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CBSE Class 12 Mathematics★ 8-11 marks · CBSE Class 12 Mathematics Chapter 7

Integrals (Integration — The Reverse of Differentiation)

Integration is the REVERSE of differentiation. Given f'(x), find f(x). This chapter covers indefinite integrals (the antiderivative), methods of integration (substitution, by parts, partial fractions), and the connection to definite integrals. CBSE Class 12 Mathematics Chapter 7.

⏱ 26 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Evaluate indefinite integrals using the standard integral table and basic rules
2Apply the substitution method: choose u = g(x), express the integral entirely in u, integrate, back-substitute
3Apply integration by parts using the ILATE priority rule: ∫u dv = uv − ∫v du
4Decompose rational functions using partial fractions and integrate each term
5Evaluate definite integrals using FTC and apply symmetry properties (odd/even function properties)

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Why this chapter matters

Integrals is the highest-weightage single chapter in Class 12 Mathematics. Integration by substitution, integration by parts (ILATE), partial fractions, and definite integral properties — particularly the odd/even symmetry property — are all guaranteed board exam questions. Mastering the technique-selection process separates high scorers from average ones.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Integrals

"Differentiation asks: given y, find dy/dx. Integration asks: given dy/dx, find y."

1. Chapter Overview

INTEGRATION is the INVERSE OPERATION of differentiation. This chapter covers: indefinite integrals (antiderivatives — a family of functions differing by a constant C), methods of integration (substitution, integration by parts, partial fractions, trigonometric identities), and the definite integral (the limit of a Riemann sum — the area under a curve). The Fundamental Theorem of Calculus LINKS differentiation and integration.

2. Indefinite Integrals — Basic Formulas

Function	Integral
xⁿ (n ≠ -1)	xⁿ⁺¹/(n+1) + C
1/x	log
eˣ	eˣ + C
aˣ	aˣ/log a + C
sin x	-cos x + C
cos x	sin x + C
sec² x	tan x + C
cosec² x	-cot x + C
sec x tan x	sec x + C
cosec x cot x	-cosec x + C

3. Methods of Integration

Substitution Method

Replace a part of the integrand with a NEW variable u = g(x). The differential: du = g'(x) dx. Substitute. Simplify. Integrate. Substitute back.

Integration by Parts

∫ u dv = uv — ∫ v du
ILATE rule (choose u in this priority): Inverse trig → Logarithmic → Algebraic → Trigonometric → Exponential
'Choose u so that its derivative is SIMPLER than u itself.'

Partial Fractions

Decompose a rational function P(x)/Q(x) into SIMPLER fractions. Integrate each separately. Used when denominator factorises.

Trigonometric Identities

Use: sin²x = (1-cos2x)/2. cos²x = (1+cos2x)/2. sin³x. sin(mx)cos(nx) products → sums.

4. Definite Integrals

Definition

∫ₐᵇ f(x) dx = lim(n→∞) Σ f(xᵢ*) Δx (Riemann sum). 'The limit of the sum of areas of rectangles as the width approaches zero.'

Fundamental Theorem of Calculus

Part 1: If F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) — F(a). 'Evaluate the antiderivative at the limits and subtract.'
Part 2: d/dx [∫ₐˣ f(t) dt] = f(x). 'The derivative of the integral is the original function.'

Properties of Definite Integrals

∫ₐᵇ f(x) dx = —∫ᵦᵃ f(x) dx
∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx
∫₋ₐᵃ f(x) dx = 0 if f is ODD. 2∫₀ᵃ f(x) dx if f is EVEN.

5. Exam Focus

Basic integration formulas — MEMORISE
Substitution method. Integration by parts (ILATE). Partial fractions.
Definite integral — FTC: ∫ₐᵇ f(x)dx = F(b) — F(a)
Properties of definite integrals — especially odd/even function shortcuts

6. Conclusion

Integration is the ART OF RECONSTRUCTION:

INDEFINITE: The antiderivative. A FAMILY of functions. 'Don't forget the +C!'
METHODS: Substitution. Parts. Partial fractions. 'Choose the right tool for the job.'
DEFINITE: The area under a curve. 'The Fundamental Theorem connects the two halves of calculus.'

