Integrals

2.1 Core concepts

• An antiderivative (primitive) of f is any function F with F′(x) = f(x); the family {F + C : C ∈ R} forms the indefinite integral, written ∫f(x)dx = F(x) + C, where C is the arbitrary constant of integration (NCERT §7.1–§7.2, p. 225–227).
• Two functions having the same derivative on an interval differ only by a constant — so the family {F + C} provides every antiderivative of f (NCERT §7.2 Remark, p. 227).
• Standard integrals derived directly from known derivatives include ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1), ∫cos x dx = sin x + C, ∫sin x dx = −cos x + C, ∫sec²x dx = tan x + C, ∫eˣ dx = eˣ + C, ∫(1/x)dx = log|x| + C, ∫aˣ dx = aˣ/log a + C (NCERT §7.2, p. 228–229).
• Properties of indefinite integrals: (i) d/dx[∫f(x)dx] = f(x) and ∫f′(x)dx = f(x) + C; (ii) ∫[f(x)+g(x)]dx = ∫f(x)dx + ∫g(x)dx; (iii) ∫k f(x)dx = k∫f(x)dx, and these generalise to any finite linear combination (NCERT §7.2.1, p. 229–231).
• Integration by substitution transforms ∫f(x)dx via x = g(t), dx = g′(t)dt, giving ∫f(g(t))g′(t)dt — chosen so that a function whose derivative also appears in the integrand becomes the new variable (NCERT §7.3.1, p. 235–236).
• Using substitution one derives ∫tan x dx = log|sec x| + C, ∫cot x dx = log|sin x| + C, ∫sec x dx = log|sec x + tan x| + C, ∫cosec x dx = log|cosec x − cot x| + C (NCERT §7.3.1, p. 237–238).
• Trigonometric identities (power-reduction, product-to-sum, sin 3x = 3 sin x − 4 sin³x) reduce integrals of cos²x, sin²x cos³x, sin³x etc. to standard forms (NCERT §7.3.2, p. 241–242).
• Six special-type integrals (§7.4, p. 243–246): ∫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.
• For ∫dx/(ax²+bx+c) and ∫dx/√(ax²+bx+c) complete the square ax² + bx + c = a[(x + b/2a)² + (c/a − b²/4a²)] to reduce to standard forms; for ∫(px+q)/(ax²+bx+c)dx write px + q = A·d/dx(ax²+bx+c) + B and split (NCERT §7.4, p. 246–247).
• Integration by partial fractions decomposes a proper rational function P(x)/Q(x) into a sum of simpler fractions; five canonical forms are tabulated in Table 7.2 for distinct linear, repeated linear, three linear, repeated-plus-linear, and linear-plus-irreducible-quadratic denominators (NCERT §7.5, Table 7.2, p. 252–253).
• Improper rational functions are first reduced by long division: P(x)/Q(x) = T(x) + P₁(x)/Q(x), then T(x) is integrated as a polynomial and P₁(x)/Q(x) by partial fractions (NCERT §7.5, p. 252).
• Integration by parts: ∫u·(dv/dx)dx = uv − ∫v·(du/dx)dx, i.e., integral of (first × second) = first × ∫second − ∫(derivative of first × ∫second). Choice of "first function" follows ILATE-style guidance — when one function is a power/polynomial it is taken as first, but inverse-trig or logarithmic functions are taken as first when paired with algebraic/exponential functions (NCERT §7.6, p. 259–260).
• A useful by-parts result: ∫eˣ[f(x) + f′(x)]dx = eˣ f(x) + C (NCERT §7.6.1, p. 262–263).
• Three by-parts square-root integrals (§7.6.2, p. 264–265): ∫√(x²−a²)dx = (x/2)√(x²−a²) − (a²/2) log|x + √(x²−a²)| + C; ∫√(x²+a²)dx = (x/2)√(x²+a²) + (a²/2) log|x + √(x²+a²)| + C; ∫√(a²−x²)dx = (x/2)√(a²−x²) + (a²/2) sin⁻¹(x/a) + C.
• The definite integral ∫ₐᵇ f(x)dx has a unique numerical value — defined either as a limit of a sum or, when F is an antiderivative of continuous f on [a, b], as F(b) − F(a) (NCERT §7.7, §7.8.3 Theorem 2, p. 267–268).
• First fundamental theorem of calculus: if A(x) = ∫ₐˣ f(x)dx is the area function for continuous f on [a, b], then A′(x) = f(x) on [a, b] (NCERT §7.8.1–§7.8.2 Theorem 1, p. 267–268).
• Second fundamental theorem of calculus enables evaluation: ∫ₐᵇ f(x)dx = [F(x)]ₐᵇ = F(b) − F(a) — the arbitrary constant C cancels in the subtraction (NCERT §7.8.3, p. 268).
• Evaluation by substitution for definite integrals: substitute, change the limits accordingly, then evaluate in the new variable without going back (NCERT §7.9 Note, p. 271–272).
• Properties of definite integrals (P0–P7, §7.10, p. 273–276): P0 ∫ₐᵇ f(x)dx = ∫ₐᵇ f(t)dt; P1 ∫ₐᵇ f(x)dx = −∫ᵦᵃ f(x)dx and ∫ₐᵃ = 0; P2 ∫ₐᵇ = ∫ₐᶜ + ∫ᶜᵇ; P3 ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b−x)dx; P4 (king property) ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx; P5 ∫₀²ᵃ f(x)dx = ∫₀ᵃ f(x)dx + ∫₀ᵃ f(2a−x)dx; P6 ∫₀²ᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f(2a−x) = f(x), and = 0 if f(2a−x) = −f(x); P7 ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f is even, = 0 if f is odd.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 7.1 (Area function, p. 267): Region bounded by y = f(x), the x-axis, and the ordinates x = a, x = b; the shaded portion from a to a moving point x represents A(x) = ∫ₐˣ f(x)dx — visual basis for the first FTC.
• Table 7.1 — Symbols/Terms (p. 227): ∫f(x)dx, integrand, variable of integration, "integrate", "an integral of f", integration, constant of integration.
• Table of standard formulae (p. 228–229): 13 derivative–integral pairs covering powers, sin, cos, sec², cosec², sec tan, cosec cot, the three inverse-trig forms, eˣ, 1/x, aˣ.
• Table 7.2 — Partial fraction templates (p. 253): five rational-function shapes mapped to their decomposition forms (A/(x−a) + B/(x−b); A/(x−a) + B/(x−a)²; three distinct linear factors; (x−a)²(x−b); linear × irreducible quadratic with Bx+C numerator).
• Six "particular function" formulae (p. 243): memorise the boxed list 7.4 (1)–(6) — the (1/2a) log forms, the (1/a) tan⁻¹ form, the log|x + √…| forms, and the sin⁻¹(x/a) form.
• Three √-integrals via by-parts (p. 264–265): ∫√(x²−a²)dx, ∫√(x²+a²)dx, ∫√(a²−x²)dx with their (x/2)√… + (a²/2)(log or sin⁻¹) structure.

