Integrals
Learn Integrals with notes, examples, and practice questions.
Chapter Overview — What Is Integration?
Think of differentiation as going downhill on a mountain path, and integration as retracing your steps back to the starting height. Both directions describe the same landscape.
Differentiation answers: "How fast is a quantity changing?" Integration answers the opposite question: "What quantity produced this rate of change?"
If you know the derivative of a function everywhere on an interval, finding the original function is called integration (or anti-differentiation). The original function is called an antiderivative or primitive.
Integration also measures area under a curve. These two ideas — reversing derivatives and computing area — are the two pillars of integral calculus. They connect through the Fundamental Theorem of Calculus.
Indefinite Integral & the Constant C
The symbol ∫ f(x) dx represents the entire family of functions whose derivative is f(x). We call f(x) the integrand and x the variable of integration.
If F′(x) = f(x), then ∫ f(x) dx = F(x) + C. Here C is an arbitrary constant — because the derivative of any constant is zero, infinitely many antiderivatives differ only by C.
In board exams, write + C after every indefinite integral unless a specific initial condition fixes C.
- Integrand → the function being integrated (f(x))
- Antiderivative → any function F with F′(x) = f(x)
- Constant of integration → C (never forget in indefinite integrals)
- d/dx [∫ f(x) dx] = f(x) — differentiation undoes integration
- ∫ f′(x) dx = f(x) + C — integration undoes differentiation
Standard Integral Formulae (Must Memorise)
Power rule
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ −1
1/x
∫ (1/x) dx = ln|x| + C
Exponential
∫ eˣ dx = eˣ + C
sin & cos
∫ sin x dx = −cos x + C | ∫ cos x dx = sin x + C
sec² & cosec²
∫ sec²x dx = tan x + C | ∫ cosec²x dx = −cot x + C
Inverse trig
∫ dx/(1+x²) = tan⁻¹x + C | ∫ dx/√(1−x²) = sin⁻¹x + C
sec & cosec
∫ sec x tan x dx = sec x + C | ∫ cosec x cot x dx = −cosec x + C
Every differentiation formula gives a reverse integration formula. These are your building blocks — most exam integrals reduce to one of these after substitution or algebraic manipulation.
Properties of Indefinite Integrals
- Linearity: ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
- Scalar multiple: ∫ k·f(x) dx = k · ∫ f(x) dx
- Combined: ∫ [k₁f₁ + k₂f₂ + …] dx = k₁∫f₁ + k₂∫f₂ + …
- Two integrals with the same derivative differ only by a constant — they represent the same family of curves
Method 1 — Integration by Inspection
Look at the integrand and ask: "Whose derivative looks like this?" Adjust for missing constant factors.
Example pattern: ∫ cos 2x dx — derivative of sin 2x is 2 cos 2x, so multiply/divide by 2 → (1/2) sin 2x + C.
Method 2 — Substitution (Change of Variable)
∫ tan x dx
= ln|sec x| + C
∫ cot x dx
= ln|sin x| + C
∫ sec x dx
= ln|sec x + tan x| + C
∫ cosec x dx
= ln|cosec x − cot x| + C
Choose u = g(x) so that g′(x) appears (up to a constant) in the integrand. Then ∫ f(g(x))·g′(x) dx = ∫ f(u) du.
Classic triggers: numerator is (or is close to) the derivative of the denominator; expressions like √(ax+b), sin(f(x)), e^(f(x)).
Method 3 — Trigonometric Identities
- cos²x = (1 + cos 2x)/2 — use for ∫ cos²x dx
- sin²x = (1 − cos 2x)/2 — use for ∫ sin²x dx
- sin x cos y = ½[sin(x+y) + sin(x−y)] — product to sum
- sin 3x = 3 sin x − 4 sin³x — reduces ∫ sin³x dx
- 1 ± cos x identities help with ∫ dx/(1 ± cos x)
Method 4 — Partial Fractions
For rational functions P(x)/Q(x) where degree of P < degree of Q, factorise Q(x) and split into simpler fractions before integrating.
If degree of P ≥ degree of Q, perform polynomial long division first.
- Distinct linear factors: A/(x−a) + B/(x−b)
- Repeated linear factor: A/(x−a) + B/(x−a)²
- Irreducible quadratic: (Ax + B)/(x² + bx + c)
- Equate coefficients or substitute convenient x-values to find A, B, C
Method 5 — Integration by Parts
By parts
∫ u dv = uv − ∫ v du
eˣ pattern
∫ eˣ[f(x) + f′(x)] dx = eˣ·f(x) + C
√(a²−x²) type
Use x = a sin θ or integration by parts with constant 1
For products ∫ u dv = uv − ∫ v du. Choose u using LIATE priority: Log → Inverse trig → Algebraic → Trig → Exponential (pick u from earlier in list).
Special type: ∫ eˣ[f(x) + f′(x)] dx = eˣ·f(x) + C — recognise this pattern directly.