'Integration is the most creative act in calculus. Differentiation has rules. Integration has techniques — and judgment.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Standard Integral Table

∫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. ∫1/√(1−x²) dx = sin⁻¹x + C. ∫1/(1+x²) dx = tan⁻¹x + C. ∫1/√(a²−x²) dx = sin⁻¹(x/a) + C. ∫1/(a²+x²) dx = (1/a)tan⁻¹(x/a) + C. ∫1/√(x²±a²) dx = ln|x + √(x²±a²)| + C.

The sign on ∫sin x dx = −cos x + C (NEGATIVE cos x) is the most common error. Similarly ∫cosec²x dx = −cot x + C (negative). Memorise by differentiating backwards: d/dx(−cos x) = sin x ✓.

Integration by Substitution

If integral has form ∫f(g(x))·g'(x) dx: let u = g(x), du = g'(x) dx → integral becomes ∫f(u) du. PROCEDURE: (1) Identify an inner function u = g(x). (2) Compute du = g'(x)dx. (3) Rewrite entire integral in terms of u. (4) Integrate with respect to u. (5) Back-substitute g(x) for u. KEY PATTERNS: ∫f'(x)/f(x) dx = ln|f(x)| + C. ∫[f(x)]ⁿ·f'(x) dx = [f(x)]ⁿ⁺¹/(n+1) + C. For ∫sin²x dx or ∫cos²x dx: use half-angle identity sin²x = (1−cos 2x)/2 first.

The factor g'(x) must already be present in the integral (or obtainable by adjusting a constant). You CANNOT introduce extra x-factors that aren't there. If the factor is off by a constant (like ∫2x/(x²+1) dx needs only u=x²+1, du=2x dx), adjust by multiplying and dividing by that constant.

Integration by Parts — ILATE

∫u dv = uv − ∫v du. ILATE priority for choosing u (first function): I = Inverse trig (sin⁻¹x, tan⁻¹x). L = Logarithm (ln x, log x). A = Algebraic (xⁿ, polynomials). T = Trigonometric (sin x, cos x). E = Exponential (eˣ, aˣ). Choose u as the function HIGHER in this list; dv as the remaining part. SPECIAL: ∫eˣ[f(x) + f'(x)] dx = eˣf(x) + C. For ∫ln x dx: u = ln x (L before A), dv = dx → ∫ln x dx = x ln x − x + C.

For ∫x²eˣ dx: u = x² (A before E), dv = eˣ dx. Apply parts twice. For cyclic integrals like ∫eˣ sin x dx: apply parts TWICE, the original integral appears on the right — rearrange algebraically to solve for I. Do NOT apply parts a third time.

Partial Fractions

For rational function P(x)/Q(x) where deg(P) < deg(Q): CASE 1 — Linear distinct factors: (2x+1)/[(x+1)(x−2)] = A/(x+1) + B/(x−2). CASE 2 — Repeated linear factor: 1/[(x+1)²(x+2)] = A/(x+1) + B/(x+1)² + C/(x+2). CASE 3 — Irreducible quadratic: (x+1)/[(x²+1)(x+2)] = (Ax+B)/(x²+1) + C/(x+2). PROCEDURE: Multiply both sides by Q(x), substitute convenient x-values (roots of each factor) to find constants A, B, C. Then integrate each simple fraction.

Always check that deg(P) < deg(Q) BEFORE applying partial fractions. If deg(P) ≥ deg(Q), first perform polynomial long division to reduce, then apply partial fractions to the remainder. Common exam error: forgetting to do long division first when degrees are equal or numerator degree is higher.

Definite Integrals and Key Properties

FTC: ∫ₐᵇ f(x) dx = F(b) − F(a) where F'(x) = f(x). KEY PROPERTIES: (P1) ∫ₐᵇ f(x) dx = −∫ᵦₐ f(x) dx. (P2) ∫ₐᵃ f(x) dx = 0. (P3) ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx. (P4) ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx. (P5 — King's Property): ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx. (P6 — Symmetry): If f(x) is even [f(−x)=f(x)]: ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx. If f(x) is odd [f(−x)=−f(x)]: ∫₋ₐᵃ f(x) dx = 0.

