Integration Formulas for JEE Main and Advanced

MathsClass 12, IntegralsFor JEE Main and Advanced

Integration formulas for JEE Main and Advanced on one page: the standard integrals, the forms you reach by completing the square, the methods (substitution, by parts, partial fractions), the properties of definite integrals, the Leibniz rule and the limit of a sum, each with the condition under which it holds. The table shows which formulas decided the most questions in recent JEE papers, and section 8 solves ten of those questions step by step. This is the Class 12 chapter Integrals.

Which formulas JEE uses most

We read all 85 questions on integration in our 48 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026) and noted the formulas each solution needs; one question can need several. The formulas used most:

Formula	Questions
Substitution: $\int f (g (x)) g^{'} (x) d x = \int f (t) d t$ with $t = g (x)$	2732%
Leibniz rule: $\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$	1619%
Splitting $\int_{a}^{b}$ where $[x]$ , $∣ x ∣$ , ${x}$ or a max/min changes	1214%
$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ $(n \neq = - 1)$ and $\int \frac{d x}{x} = ln ∣ x ∣$ , also with $a x + b$	1113%
By parts: $\int u v d x = u \int v d x - \int (u^{'} \int v d x) d x$	1113%
King's property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$	1113%
Symmetric limits: $\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x$	911%
Newton-Leibniz: $\int_{a}^{b} f (x) d x = F (b) - F (a)$	67%

Show the other 15

Formula	Questions
Inverse-trig forms: $\int \frac{d x}{a ^{2} + x ^{2}}$ , $\int \frac{d x}{a ^{2} - x ^{2}}$	67%
Log forms: $\int \frac{d x}{x ^{2} - a ^{2}}$ , $\int \frac{d x}{a ^{2} - x ^{2}}$ , $\int \frac{d x}{x ^{2} \pm a ^{2}}$	56%
Basic trigonometric integrals and power reduction, for example $sin^{2} x = \frac{1 - cos 2 x}{2}$	45%
Periodic $f$ with period $T$ : $\int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x$	34%
$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$	34%
Linear over quadratic: write the numerator as $A \cdot (denominator)^{'} + B$	34%
Square-root forms: $\int a^{2} - x^{2} d x$ and $\int x^{2} \pm a^{2} d x$	34%
Limit of a sum: $n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$	22%
$f$ and $f^{- 1}$ : $\int_{a}^{b} f (x) d x + \int_{f (a)}^{f (b)} f^{- 1} (y) d y = b f (b) - a f (a)$	22%
A definite integral inside an equation is a constant: call it $λ$	11%
$sin^{m} x cos^{n} x$ with $m + n$ a negative even integer: put $t = tan x$	11%
$\int tan x d x = ln ∣ sec x ∣$ , $\int sec x d x = ln ∣ sec x + tan x ∣$ and the cot, cosec pair	11%
$\int e^{x} d x = e^{x}$ , $\int a^{x} d x = \frac{a ^{x}}{ln a}$ , $\int ln x d x = x ln x - x$	11%
Splitting the integrand into simpler parts (linearity)	11%
$\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ when $f (2 a - x) = f (x)$	11%

1. Standard integrals

Add a constant $C$ to every indefinite integral. Each result is checked by differentiating it back to the integrand.

Integral	Result (+ C)	Valid for
$\int x^{n} d x$	$\frac{x ^{n + 1}}{n + 1}$	$n \neq = - 1$
$\int \frac{1}{x} d x$	$ln ∣ x ∣$	$x \neq = 0$
$\int e^{x} d x$	$e^{x}$	all $x$
$\int a^{x} d x$	$\frac{a ^{x}}{ln a}$	$a > 0, a \neq = 1$
$\int sin x d x$	$- cos x$	all $x$
$\int cos x d x$	$sin x$	all $x$
$\int sec^{2} x d x$	$tan x$	$cos x \neq = 0$
$\int cosec^{2} x d x$	$- cot x$	$sin x \neq = 0$
$\int sec x tan x d x$	$sec x$	$cos x \neq = 0$
$\int cosec x cot x d x$	$- cosec x$	$sin x \neq = 0$
$\int tan x d x$	$ln ∣ sec x ∣$	$cos x \neq = 0$
$\int cot x d x$	$ln ∣ sin x ∣$	$sin x \neq = 0$
$\int sec x d x$	$ln ∣ sec x + tan x ∣$	$cos x \neq = 0$
$\int cosec x d x$	$ln ∣ cosec x - cot x ∣$	$sin x \neq = 0$
$\int ln x d x$	$x ln x - x$	$x > 0$

Linear argument

\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C, a \neq = 0

$F$ is an antiderivative of $f$ . So $\int e^{3 x + 1} d x = \frac{e ^{3 x + 1}}{3} + C$ and $\int \frac{d x}{2 x + 5} = \frac{1}{2} ln ∣2 x + 5∣ + C$ .