Special Integral Forms (Board Favourites)
∫ 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
Definite Integrals
∫ₐᵇ f(x) dx gives a single number — the signed area under y = f(x) from x = a to x = b. Lower limit a, upper limit b.
Fundamental Theorem (Part 2): If F′(x) = f(x) on [a, b], then ∫ₐᵇ f(x) dx = F(b) − F(a). Find the antiderivative first, then substitute limits — no + C needed.
For substitution in definite integrals, either resubstitute x at the end, or change the limits to match the new variable (faster).
Properties of Definite Integrals (Exam Shortcuts)
P1 — Reversal
∫ₐᵇ f dx = − ∫ᵇₐ f dx
P2 — Additivity
∫ₐᵇ f dx = ∫ₐᶜ f dx + ∫ᶜᵇ f dx
P3 — Mirror
∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx
P4 — Even / Odd
Even f: ∫₋ₐᵃ f = 2∫₀ᵃ f | Odd f: ∫₋ₐᵃ f = 0
P5 — Zero width
∫ₐᵃ f(x) dx = 0
Property P3 is powerful for integrals like ∫₀^(π/2) sinⁿx / (sinⁿx + cosⁿx) dx — add the integral to its mirror form to get π/4.
Solved Examples
Step-by-step solutions — read each step before checking the final answer.
Example 1: Inspection — cos 2x
Find ∫ cos 2x dx.
- 1Recall: d/dx (sin 2x) = 2 cos 2x.
- 2So cos 2x is half the derivative of sin 2x.
- 3∫ cos 2x dx = (1/2) sin 2x + C.
Answer
(1/2) sin 2x + C
Example 2: Substitution — x√(x² + 1)
Evaluate ∫ x√(x² + 1) dx.
- 1Let t = x² + 1 → dt = 2x dx → x dx = dt/2.
- 2Integral becomes ∫ √t · (dt/2) = (1/2) ∫ t^(1/2) dt.
- 3= (1/2) · t^(3/2) / (3/2) + C = (1/3)(x² + 1)^(3/2) + C.
Answer
(1/3)(x² + 1)^(3/2) + C
Example 3: Partial Fractions
Evaluate ∫ dx / [(x+1)(x+2)].
- 1Write 1/[(x+1)(x+2)] = A/(x+1) + B/(x+2).
- 21 = A(x+2) + B(x+1). Put x = −1 → A = 1. Put x = −2 → B = −1.
- 3∫ = ∫ [1/(x+1) − 1/(x+2)] dx = ln|x+1| − ln|x+2| + C.
Answer
ln|(x+1)/(x+2)| + C
Example 4: Integration by Parts
Find ∫ x eˣ dx.
- 1Let u = x, dv = eˣ dx → du = dx, v = eˣ.
- 2∫ x eˣ dx = x eˣ − ∫ eˣ dx.
- 3= x eˣ − eˣ + C = eˣ(x − 1) + C.
Answer
eˣ(x − 1) + C
Example 5: Definite Integral
Evaluate ∫₀² x² dx.
- 1Antiderivative of x² is x³/3.
- 2Apply limits: [x³/3]₀² = 8/3 − 0.
Answer
8/3
Example 6: Odd Function Property
Evaluate ∫₋₁¹ x³ dx without full calculation.
- 1f(x) = x³ is odd: f(−x) = −f(x).
- 2For odd functions on symmetric interval [−a, a], the integral is 0.
Answer
0
Key Points to Remember
- ✓ Integration reverses differentiation; always add + C in indefinite integrals
- ✓ Memorise standard formulae — they are the alphabet of integration
- ✓ Substitution: look for a function whose derivative is present in the integrand
- ✓ Partial fractions work for rational functions; divide first if improper
- ✓ LIATE rule helps choose u in integration by parts
- ✓ Definite integral = F(b) − F(a); constant C cancels out
- ✓ Use even/odd and mirror properties to solve ∫₀^(π/2) type problems quickly
Exam Tips
- • Write + C in every indefinite integral — easy marks lost if forgotten
- • Show substitution clearly: let u = …, du = …
- • For definite integrals, bracket notation [F(x)]ₐᵇ is preferred
- • Check answer by differentiating — derivative should match integrand
- • Board often asks ∫₀^(π/2) f(sin x, cos x) dx — try x → π/2 − x
- • Practice Exercise 7.1–7.10 style problems from NCERT for pattern recognition
Formula Cheat Sheet
Quick reference — NCERT board exam formulas
Standard Integrals
- ∫ xⁿ dx = xⁿ⁺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
- ∫ sec²x dx = tan x + C ; ∫ cosec²x dx = −cot x + C
- ∫ sec x tan x dx = sec x + C
- ∫ cosec x cot x dx = −cosec x + C
- ∫ dx/(1+x²) = tan⁻¹x + C
- ∫ dx/√(1−x²) = sin⁻¹x + C
Practice MCQs — Integrals
Test your understanding with topic-wise multiple choice questions. Explanations appear after each answer.
∫ (1/x) dx equals: