Integrals Class 12: Notes, Methods and Solved Questions

What Integrals is about

Weightage: This chapter is part of Unit III, Calculus, which carries 35 of the 80 theory marks together with Continuity and Differentiability, Application of Derivatives, Application of Integrals and Differential Equations.Source: CBSE Mathematics syllabus 2026-27, Unit III Calculus

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.6 marks per paper. See the questions

Key concepts

Integration as the inverse of differentiation

NCERT section 7.2

Since $\frac{d}{dx}(F(x)+C)=f(x)$ for every constant C, the indefinite integral is a whole family of functions: $\int f(x)dx=F(x)+C$.

Reading the derivative table backwards gives the standard integrals: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ ($n\ne -1$), $\int \frac{1}{x}dx=\log |x|+C$, $\int e^{x}dx=e^x+C$, $\int a^{x}dx=\frac{a^x}{\log a}+C$, $\int \sin xdx=-\cos x+C$, $\int \cos xdx=\sin x+C$, $\int {\sec }^{2}xdx=\tan x+C$, $\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+C$, $\int \frac{dx}{1+x^2}={\tan }^{-1}x+C$.

Properties: $\int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx$ and $\int kf(x)dx=k\int f(x)dx$. You can always check an answer by differentiating it.

Worked example

Find $\int \left(3x^2-\frac{4}{x}+2e^x+{\sec }^{2}x\right)dx$. [1 mark]

Try it first, then show the solutionSolution

1. Integrate term by term: $x^3-4\log |x|+2e^x+\tan x+C$

Answer: $x^3-4\log |x|+2e^x+\tan x+C$

Integration by substitution

NCERT section 7.3.1

If the integrand contains a function and (a constant multiple of) its derivative, put that function equal to t. Then $\int f(g(x))g^{\prime }(x)dx=\int f(t)dt$ with $t=g(x)$, $dt=g^{\prime }(x)dx$. Finish by writing the answer back in x.

Useful results proved by substitution:

• $\int \tan xdx=\log |\sec x|+C$, $\int \cot xdx=\log |\sin x|+C$
• $\int \sec xdx=\log |\sec x+\tan x|+C$, $\int \mathrm{cosec}xdx=\log |\mathrm{cosec}x-\cot x|+C$
• $\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C$

Worked example

Find $\int \frac{\sin x}{(1+\cos x{)}^2}dx$. [2 marks]

Try it first, then show the solutionSolution

1. Put $t=1+\cos x$, so $dt=-\sin xdx$.
2. $\int \frac{-dt}{t^2}=\frac{1}{t}+C=\frac{1}{1+\cos x}+C$

Answer: $\frac{1}{1+\cos x}+C$

Integration using trigonometric identities

NCERT section 7.3.2

Powers and products of sines and cosines are first rewritten as sums using identities:

• ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, ${\cos }^{2}x=\frac{1+\cos 2x}{2}$
• ${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$, ${\cos }^{3}x=\frac{3\cos x+\cos 3x}{4}$
• $2\sin A\cos B=\sin (A+B)+\sin (A-B)$, $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, $2\sin A\sin B=\cos (A-B)-\cos (A+B)$

For an odd power such as ${\sin }^{3}x{\cos }^{2}x$, keep one sin x aside and substitute $t=\cos x$.

Worked example

Find $\int \sin 3x\cos 2xdx$. [2 marks]

Try it first, then show the solutionSolution

1. $\sin 3x\cos 2x=\frac{1}{2}(\sin 5x+\sin x)$
2. $\int =\frac{1}{2}\left(-\frac{\cos 5x}{5}-\cos x\right)+C$

Answer: $-\frac{\cos 5x}{10}-\frac{\cos x}{2}+C$

Integrals of some particular functions

NCERT section 7.4

Six standard results:

• $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$
• $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$
• $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C$
• $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$

For $\int \frac{dx}{a{x}^2+bx+c}$ or $\int \frac{dx}{\sqrt{a{x}^2+bx+c}}$, take out a, complete the square, and match one of these forms.

Worked example

Find $\int \frac{dx}{x^2+6x+13}$. [2 marks]

Try it first, then show the solutionSolution

1. $x^2+6x+13=(x+3{)}^2+2^2$
2. $\int \frac{dx}{(x+3{)}^2+2^2}=\frac{1}{2}{\tan }^{-1}\frac{x+3}{2}+C$

Answer: $\frac{1}{2}{\tan }^{-1}\left(\frac{x+3}{2}\right)+C$

Linear over quadratic: (px + q)/(ax² + bx + c)

NCERT section 7.4

For $\int \frac{px+q}{a{x}^2+bx+c}dx$ or $\int \frac{px+q}{\sqrt{a{x}^2+bx+c}}dx$, write the numerator as a multiple of the derivative of the quadratic plus a constant:

$px+q=A\frac{d}{dx}(a{x}^2+bx+c)+B=A(2ax+b)+B$

Find A and B by comparing coefficients. The first part integrates to $A\log |a{x}^2+bx+c|$ (or $2A\sqrt{a{x}^2+bx+c}$ under a root); the second is a special integral after completing the square.

