Integrals Class 12 Notes: Anti-derivatives, Substitution & Formulas

These integrals class 12 notes condense Chapter 7 of NCERT Mathematics Part II into a revision-ready page: what integration means, the full standard-formula table, substitution, trigonometric identities, three worked examples with original numbers, and the errors that quietly eat marks.

The sequence follows the rationalised NCERT text in use for the 2026-27 session, and every page number points at the printed textbook so you can verify a formula in seconds. This page is shorter than the chapter but complete enough to revise from alone.

It covers the grounded core — sections 7.1 to 7.3.2. The later methods (partial fractions, integration by parts) and the definite integral build directly on this foundation, as do the Class 12 Mathematics notes that stand alongside this page.

Everything below is checked against the printed text. The official NCERT PDF (lemh201.pdf) on ncert.nic.in contains the whole chapter — open it to verify any formula that does not feel automatic before the exam.

Why Integrals Exist: The Two Problems That Built Chapter 7

Chapter 7 begins from what differentiation leaves unanswered (NCERT, p. 226). The derivative solved the tangent-and-slope problem; integration addresses two fresh questions, which the text states as:

• (a) given the derivative of a function, recover the function itself;
• (b) find the area bounded by the graph of a function.

Problem (a) produces the indefinite integral — a family of anti-derivatives. Problem (b) produces the definite integral, a single number. The Fundamental Theorem of Calculus is the bridge between the two (p. 226).

A revision page needs to know where it stands. The chapter climbs in this order:

This page grounds exactly that core. The methods of partial fractions and integration by parts are named in 7.3 but developed from here (p. 236). Chapter 6, covered in the Application of Derivatives notes, is the slope-problem companion to this chapter, and the area side continues in the Application of Integrals notes.

Anti-derivatives and the Constant C: What Integration Actually Means

If \( \frac{d}{dx}F(x) = f(x) \) on an interval, then F is an anti-derivative (primitive) of f (p. 227). Because the derivative of a constant is zero, each result like \( \frac{d}{dx}(\sin x) = \cos x \) actually hides infinitely many anti-derivatives.

So \( \sin x \) is one anti-derivative of \( \cos x \), and \( \sin x + C \) gives them all, where C is any real number — the arbitrary constant of integration. The set \( \{F + C : C \in \mathbb{R}\} \) is the whole family of anti-derivatives of f (p. 227).

Why adding C covers everything: if g and h both have derivative f, then \( (g – h)’ = 0 \), so \( g – h \) is a constant (p. 227). One sentence — but it is the reason every answer below ends in + C.

Property I makes integration self-checking (p. 230):

\[ \frac{d}{dx}\int f(x)\,dx = f(x), \qquad \int f'(x)\,dx = f(x) + C \]

A way to hold onto the meaning: integration is rewinding a recording. Differentiation records — it turns a function into its rate of change. Integration plays it backwards, recovering the function — almost. A recording cannot recover its own starting point, and that missing starting value is exactly C. Two rewinds of the same track are identical except for where they begin.

The introduction names the real question behind all of it: knowing an object’s instantaneous velocity, can we find its position (p. 226)? Try it with new numbers. A particle has velocity \( v(t) = 12 – 4t \) m/s and is at \( s(0) = 5 \) m. Position is the anti-derivative of velocity:

\[ s(t) = \int (12 – 4t)\,dt = 12t – 2t^2 + C \]

The initial condition fixes C: \( 5 = 0 – 0 + C \), so \( C = 5 \) and \( s(t) = 12t – 2t^2 + 5 \). Check: \( s'(t) = 12 – 4t = v(t) \). That is the whole story — anti-differentiate, then let a condition choose the one curve you need.

Definitions: Integrand, Variable of Integration, and the Constant C

Table 7.1 (p. 228) fixes the vocabulary of every integral. It is reworded below, with a concrete example added to each row.

Standard Integrals Table: The Formulas You Must Memorise

Every row below is a derivative read backwards — each comes from the derivative list in the text (p. 229). Learn the table cold; inspection and substitution only work when you know what you are aiming at.

The text adds a practical warning (p. 229): the interval over which each function is defined is normally not written down, but you must keep it in mind in a specific problem.

The NCERT text never lays the two operations side by side, so here is the comparison that makes integration feel familiar:

Five Properties of Indefinite Integrals: Why You Can Split and Factor

Properties I–V (pp. 230–232) are what let you break a hard integral into table rows. The proof of each is the same idea: differentiate both sides and compare. What matters in the exam is what each one allows.

Splitting is what made \( \int \frac{x^3 – 1}{x^2}\,dx = \int x\,dx – \int x^{-2}\,dx \) one line of table work (p. 233). Always recombine the separate constants \( C_1, C_2 \) into a single C in the final answer (p. 233).

