Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Integration is the reverse of differentiation — it finds areas, volumes, and accumulations. Class 11 introduces indefinite and definite integrals, the standard formulas, and three major techniques: substitution, integration by parts, and partial fractions. These are extended further in Class 12.
Indefinite Integrals — Standard Forms
| ∫f(x)dx | Result | ||
|---|---|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹/(n+1) + C | ||
| 1/x | ln\ | x\ | + C |
| eˣ | eˣ + C | ||
| aˣ | aˣ/ln a + C | ||
| sin x | −cos x + C | ||
| cos x | sin x + C | ||
| sec²x | tan x + C | ||
| cosec²x | −cot x + C | ||
| sec x tan x | sec x + C | ||
| cosec x cot x | −cosec x + C | ||
| 1/√(1−x²) | sin⁻¹x + C | ||
| 1/(1+x²) | tan⁻¹x + C | ||
| 1/√(x²−a²) | ln\ | x+√(x²−a²)\ | + C |
| 1/(x²+a²) | (1/a)tan⁻¹(x/a) + C | ||
| 1/(x²−a²) | (1/2a)ln\ | (x−a)/(x+a)\ | + C |
Integration by Substitution
If the integrand is of the form f(g(x)) · g'(x):
Let u = g(x), du = g'(x)dx → ∫f(g(x))g'(x)dx = ∫f(u)du
Example: ∫2x·cos(x²)dx → u = x², du = 2x dx → ∫cos u du = sin(x²) + C
Integration by Parts (ILATE Rule)
∫u dv = uv − ∫v du
ILATE priority for choosing u: 1. I — Inverse trigonometric 2. L — Logarithmic 3. A — Algebraic (polynomials) 4. T — Trigonometric 5. E — Exponential
Example: ∫x eˣ dx → u = x, dv = eˣ dx = x eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C
Integration by Partial Fractions
Used when integrand is a rational function.
Example: ∫1/((x+1)(x−2))dx
| = ∫[(1/3)/(x−2) − (1/3)/(x+1)]dx = **(1/3)ln | x−2 | − (1/3)ln | x+1 | + C** |
|---|
Definite Integrals
∫[a,b] f(x)dx = F(b) − F(a) (Fundamental Theorem of Calculus)
Properties
| Property | Statement |
|---|---|
| Reversal | ∫[a,b]f = −∫[b,a]f |
| Zero width | ∫[a,a]f = 0 |
| Additivity | ∫[a,b]f + ∫[b,c]f = ∫[a,c]f |
| Even function | ∫[−a,a]f(x)dx = 2∫[0,a]f(x)dx |
| Odd function | ∫[−a,a]f(x)dx = 0 |
| Substitution | ∫[a,b]f(x)dx = ∫[a,b]f(a+b−x)dx |
Worked Examples
Example 1: Evaluate ∫(3x² + 2x − 1)dx. = x³ + x² − x + C
Example 2: Evaluate ∫sin²x dx. Use identity: sin²x = (1−cos2x)/2 = ∫(1/2 − cos2x/2)dx = x/2 − sin2x/4 + C
Example 3: Evaluate ∫₀^(π/2) sin x dx. = [−cos x]₀^(π/2) = −cos(π/2) + cos(0) = 0 + 1 = 1
Example 4: Evaluate ∫x ln x dx. ILATE: u = ln x, dv = x dx → du = dx/x, v = x²/2 = (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − x²/4 + C
Important Questions
One Mark
- Evaluate ∫(1/√x)dx.
- State the ILATE rule.
- Find ∫eˣ(sin x + cos x)dx.
- Evaluate ∫₀^1 x dx.
Short Answer (3 marks)
- Evaluate ∫(x+1)/√(x+2) dx using substitution.
- Evaluate ∫x² e^(−x) dx using integration by parts.
- Evaluate ∫2x/(x²+1) dx.
Long Answer (5 marks)
- Evaluate ∫₀^(π/2) sin²x dx using the reduction formula or properties.
- Evaluate ∫(3x+2)/((x+1)(x+2)(x+3)) dx using partial fractions.
Common Mistakes
- Forgetting the constant of integration (+C) in indefinite integrals.
- Integration by parts: choosing u and dv against the ILATE order, leading to a harder integral.
- ∫sin²x dx ≠ −cos²x — must use the half-angle identity first.
Board Exam Tips
- Integration by parts with ln x or inverse trig functions appears almost every exam as a 3-mark question.
- Definite integral using substitution (change limits!) is a 3-mark standard question.
- Properties of definite integrals (even/odd function, f(a+b−x)) are used to simplify otherwise hard integrals — practise these.
- JEE: Reduction formulas and Walli's formula (∫₀^(π/2) sinⁿx dx) are frequently tested.