2.4 Common confusions / NTA trap points

• Forgetting the constant of integration C in indefinite integrals, or wrongly adding C in definite integrals where it cancels (NCERT §7.8.3 Remark, p. 268).
• Confusing ∫dx/(x²+a²) = (1/a) tan⁻¹(x/a) + C with ∫dx/(x²−a²) = (1/2a) log|(x−a)/(x+a)| + C — students swap signs and the 1/a vs 1/2a factor (NCERT §7.4, p. 243).
• Misreading ∫dx/√(a²−x²) = sin⁻¹(x/a) + C as cos⁻¹(x/a); both are antiderivatives of related functions but differ in sign — only sin⁻¹(x/a) matches the standard listing (NCERT §7.4 (5), p. 244).
• Applying integration by parts with the wrong "first function" — NCERT explicitly warns ∫x cos x dx with cos x as first gives a worse integral, so choosing the power/polynomial as first (or log/inverse-trig as first when paired with algebraic) matters (NCERT §7.6, p. 260).
• Forgetting to change limits when using substitution in a definite integral, or changing limits but then re-substituting back to x (NCERT §7.9 Note, p. 272).
• Using property P7 without first checking whether f is even or odd — e.g., ∫₋₁¹ sin⁵x cos⁴x dx = 0 because the integrand is odd (NCERT Example 31, p. 277).
• Mis-applying partial fractions when the rational function is improper (degree of numerator ≥ degree of denominator); first do polynomial long division.
• Forgetting that for ∫dx/√(x²+a²) the answer is log|x + √(x²+a²)|, not sin⁻¹(x/a).
• Misapplying ILATE: log and inverse-trig are usually first; algebraic and exponential are usually second; trigonometric depends on context.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 5(ii), p. 236). Compute ∫ 2x sin(x² + 1) dx.

Step 1 — substitute t = x² + 1: dt = 2x dx. Step 2 — rewrite: ∫ sin t dt. Step 3 — integrate: −cos t + C = −cos(x² + 1) + C.

Example B (NCERT Example 8(i), p. 247). ∫ dx/(x² − 16).

Step 1 — identify a: a = 4 since a² = 16. Step 2 — apply formula: (1/(2·4)) log|(x − 4)/(x + 4)| + C. Step 3 — simplify: (1/8) log|(x − 4)/(x + 4)| + C.

Example C (NCERT Example 19, p. 261). ∫ x eˣ dx by parts.

Step 1 — choose u = x, dv = eˣ dx: du = dx, v = eˣ. Step 2 — apply by-parts: x eˣ − ∫ eˣ dx. Step 3 — finish: x eˣ − eˣ + C = (x − 1) eˣ + C.

Example D (NCERT Example 18, p. 261). ∫ log x dx.

Step 1 — choose u = log x, dv = dx: du = (1/x) dx, v = x. Step 2 — apply by-parts: x log x − ∫ (1/x)·x dx. Step 3 — finish: x log x − x + C = x(log x − 1) + C.

Example E (NCERT Example 31, p. 277). Show ∫_{−1}^{1} sin⁵ x cos⁴ x dx = 0.

Step 1 — check parity: f(−x) = sin⁵(−x) cos⁴(−x) = −sin⁵ x · cos⁴ x = −f(x). f is odd. Step 2 — apply P7(ii): odd function over symmetric interval has integral 0. Step 3 — conclude: integral = 0.