King's Property (P5) is the most exam-tested property: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx. Technique: let I = ∫₀ᵃ f(x) dx and also I = ∫₀ᵃ f(a−x) dx (same value). Add: 2I = ∫₀ᵃ [f(x) + f(a−x)] dx — the sum often simplifies to a constant, making the integral trivial.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Writing ∫sin x dx = cos x + C (positive)

✓ ∫sin x dx = −cos x + C (NEGATIVE). Verify by differentiating: d/dx(−cos x) = −(−sin x) = sin x ✓. A reliable check: the derivative of your answer should equal the integrand. Always verify the sign of trig integrals.

WATCH OUT

✗ Using long division for partial fractions when the degree of numerator is lower

✓ Long division is needed ONLY when deg(numerator) ≥ deg(denominator). For (2x+1)/[(x+1)(x−2)]: numerator degree is 1, denominator degree is 2 → directly apply partial fractions. The trap is (x²+1)/(x²−1): numerator degree = denominator degree → must do long division first to get 1 + 2/(x²−1), then partial fractions.

WATCH OUT

✗ Forgetting the constant of integration C in indefinite integrals

✓ Every indefinite integral MUST end with + C. The constant represents the family of all antiderivatives. Omitting C loses half a mark per question in CBSE boards. The definite integral ∫ₐᵇ has NO constant C (bounds eliminate it via F(b)−F(a)).

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 7.1–7.11

Exercises 7.1–7.11

Basic integrals; substitution; integration by parts; special integrals (√(a²±x²) type); partial fractions; definite integrals; FTC applications; properties proofs

120

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· integration-by-parts

Evaluate: ∫x·eˣ dx.

▶ Show solution

Apply integration by parts: ∫u dv = uv − ∫v du. By ILATE: u = x (Algebraic), dv = eˣ dx (Exponential). Then du = dx, v = eˣ. ∫x·eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C. VERIFY: d/dx[eˣ(x−1)] = eˣ(x−1) + eˣ(1) = eˣ(x−1+1) = xeˣ ✓.

Q2MEDIUM· definite-king-property

Evaluate: ∫₀^(π/2) [sin x / (sin x + cos x)] dx.

▶ Show solution

Let I = ∫₀^(π/2) sin x/(sin x + cos x) dx. Apply King's Property: replace x with (π/2 − x): I = ∫₀^(π/2) sin(π/2−x)/[sin(π/2−x) + cos(π/2−x)] dx = ∫₀^(π/2) cos x/(cos x + sin x) dx. Adding the two expressions for I: 2I = ∫₀^(π/2) [sin x/(sin x + cos x) + cos x/(cos x + sin x)] dx = ∫₀^(π/2) [(sin x + cos x)/(sin x + cos x)] dx = ∫₀^(π/2) 1 dx = [x]₀^(π/2) = π/2. Therefore I = π/4.

Q3HARD· partial-fractions

Evaluate: ∫ (x² + 1)/[(x² + 2)(x² + 3)] dx.

▶ Show solution

Let I = ∫ (x²+1)/[(x²+2)(x²+3)] dx. PARTIAL FRACTIONS: Let t = x². Then (t+1)/[(t+2)(t+3)] = A/(t+2) + B/(t+3). Multiply: t+1 = A(t+3) + B(t+2). At t = −2: −2+1 = A(−2+3) → −1 = A. At t = −3: −3+1 = B(−3+2) → −2 = −B → B = 2. So (x²+1)/[(x²+2)(x²+3)] = −1/(x²+2) + 2/(x²+3). I = −∫1/(x²+2) dx + 2∫1/(x²+3) dx. = −(1/√2)tan⁻¹(x/√2) + 2·(1/√3)tan⁻¹(x/√3) + C [using ∫1/(x²+a²)dx = (1/a)tan⁻¹(x/a) + C]. = −(1/√2)tan⁻¹(x/√2) + (2/√3)tan⁻¹(x/√3) + C.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•∫sin x dx = −cos x + C (negative). ∫cos x dx = sin x + C (positive).
•∫1/(a²+x²) dx = (1/a)tan⁻¹(x/a) + C. ∫1/√(a²−x²) dx = sin⁻¹(x/a) + C.
•Substitution: u = inner function, ensure g'(x) is present in numerator/coefficient.
•ILATE: Inverse trig > Log > Algebraic > Trig > Exponential. u = higher priority function.
•∫eˣ[f(x) + f'(x)] dx = eˣf(x) + C (direct formula).
•Partial fractions: check deg(P) < deg(Q). Linear factors: A/(ax+b). Repeated: A/() + B/()² form.
•King's Property: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx. Add both forms → 2I.
•Odd function on [−a,a]: integral = 0. Even function: integral = 2×∫₀ᵃ.
•FTC: ∫ₐᵇ f(x) dx = F(b) − F(a). No +C needed for definite integrals.
•∫ln x dx = x ln x − x + C (parts with u=ln x, dv=dx).

