1. Chapter at a Glance
Differentiation answers "how fast is it changing?". Integration answers the reverse questions: "given the rate, what is the total?" and "what is the area under this curve?". Given velocity, integration recovers distance; given marginal cost, it recovers total cost. Ideas of summing infinitely many small pieces go back to Archimedes, and the Kerala school mathematicians worked with infinite series closely tied to integration; Newton and Leibniz unified everything through the Fundamental Theorem of Calculus. This is the largest chapter of Class 12, covering indefinite integrals, the main methods of integration, and definite integrals with their properties.
- Integration as the inverse of differentiation; constant of integration.
- Standard integrals and properties.
- Integration by substitution (including trigonometric identities).
- Integration using partial fractions.
- Integration by parts; ∫eˣ[f(x) + f′(x)]dx.
- Special integrals: 1/(x² ± a²), 1/√(x² ± a²), √(a² − x²), etc.
- Definite integrals; Fundamental Theorem; properties P₀–P₆.
2. Integration as Anti-differentiation
If d/dx F(x) = f(x), then ∫ f(x) dx = F(x) + C. F is an antiderivative (primitive) and C is the arbitrary constant — antiderivatives of the same function differ only by a constant.
| ∫ f(x) dx | Result |
|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹/(n + 1) + C |
| 1/x | log|x| + C |
| eˣ | eˣ + C |
| aˣ | aˣ/log a + C |
| sin x / cos x | −cos x + C / sin x + C |
| sec²x / cosec²x | tan x + C / −cot x + C |
| sec x tan x / cosec x cot x | sec x + C / −cosec x + C |
| 1/√(1 − x²) | sin⁻¹x + C |
| 1/(1 + x²) | tan⁻¹x + C |
| tan x / cot x | log|sec x| + C / log|sin x| + C |
| sec x / cosec x | log|sec x + tan x| + C / log|cosec x − cot x| + C |
Linearity: ∫[k f(x) ± g(x)] dx = k∫f dx ± ∫g dx.
Find ∫(x³ + 5x² − 4)/x² dx.
Solution: ∫(x + 5 − 4x⁻²) dx = x²/2 + 5x + 4/x + C.
3. Integration by Substitution
If the integrand contains a function and (a multiple of) its derivative, substitute t = that function: ∫f(g(x))g′(x)dx = ∫f(t)dt.
Find (a) ∫2x sin(x² + 1) dx (b) ∫ (tan⁻¹x)/(1 + x²) dx (c) ∫ sin³x dx.
Solution: (a) t = x² + 1: ∫sin t dt = −cos(x² + 1) + C. (b) t = tan⁻¹x: ∫t dt = (tan⁻¹x)²/2 + C. (c) sin³x = (1 − cos²x)sin x; t = cos x: −∫(1 − t²)dt = −cos x + cos³x/3 + C.
- ∫ f′(x)/f(x) dx = log|f(x)| + C.
- ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + C.
- sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2 for even powers.
4. Special Integrals
| Integral | Result |
|---|---|
| ∫ dx/(x² − a²) | (1/2a) log|(x − a)/(x + a)| + C |
| ∫ dx/(a² − x²) | (1/2a) log|(a + x)/(a − x)| + C |
| ∫ dx/(x² + a²) | (1/a) tan⁻¹(x/a) + C |
| ∫ dx/√(x² − a²) | log|x + √(x² − a²)| + C |
| ∫ dx/√(a² − x²) | sin⁻¹(x/a) + C |
| ∫ dx/√(x² + a²) | 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 |
For ∫dx/(ax² + bx + c), complete the square in the denominator and use one of these.
Find ∫ dx/(x² − 6x + 13).
Solution: x² − 6x + 13 = (x − 3)² + 2². Result ½ tan⁻¹((x − 3)/2) + C.
5. Partial Fractions
For a proper rational function P(x)/Q(x), split into simpler fractions:
| Factor in Q(x) | Partial fraction form |
|---|---|
| (x − a)(x − b) | A/(x − a) + B/(x − b) |
| (x − a)² | A/(x − a) + B/(x − a)² |
| (x − a)(x² + bx + c), irreducible quadratic | A/(x − a) + (Bx + C)/(x² + bx + c) |
Find ∫ dx/[(x + 1)(x + 2)].
Solution: 1/[(x + 1)(x + 2)] = 1/(x + 1) − 1/(x + 2). Result log|(x + 1)/(x + 2)| + C.
6. Integration by Parts
Choose u by ILATE order: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential.
Special: ∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C.
Find (a) ∫ x cos x dx (b) ∫ log x dx (c) ∫ eˣ(sin x + cos x) dx.
Solution: (a) u = x: x sin x − ∫sin x dx = x sin x + cos x + C. (b) u = log x, v = 1: x log x − ∫1 dx = x log x − x + C. (c) f = sin x, f′ = cos x: eˣ sin x + C.
7. Definite Integrals
If F′(x) = f(x) and f is continuous on [a, b], then
Geometrically, for f ≥ 0 this is the area between y = f(x), the x-axis and x = a, x = b. The area function A(x) = ∫ₐˣ f(t)dt satisfies A′(x) = f(x).
Evaluate (a) ∫₀² (x² + 1) dx (b) ∫₀^(π/2) sin x dx.
Solution: (a) [x³/3 + x]₀² = 8/3 + 2 = 14/3. (b) [−cos x]₀^(π/2) = 0 + 1 = 1.
With substitution in a definite integral, change the limits too: ∫₀¹ 2x/(1 + x²) dx, t = 1 + x²: ∫₁² dt/t = log 2.
8. Properties of Definite Integrals
| Property | Statement |
|---|---|
| P₀ | ∫ₐᵇ f(x)dx = ∫ₐᵇ f(t)dt |
| P₁ | ∫ₐᵇ f = −∫ᵦₐ f; ∫ₐₐ f = 0 |
| P₂ | ∫ₐᵇ f = ∫ₐᶜ f + ∫ᶜᵇ f |
| P₃ | ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a + b − x)dx |
| P₄ | ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx |
| P₅ | ∫₀²ᵃ f = 2∫₀ᵃ f if f(2a − x) = f(x); 0 if f(2a − x) = −f(x) |
| P₆ | ∫₋ₐᵃ f = 2∫₀ᵃ f if f even; 0 if f odd |
Evaluate I = ∫₀^(π/2) √(sin x)/(√(sin x) + √(cos x)) dx.
Solution: By P₄, I = ∫₀^(π/2) √(cos x)/(√(cos x) + √(sin x)) dx. Adding: 2I = ∫₀^(π/2) 1 dx = π/2 → I = π/4.
Evaluate ∫₋₁¹ sin⁵x cos⁴x dx.
Solution: The integrand is odd → 0 (P₆).
9. Solved Problems (Board Pattern)
Find ∫ (3x − 2)/[(x + 1)²(x + 3)] dx.
Solution: = A/(x + 1) + B/(x + 1)² + C/(x + 3). 3x − 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)². x = −1: −5 = 2B → B = −5/2. x = −3: −11 = 4C → C = −11/4. x² coefficient: 0 = A + C → A = 11/4. Result: (11/4)log|x + 1| + 5/(2(x + 1)) − (11/4)log|x + 3| + C.
Find ∫ x² eˣ dx.
Solution: x²eˣ − ∫2x eˣ dx = x²eˣ − 2(xeˣ − eˣ) + C = eˣ(x² − 2x + 2) + C.
Evaluate ∫₋₅⁵ |x + 2| dx.
Solution: ∫₋₅⁻² −(x + 2)dx + ∫₋₂⁵ (x + 2)dx = 9/2 + 49/2 = 29.
10. Master Formula Table
| Method / result | Formula |
|---|---|
| Power | ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C |
| Log form | ∫f′/f dx = log|f| + C |
| By parts | ∫uv = u∫v − ∫(u′∫v) |
| eˣ special | ∫eˣ(f + f′) = eˣf + C |
| Inverse tan type | ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) |
| Inverse sin type | ∫dx/√(a² − x²) = sin⁻¹(x/a) |
| FTC | ∫ₐᵇ f = F(b) − F(a) |
| King's property | ∫₀ᵃ f(x) = ∫₀ᵃ f(a − x) |
| Even / odd | ∫₋ₐᵃ f = 2∫₀ᵃ f / 0 |
11. Master Summary Table
| Integrand looks like | Method |
|---|---|
| Function × its derivative | Substitution |
| Rational function | Partial fractions (after division if improper) |
| Product of different types (x eˣ, x sin x, log x) | By parts (ILATE) |
| 1/(quadratic) or 1/√(quadratic) | Complete the square → special integrals |
| Powers of sin/cos | Identities / substitution |
| Symmetric definite integral | Properties P₃–P₆ |
12. Mind Map
- Anti-derivative + C
- Standard table
- Substitution
- Partial fractions, parts
- x² ± a² forms
- Complete square
- FTC: F(b) − F(a)
- Properties P₀–P₆
13. Tips and Tricks
- Differentiate your answer to check an indefinite integral.