Worked example

Find $\int \frac{2x+3}{x^2+4x+5}dx$. [3 marks]

Try it first, then show the solutionSolution

1. $2x+3=(2x+4)-1$, and $2x+4$ is the derivative of $x^2+4x+5$.
2. $\int \frac{2x+4}{x^2+4x+5}dx=\log (x^2+4x+5)$ (the quadratic is always positive).
3. $\int \frac{dx}{(x+2{)}^2+1}={\tan }^{-1}(x+2)$

Answer: $\log (x^2+4x+5)-{\tan }^{-1}(x+2)+C$

Integration by partial fractions

NCERT section 7.5

A proper rational function $\frac{P(x)}{Q(x)}$ (degree of P less than degree of Q) is split into simpler fractions. If it is improper, divide first. Forms to use:

• $\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$
• $\frac{px+q}{(x-a{)}^2}=\frac{A}{x-a}+\frac{B}{(x-a{)}^2}$
• $\frac{p{x}^2+qx+r}{(x-a{)}^2(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a{)}^2}+\frac{C}{x-b}$
• $\frac{p{x}^2+qx+r}{(x-a)(x^2+bx+c)}=\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}$ (quadratic not factorisable)

Find the constants by substituting convenient x values and comparing coefficients. When only $x^2$ appears (like $\frac{x^2}{(x^2+1)(x^2+4)}$), put $x^2=y$ just for the splitting.

Worked example

Find $\int \frac{3x+5}{(x-1{)}^2(x+2)}dx$. [3 marks]

Try it first, then show the solutionSolution

1. $\frac{3x+5}{(x-1{)}^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}$, so $3x+5=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2$.
2. x = 1: 8 = 3B, $B=\frac{8}{3}$. x = −2: −1 = 9C, $C=-\frac{1}{9}$. Coefficient of $x^2$: A + C = 0, $A=\frac{1}{9}$.
3. $\int =\frac{1}{9}\log |x-1|-\frac{8}{3(x-1)}-\frac{1}{9}\log |x+2|+C$

Answer: $\frac{1}{9}\log \left|\frac{x-1}{x+2}\right|-\frac{8}{3(x-1)}+C$

Integration by parts

NCERT section 7.6

For a product of two functions:

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int \left[f^{\prime }(x)\int g(x)dx\right]dx$

In words: first × integral of second − integral of (derivative of first × integral of second). Choose the first function by the order ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. For $\int \log xdx$ or $\int {\tan }^{-1}xdx$, take 1 as the second function.

Sometimes the original integral reappears (as in $\int e^x\sin xdx$); then solve for it like an equation.

Worked example

Find $\int x{\sec }^{2}xdx$. [2 marks]

Try it first, then show the solutionSolution

1. First function x, second ${\sec }^{2}x$: $x\tan x-\int 1\cdot \tan xdx$
2. $=x\tan x-\log |\sec x|+C$

Answer: $x\tan x-\log |\sec x|+C$

Integrals of the form eˣ[f(x) + f′(x)]

NCERT section 7.6.1

Integration by parts gives a shortcut:

$\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$

To use it, split the bracket into a function and its derivative. Typical pairs: $\frac{1}{x}$ and $-\frac{1}{x^2}$; $\tan x$ and ${\sec }^{2}x$; ${\sin }^{-1}x$ and $\frac{1}{\sqrt{1-x^2}}$; $\frac{1}{x+1}$ and $-\frac{1}{(x+1{)}^2}$.

Worked example

Find $\int e^x\frac{x}{(x+1{)}^2}dx$. [2 marks]

Try it first, then show the solutionSolution

1. $\frac{x}{(x+1{)}^2}=\frac{(x+1)-1}{(x+1{)}^2}=\frac{1}{x+1}-\frac{1}{(x+1{)}^2}$
2. With $f(x)=\frac{1}{x+1}$, $f^{\prime }(x)=-\frac{1}{(x+1{)}^2}$, so the integral is $e^{x}f(x)+C$.

Answer: $\frac{e^x}{x+1}+C$

Integrals of √(a² − x²), √(x² − a²) and √(x² + a²)

NCERT section 7.6.2

Using integration by parts (with 1 as the second function):

• $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
• $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log \left|x+\sqrt{x^2-a^2}\right|+C$
• $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+C$

For $\int \sqrt{a{x}^2+bx+c}dx$, complete the square first. The pattern is always "half x times the root, plus or minus half $a^2$ times the matching special integral".

Worked example

Find $\int \sqrt{x^2+4x+8}dx$. [3 marks]

Try it first, then show the solutionSolution

1. $x^2+4x+8=(x+2{)}^2+2^2$; put $u=x+2$, a = 2.
2. $\int \sqrt{u^2+4}du=\frac{u}{2}\sqrt{u^2+4}+2\log \left|u+\sqrt{u^2+4}\right|+C$

Answer: $\frac{x+2}{2}\sqrt{x^2+4x+8}+2\log \left|x+2+\sqrt{x^2+4x+8}\right|+C$

Definite integrals and the Fundamental Theorem of Calculus

NCERT section 7.8

The definite integral ${\int }_a^{b}f(x)dx$ (a is the lower limit, b the upper limit) is a number. For $f\ge 0$ it is the area under y = f(x) between x = a and x = b.

Area function: $A(x)={\int }_a^{x}f(t)dt$. First fundamental theorem: $A^{\prime }(x)=f(x)$. Second fundamental theorem: if F is any antiderivative of f on [a, b], then

${\int }_a^{b}f(x)dx=[F(x){]}_a^b=F(b)-F(a)$

Both are stated without proof in the syllabus. No constant C is needed, because it cancels.

Worked example

Evaluate ${\int }_1^2\left(x^2+\frac{1}{x}\right)dx$. [2 marks]

Try it first, then show the solutionSolution

1. Antiderivative: $\frac{x^3}{3}+\log x$
2. $\left(\frac{8}{3}+\log 2\right)-\left(\frac{1}{3}+0\right)=\frac{7}{3}+\log 2$

Answer: $\frac{7}{3}+\log 2$

Evaluating definite integrals by substitution

NCERT section 7.9

When you substitute t = g(x) in a definite integral, change the limits too: x = a becomes t = g(a), and x = b becomes t = g(b). Then evaluate in t and never go back to x.

This is quicker and avoids the error of putting x-limits into a t-expression.

Worked example

Evaluate ${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$. [2 marks]

Try it first, then show the solutionSolution

1. Put $t={\tan }^{-1}x$, $dt=\frac{dx}{1+x^2}$; x = 0 → t = 0, x = 1 → $t=\frac{\pi }{4}$.
2. ${\int }_0^{\pi /4}tdt={\left[\frac{t^2}{2}\right]}_0^{\pi /4}=\frac{{\pi }^2}{32}$

Answer: $\frac{{\pi }^2}{32}$

Properties of definite integrals: the a + b − x property

NCERT section 7.10

The key properties (NCERT P0 to P7):

• P1: ${\int }_a^{b}f=-{\int }_b^{a}f$; P2: ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$
• P3: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$; P4: ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$
• P5: ${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{a}f(2a-x)dx$
• P6: ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ if $f(2a-x)=f(x)$, and 0 if $f(2a-x)=-f(x)$

Method with P3/P4: call the integral I, rewrite it with $a+b-x$ in place of x, then add the two forms. The awkward parts usually combine into something simple.

Worked example

Evaluate ${\int }_0^{\pi /2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$. [3 marks]

Try it first, then show the solutionSolution

1. Let I be the integral. By P4, replacing x by $\frac{\pi }{2}-x$: $I={\int }_0^{\pi /2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$.
2. Adding: $2I={\int }_0^{\pi /2}1dx=\frac{\pi }{2}$

Answer: $I=\frac{\pi }{4}$

Even and odd functions, and integrals with modulus

NCERT section 7.10

P7: on a symmetric interval [−a, a],

• ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$ if f is even ($f(-x)=f(x)$);
• ${\int }_{-a}^{a}f(x)dx=0$ if f is odd ($f(-x)=-f(x)$).

Split a sum into its even and odd parts and drop the odd ones. Examples of odd functions: $x^3$, $\sin x$, ${\tan }^{5}x$, $x\cos x$, $\log \frac{2-x}{2+x}$.

For a modulus, use P2: find where the expression inside changes sign, split the interval there, and remove the modulus with the correct sign on each piece.