Method of Inspection: Reverse-Engineering a Derivative

Inspection is the search strategy for an anti-derivative: look for a function whose derivative is the integrand, then fix any mismatch of a constant factor (p. 232). The two-step logic is guess the function, then divide by the factor that appears on differentiation.

• \( \int \cos 5x\,dx \): guess \( \sin 5x \); since \( \frac{d}{dx}\sin 5x = 5\cos 5x \), divide by 5 → \( \frac{1}{5}\sin 5x + C \).
• \( \int (6x^5 – 8x^3)\,dx \): guess \( x^6 – 2x^4 \); its derivative is exactly \( 6x^5 – 8x^3 \) → \( x^6 – 2x^4 + C \).

Fixing C from an extra condition: if \( F'(x) = 3x^2 – 4 \) and \( F(1) = 5 \), first \( F(x) = x^3 – 4x + C \). Then \( 5 = 1 – 4 + C \), so \( C = 8 \), giving the unique function \( F(x) = x^3 – 4x + 8 \) (method of Example 4, p. 234).

Inspection has an honest limit. No elementary anti-derivative (in terms of polynomials, logs, exponentials, trig and their inverses) exists for \( e^{-x^2} \) (p. 235) — knowing this saves you from wasting a page of search in an exam.

Integration by Substitution: Spot the Function, Change the Variable

When the integrand hides a function and its derivative, substitute. Put \( x = g(t) \), write \( dx = g'(t)\,dt \), then (p. 236):

\[ \int f(x)\,dx = \int f(g(t))\,g'(t)\,dt \]

Rule for choosing the substitution: substitute for the function whose derivative also sits in the integrand (p. 236). That guarantees the x’s cancel and the integral becomes one in t.

Worked with new numbers — integrate \( \int x\cos(x^2 + 3)\,dx \):

Four logarithmic results appear so often that they are used without reference (pp. 238–239). The trick behind each is the same — set the denominator as t so the numerator becomes \( \pm dt \):

For sec and cosec, the move is the conjugate-multiply trick: multiply numerator and denominator by the partner — \( \sec x + \tan x \) for sec, \( \cosec x + \cot x \) for cosec — so the new numerator is exactly the derivative of the new denominator (p. 239). A minus sign appears for cosec, cleaned up using \( \cosec^2 x – \cot^2 x = 1 \).

Memory device: anchor the four results by their shapes — tan gives sec, cot gives sin (single, no extra term); sec gives sec + tan, cosec gives cosec − cot (they carry their partner inside the log). For the power rule keep the chant “add one, divide by the new power” — it catches the \( n \neq -1 \) exception before you start.

Integration Using Trigonometric Identities: Rewrite Before You Integrate

When the integrand has powers or products of sine and cosine, rewrite with known identities first, then integrate row by row (p. 242). The text’s toolkit:

• \( \cos^2 x = \frac{1 + \cos 2x}{2} \) — for even powers;
• \( 2\sin A\cos B = \sin(A+B) + \sin(A-B) \) — for products;
• odd powers split: \( \sin^3 x = \sin x\,(1 – \cos^2 x) \), leaving a \( \sin x \) ready for substitution.

Worked with new numbers — integrate \( \int \sin 5x \cos 3x\,dx \):

Pattern to reuse: odd power — keep one factor out for substitution and convert the rest; even power — use the double-angle identity to drop the exponent.

Worked Examples: Three Solved Problems With Fresh Numbers

Example 1 — Inspection with an initial condition: find F given \( F’ = 6x^2 – 8x + 5 \) and \( F(1) = 4 \)

Example 2 — Substitution: integrate \( \int x\,e^{x^2+1}\,dx \)

Example 3 — Trigonometric identity: integrate \( \int \sin 5x \cos 3x\,dx \)

Common Mistakes in Integrals: Error → Correction Pairs

Misconception autopsy: the two silent mark-killers

Dropped + C. The diagnosis: students treat \( \int f(x)\,dx \) as a single function instead of a family. The cure is Property I: \( \int f'(x)\,dx = f(x) + C \) (p. 230). Whenever the answer you write down can be shifted vertically by any constant and still differentiate to the same integrand, that freedom is the C you must write.

Dropped modulus in \( \log x \). The diagnosis: \( \log x \) only makes sense for \( x \gt 0 \), but \( 1/x \) is defined for negative x as well. The cure is in Example 1(iii): \( \frac{d}{dx}\log(-x) = \frac{1}{x} \) for \( x \lt 0 \), so the two branches combine into \( \log|x| \) (p. 232).

If a problem allows x negative, the modulus is not decoration — it is the whole answer.