- When a definite integral looks hard but has limits 0 to π/2 (or 0 to a), try P₄ and add the two forms.
- For ∫(px + q)/(ax² + bx + c), write px + q = A(derivative of denominator) + B.
- Forgetting + C in indefinite integrals.
- Forgetting to change limits after substitution in definite integrals.
- ∫1/x dx = log x (needs |x|).
14. Important One-liners
- Integration is the reverse of differentiation.
- Any two antiderivatives differ by a constant.
- The definite integral of a non-negative function is an area.
- The integral of an odd function over [−a, a] is 0.
- ∫₀^(π/2) log sin x dx = −(π/2) log 2 (a classic result using P₄).
15. Exam-pattern Questions with Detailed Answers
∫ (sec²x / tan x) dx equals:
- log|tan x| + C
- tan²x + C
- log|sec x| + C
- sec x + C
Show answer
f′/f form with f = tan x.
Answer: (A)
Evaluate ∫₁² (4x³ − 5x² + 6x + 9) dx.
Show answer
[x⁴ − 5x³/3 + 3x² + 9x]₁² = (16 − 40/3 + 12 + 18) − (1 − 5/3 + 3 + 9) = (46 − 40/3) − (13 − 5/3) = 33 − 35/3 = 64/3.
Answer: 64/3
Find ∫ (2x + 3)⁵ dx.
Show answer
(2x + 3)⁶/12 + C.
Answer: (2x + 3)⁶/12 + C
Find ∫ x/(x² + 1) dx and ∫₀¹ x/(x² + 1) dx.
Show answer
½ log(x² + 1) + C; definite: ½ log 2.
Answer: ½ log(x² + 1) + C; ½ log 2
Find ∫ dx/√(9 − 25x²).
Show answer
= (1/5)∫ dx/√((3/5)² − x²) = (1/5) sin⁻¹(5x/3) + C.
Answer: (1/5) sin⁻¹(5x/3) + C
Find ∫ x² log x dx.
Show answer
u = log x: (x³/3)log x − ∫(x³/3)(1/x)dx = (x³/3)log x − x³/9 + C.
Answer: (x³/3)log x − x³/9 + C
Find ∫ sin²x dx.
Show answer
∫(1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C.
Answer: x/2 − (sin 2x)/4 + C
Find ∫ eˣ (1/x − 1/x²) dx.
Show answer
f = 1/x, f′ = −1/x² → eˣ/x + C.
Answer: eˣ/x + C
Evaluate ∫₀^π x sin x/(1 + cos²x) dx.
Show answer
P₄: I = ∫₀^π (π − x) sin x/(1 + cos²x) dx. Add: 2I = π∫₀^π sin x/(1 + cos²x) dx. With t = cos x: π∫₋₁¹ dt/(1 + t²) = π(π/2). So I = π²/4.
Answer: π²/4
Find ∫ (x + 2)/(2x² + 6x + 5) dx.
Show answer
x + 2 = ¼(4x + 6) + ½. Integral = ¼ log|2x² + 6x + 5| + ½∫dx/(2x² + 6x + 5). 2x² + 6x + 5 = 2[(x + 3/2)² + ¼] → ½ × ½ × 2 tan⁻¹(2x + 3) = ½ tan⁻¹(2x + 3). Answer: ¼ log|2x² + 6x + 5| + ½ tan⁻¹(2x + 3) + C.
Answer: ¼ log(2x² + 6x + 5) + ½ tan⁻¹(2x + 3) + C
Evaluate ∫₀^(π/2) log sin x dx.
Show answer
I = ∫ log cos x (P₄). 2I = ∫₀^(π/2) log(sin x cos x) = ∫ log(sin 2x) dx − (π/2) log 2. With t = 2x, ∫₀^(π/2) log sin 2x dx = ½∫₀^π log sin t dt = ∫₀^(π/2) log sin t dt = I (P₅). So 2I = I − (π/2) log 2 → I = −(π/2) log 2.
Answer: −(π/2) log 2
Find ∫ √(x² + 4x + 6) dx.
Show answer
= ∫√((x + 2)² + (√2)²) dx = ((x + 2)/2)√(x² + 4x + 6) + log|x + 2 + √(x² + 4x + 6)| + C.
Answer: As above
16. Quick Revision Summary
- ∫f dx = F + C where F′ = f; learn the standard table.
- Substitution, partial fractions and parts (ILATE) cover most integrals.
- Special integrals via completing the square.
- FTC: ∫ₐᵇ f = F(b) − F(a); change limits with substitution.
- Properties P₀–P₆, especially f(a − x) and even/odd, make hard integrals easy.