Worked example

Evaluate ${\int }_{-\pi /4}^{\pi /4}(x^3+x\cos x+{\tan }^{5}x+1)dx$. [2 marks]

Try it first, then show the solutionSolution

1. $x^3$, $x\cos x$ and ${\tan }^{5}x$ are odd, so their integrals over $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$ are 0.
2. ${\int }_{-\pi /4}^{\pi /4}1dx=\frac{\pi }{2}$

Answer: $\frac{\pi }{2}$

Maths in real life: Integrals

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Integration by Substitution
• Integration by Partial Fractions
• Properties of Definite Integrals

Integrals formula sheet

Power and log

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C(n\ne -1),\int \frac{dx}{x}=\log |x|+C$

Exponential

$\int e^{x}dx=e^x+C,\int a^{x}dx=\frac{a^x}{\log a}+C$

Trigonometric

$\int \sin xdx=-\cos x+C,\int \cos xdx=\sin x+C,\int {\sec }^{2}xdx=\tan x+C,\int \sec x\tan xdx=\sec x+C$

tan, cot, sec, cosec

$\int \tan xdx=\log |\sec x|+C,\int \cot xdx=\log |\sin x|+C,\int \sec xdx=\log |\sec x+\tan x|+C$

Inverse trigonometric results

$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+C,\int \frac{dx}{1+x^2}={\tan }^{-1}x+C$

Special integrals (squares)

$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C,\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C,\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$

Special integrals (roots)

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C,\int \frac{dx}{\sqrt{x^2\pm a^2}}=\log \left|x+\sqrt{x^2\pm a^2}\right|+C$

Integrals of roots

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C,\int \sqrt{x^2\pm a^2}dx=\frac{x}{2}\sqrt{x^2\pm a^2}\pm \frac{a^2}{2}\log \left|x+\sqrt{x^2\pm a^2}\right|+C$

Integration by parts

$\int fgdx=f\int gdx-\int \left(f^{\prime }\int gdx\right)dx$

eˣ shortcut

$\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$

Fundamental Theorem of Calculus

${\int }_a^{b}f(x)dx=F(b)-F(a),\text{where }{F}^{\prime }=f$

Key properties

${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx,{\int }_{-a}^{a}f=\{\begin{array}{ll}2{\int }_0^{a}f,& f\text{ even}\\ 0,& f\text{ odd}\end{array}$

How the board asks Integrals

The question patterns to expect, and the method that works for each.

Tricks to remember Integrals

Short memory aids. Say each one aloud once, then use it on a question the same day.

• ILATE

  Inverse trig, Log, Algebraic, Trig, Exponential: whichever comes first is the first function in integration by parts.

  Use it when: Choosing f in ∫f g dx.

• Spot f and f′ next to eˣ

  If $e^x$ multiplies a bracket, look for a function and its derivative inside; the answer is $e^x$ times the function.

  Use it when: eˣ[f + f′] questions.

• Plus a² gives tan⁻¹, root of a² minus x² gives sin⁻¹

  After completing the square: sum of squares in a denominator → ${\tan }^{-1}$; $\sqrt{a^2-x^2}$ → ${\sin }^{-1}$; everything else → log.

  Use it when: Picking the special integral.

• Odd vanishes, even doubles

  On [−a, a], odd parts integrate to 0 and even parts give twice the integral from 0 to a.

  Use it when: Symmetric-limit definite integrals.

• Flip and add

  For awkward definite integrals, replace x by a + b − x and add to the original.

  Use it when: P3/P4 questions.

Integrals: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Integrals

The slips examiners see most often, and what to write instead.

• ✗Mistake: Dropping the constant C or the modulus in log

  Writing $\int \frac{dx}{x}=\log x$.

  Wrong working

  $\int \frac{dx}{x-2}=\log (x-2)$

  Right working

  $\int \frac{dx}{x-2}=\log |x-2|+C$

  ✓Do this: Every indefinite integral ends with + C; use |…| inside log unless the expression is clearly positive.

  Usually costs: ½ mark each time.

• ✗Mistake: Keeping x-limits after substituting

  Changing the variable but not the limits.

  ✓Do this: Convert the limits to t at the same step as dx to dt.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong partial fraction form

  Writing only $\frac{B}{(x-1{)}^2}$ for a repeated factor, or a constant over a quadratic factor.

  Wrong working

  $\frac{x}{(x-1{)}^2(x+2)}=\frac{A}{(x-1{)}^2}+\frac{B}{x+2}$

  Right working

  $\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}$

  ✓Do this: Repeated factors need every power; irreducible quadratics need Bx + C on top.

  Usually costs: Most of the marks.

• ✗Mistake: Ignoring the modulus in a definite integral

  Treating ${\int }_0^3|x-1|dx$ as ${\int }_0^3(x-1)dx$.

  ✓Do this: Split at the point where the inside is zero and use the right sign on each part.

  Usually costs: 2-3 marks.

• ✗Mistake: Applying P4 and not adding

  Rewriting the integral with a − x and then trying to evaluate that form directly.

  ✓Do this: Write I twice (original and transformed) and add them so the awkward parts cancel or combine.

  Usually costs: 2 marks.

NCERT exercise guide: Integrals

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.