (a²/2) ln|x + √(x² - a²)| + C
2.6 Applications of Special Substitutions
| Integral Form | Substitution |
|---|---|
| ∫ f(√(a² - x²)) dx | x = a sin θ |
| ∫ f(√(a² + x²)) dx | x = a tan θ |
| ∫ f(√(x² - a²)) dx | x = a sec θ |
| ∫ 1/(x² + a²) dx | x = a tan θ → (1/a)tan^{-1}(x/a) |
| ∫ 1/(x² - a²) dx | Partial fractions → (1/2a)ln |
| ∫ 1/√(a² - x²) dx | sin^{-1}(x/a) |
| ∫ 1/√(x² + a²) dx | ln |
'Always check the denominator for the appropriate substitution. When in doubt, try writing the integrand in a standard form by completing the square or using algebraic manipulation.'
3. Definite Integrals
3.1 Fundamental Theorem of Calculus
∫_a^b f(x) dx = F(b) - F(a), where F'(x) = f(x).
3.2 Properties
P0: ∫_a^b f(x) dx = ∫_a^b f(t) dt P1: ∫_a^b f(x) dx = -∫_b^a f(x) dx P2: ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx P3: ∫_a^b f(x) dx = ∫_a^b f(a + b - x) dx P4: ∫_0^a f(x) dx = ∫0^a f(a - x) dx P5: ∫{-a}^a f(x) dx = 2∫_0^a f(x) dx if f is even; 0 if f is odd. P6: ∫_0^{2a} f(x) dx = 2∫_0^a f(x) dx if f(2a - x) = f(x).
4. Area Under Curves
Area between y = f(x) and the x-axis from x = a to x = b: A = ∫_a^b |f(x)| dx. Area between two curves y = f(x) and y = g(x): A = ∫_a^b |f(x) - g(x)| dx.
5. Worked Problems
Problem 1: Evaluate ∫ x² eˣ dx. Solution: Using ILATE, let u = x², dv = eˣ dx. Then du = 2x dx, v = eˣ. ∫ x² eˣ dx = x²eˣ - ∫ 2x eˣ dx = x²eˣ - 2[x eˣ - ∫ eˣ dx] = x²eˣ - 2x eˣ + 2eˣ + C.
Problem 2: Evaluate ∫_0^{π/2} sin x/(sin x + cos x) dx. Solution: Let I = ∫_0^{π/2} sin x/(sin x + cos x) dx. Using P4, I = ∫_0^{π/2} cos x/(cos x + sin x) dx. Adding: 2I = ∫_0^{π/2} (sin x + cos x)/(sin x + cos x) dx = π/2. So I = π/4.
Problem 3: ∫ 1/(x² - 5x + 6) dx. Solution: 1/((x - 2)(x - 3)) = 1/(x - 3) - 1/(x - 2). So integral = ln|(x - 3)/(x - 2)| + C.
6. Common Mistakes
'Students often forget the constant of integration C in indefinite integrals — a costly error in exams.'
'When using partial fractions, ensure the numerator degree is less than the denominator degree. If not, perform polynomial division first.'
7. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Standard integrals | 3 | 2 |
| Substitution and partial fractions | 4 | 2 |
| Integration by parts | 4 | 2 |
| Definite integrals and properties | 5 | 3 |
| Area under curves | 3 | 3 |
8. Self-Test Questions
- Evaluate ∫ x sin^{-1} x dx.
- Evaluate ∫ eˣ(sin x + cos x)/(cos² x) dx.
- Prove that ∫_0^{π/2} √(tan x) dx = π/√2.
- Find the area bounded by y² = 4ax and x² = 4ay.
- Evaluate ∫_0^π x sin² x/(1 + cos² x) dx.