Exam Notes: Where Integration Marks Are Won

• Exercise 7.1 (p. 235) drills inspection and direct table rows. Q1–5 demand the inspect-and-adjust method (an anti-derivative of \( \cos 3x \) is \( \frac{1}{3}\sin 3x \), not \( \sin 3x \)). Q6–20 are split-and-integrate: expand \( (\sqrt{x} – 1/\sqrt{x})^2 \) etc. before touching the table. Q21 and Q22 are MCQs — the distractor options exist precisely to punish a dropped coefficient or a wrong constant.
• Exercise 7.2 (pp. 241–243) drills substitution where the inner function’s derivative is visible — e.g. \( \frac{2x}{1+x^2} \) with \( 1 + x^2 = t \). Q38 and Q39 are MCQs testing whether you recognise the derivative sitting in the numerator.
• Exercise 7.4 (p. 252) drills forms that reduce to the inverse-trig rows of the standard table — \( \frac{1}{\sqrt{9 – 25x^2}} \) and \( \frac{1}{\sqrt{x^2 + 2x + 2}} \) are the pattern. Q24 and Q25 are MCQs on exactly this reduction.

The visible lines that earn method marks in substitution: writing the substitution, converting \( dx \) to \( g'(t)\,dt \), integrating correctly in t, and resubstituting to x. A full-marks answer shows all four lines.

One constant C in the final answer is enough — NCERT states this convention explicitly (p. 234), and a flood of \( C_1, C_2 \) is only a chance to slip signs.

No year-based predictions are made here, but two behaviours are safe to adopt: always end by differentiating your answer (Property I makes it a free check), and never start a page-long search where inspection provably fails, as with \( e^{-x^2} \) (p. 235).

Integrals Class 12 Notes: One-Page Revision Summary

Last-night recap — read this list top to bottom, then attempt the exercise map.

• Definition: integration is the inverse of differentiation; an anti-derivative F of f satisfies \( F’ = f \) (p. 227).
• The + C rule: every indefinite answer is a family \( \{F + C\} \); fix C only when a condition like \( F(1) = 4 \) is given.
• Power rule: \( \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \), \( n \neq -1 \); at \( n = -1 \) use \( \log|x| + C \) (p. 229).
• Properties in one line: split sums, factor constants, and check every answer by differentiating (pp. 230–232).
• Substitution algorithm: set t = inner function whose derivative is visible, convert \( dx \), integrate in t, resubstitute to x (p. 236).
• Four trig results: \( \int \tan x = \log|\sec x| \); \( \int \cot x = \log|\sin x| \); \( \int \sec x = \log|\sec x + \tan x| \); \( \int \cosec x = \log|\cosec x – \cot x| \) (pp. 238–239).
• Trig identities: odd power → keep one factor for substitution; even power → double-angle identity; products → product-to-sum (p. 242).

The complement of this chapter lives in the Class 12 notes hub and the wider CBSE notes library, and the chain-rule backbone of substitution is covered in the Continuity and Differentiability notes.

Frequently Asked Questions About Integrals

Why does every indefinite integral carry a plus C at the end?

Because the derivative of any constant is zero. If F is an anti-derivative of f, then \( F + C \) is also one for every real C (p. 227). The integral stands for the whole family \( \{F + C\} \), not a single function, so omitting C ignores infinitely many correct answers.

You only pin C down to one value when a condition like \( F(1) = 4 \) is supplied.

Why is the integral of 1 over x written as log modulus x plus C and not log x plus C?

Because \( \frac{d}{dx}\log x = \frac{1}{x} \) works only for \( x \gt 0 \), while \( \frac{d}{dx}\log(-x) = \frac{1}{x} \) covers \( x \lt 0 \) (p. 232). Combining the two branches gives \( \frac{d}{dx}\log|x| = \frac{1}{x} \) for every \( x \neq 0 \), so the modulus is what makes the formula valid on both sides of zero (p. 229).

When does the power rule for integration fail?

Exactly at \( n = -1 \), i.e. for \( 1/x \). The formula \( \frac{x^{n+1}}{n+1} \) would divide by zero at that exponent, so the text excludes it explicitly (p. 229) and \( \int \frac{1}{x}\,dx = \log|x| + C \) takes over. Every other exponent, including fractions and negatives like \( -2 \), obeys the power rule.

How do I choose the right substitution in integration by substitution?

Substitute for the function whose derivative also appears in the integrand (p. 236). In \( \int x\cos(x^2+3)\,dx \), the derivative of \( x^2+3 \) is \( 2x \), which is present, so \( t = x^2+3 \) works cleanly. When the substitution is right, the x’s cancel completely and the new integral is a standard table row.

What is the difference between an indefinite integral and a definite integral?

An indefinite integral \( \int f(x)\,dx \) answers problem (a) — it returns a family of anti-derivatives \( F(x) + C \). A definite integral answers problem (b) — it returns a single number giving the area under the curve (p. 226). The Fundamental Theorem of Calculus is the connection that lets a definite integral be evaluated through an anti-derivative.

Reference: NCERT Class 12 Mathematics textbook, chapter 7 Integrals.