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-11 marks

Question type	Marks each	Typical count	What it tests
Short — Indefinite Integral	3-4	1-2	Integration by substitution; integration by parts (one-step); standard special forms
Long — Definite Integral with Properties	5-6	1	King's property; odd/even symmetry; definite integral evaluation using FTC

Prep strategy

Technique selection: match the integrand type to the method BEFORE computing. Rational function → partial fractions. Product of two function types → integration by parts. Composite function with inner derivative present → substitution. Trig powers → half-angle identities first.
King's Property for definite integrals: when you see ∫₀^a f(x)dx where f involves sin x and cos x, ALWAYS try applying f(a−x). If f(x) + f(a−x) = 1 or a constant, use 2I = constant × a.
Integration by parts: memorise ILATE. Write u and dv explicitly at the start of every IBP problem. For ∫eˣ[f(x)+f'(x)]dx, directly write eˣf(x) + C without expansion.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Area Under Probability Distributions

In statistics and machine learning, the probability that a random variable falls in a range [a, b] is computed as ∫ₐᵇ f(x) dx, where f is the probability density function. The normal distribution's bell curve area is computed using definite integrals. Every p-value in hypothesis testing, every confidence interval in medicine, and every risk model in finance relies on evaluating definite integrals of probability density functions — the direct application of the Fundamental Theorem of Calculus.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For every integration problem, write the technique you're using at the top: 'Method: Integration by parts (ILATE)' or 'Method: Substitution u = ...' — this earns method marks even if a numerical error appears later.
For definite integrals with property questions (King's, symmetry): write I = [first form], then I = [property-substituted form], then add to get 2I = [simplified form], then I = [answer]. This structured approach earns all intermediate marks.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Study the GAMMA FUNCTION: Γ(n) = ∫₀^∞ xⁿ⁻¹e⁻ˣ dx — a generalisation of factorial to non-integers. Γ(n+1) = n·Γ(n) and Γ(1/2) = √π. This integral, which cannot be evaluated in closed form for arbitrary n, connects definite integrals to combinatorics and quantum physics
Explore FEYNMAN'S TRICK (differentiation under the integral sign): introduce a parameter α into the integrand, differentiate with respect to α to get a simpler integral, solve, then integrate back in α. Richard Feynman's favourite technique, and the most powerful non-standard integration method in competitive mathematics

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Integral Calculus)Very High

CUET (Mathematics)High

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Use this decision tree: (1) Is the integrand a standard form? → Direct integration. (2) Is it a rational function (polynomial/polynomial)? → Partial fractions (after long division if needed). (3) Is it a product of two different function types (xeˣ, x²sin x, xln x)? → Integration by parts (ILATE). (4) Is there a composite function where the inner derivative is present? → Substitution. (5) Does it involve sin²x, cos²x, sin x cos x? → Half-angle identity first, then substitute. Multiple techniques may apply; choose the one that simplifies fastest.

Apply parts TWICE when: (1) The integrand is xⁿ times trig or exponential (e.g., ∫x²eˣ dx — parts reduces x² to x, then parts again to remove x). (2) The integral is cyclic: ∫eˣ sin x dx — applying parts twice gives I = ... − I, so 2I = ... → solve for I algebraically. For cyclic integrals, NEVER apply parts a third time — instead, collect the I terms and solve the resulting equation.

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