2. Inverse-trig, log and square-root forms

In every row $a > 0$ ; add $C$ . Bring the integrand to one of these forms by completing the square:

a x^{2} + b x + c = a [(x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a ^{2}}]

Integral	Result (+ C)	Valid for
$\int \frac{d x}{a ^{2} - x ^{2}}$	$sin^{- 1} \frac{x}{a}$	$∣ x ∣ < a$
$\int \frac{d x}{a ^{2} + x ^{2}}$	$\frac{1}{a} tan^{- 1} \frac{x}{a}$	all $x$
$\int \frac{d x}{x x ^{2} - a ^{2}}$	$\frac{1}{a} sec^{- 1} \frac{x}{a}$	$x > a$
$\int \frac{d x}{x ^{2} - a ^{2}}$	$\frac{1}{2 a} ln \frac{x - a}{x + a}$	$∣ x ∣ \neq = a$
$\int \frac{d x}{a ^{2} - x ^{2}}$	$\frac{1}{2 a} ln \frac{a + x}{a - x}$	$∣ x ∣ \neq = a$
$\int \frac{d x}{x ^{2} + a ^{2}}$	$ln x + x^{2} + a^{2}$	all $x$
$\int \frac{d x}{x ^{2} - a ^{2}}$	$ln x + x^{2} - a^{2}$	$∣ x ∣ > a$
$\int a^{2} - x^{2} d x$	$\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$	$∣ x ∣ \leq a$
$\int x^{2} + a^{2} d x$	$\frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln x + x^{2} + a^{2}$	all $x$
$\int x^{2} - a^{2} d x$	$\frac{x}{2} x^{2} - a^{2} - \frac{a ^{2}}{2} ln x + x^{2} - a^{2}$	$∣ x ∣ \geq a$

3. Methods of integration

3.1 Substitution

Substitution

\int f (g (x)) g^{'} (x) d x = \int f (t) d t

Put $t = g (x)$ . Use it when the integrand holds a function and its derivative. In a definite integral, change the limits too.

The substitutions JEE expects you to spot:

The integrand has	Put
$f (a x + b)$	$t = a x + b$
$\frac{f ^{'} ( x )}{f ( x )}$	$t = f (x)$ : the integral is $ln ∣ f (x) ∣$
$[f (x)]^{n} f^{'} (x)$	$t = f (x)$ : the integral is $\frac{[ f ( x ) ] ^{n + 1}}{n + 1}$ , $n \neq = - 1$
$a^{2} - x^{2}$	$x = a sin θ$
$x^{2} + a^{2}$	$x = a tan θ$
$x^{2} - a^{2}$	$x = a sec θ$
$sin^{m} x cos^{n} x$ with $m + n$ a negative even integer	$t = tan x$ : write $sin^{m} x cos^{n} x = tan^{m} x sec^{- (m + n)} x$ , then $sec^{2} x d x = d t$
$\frac{1}{a sin ^{2} x + b cos ^{2} x}$	divide top and bottom by $cos^{2} x$ , then $t = tan x$
$\frac{1}{a + b cos x}$ , $\frac{1}{a + b sin x}$	$t = tan \frac{x}{2}$ : $cos x = \frac{1 - t ^{2}}{1 + t ^{2}}$ , $sin x = \frac{2 t}{1 + t ^{2}}$ , $d x = \frac{2 d t}{1 + t ^{2}}$
$sin x + cos x$ with $sin 2 x$	$t = sin x - cos x$ , since $sin 2 x = 1 - t^{2}$
$\frac{x ^{2} + 1}{x ^{4} + k x ^{2} + 1}$	divide by $x^{2}$ ; $t = x - \frac{1}{x}$ (for $x^{2} - 1$ on top: $t = x + \frac{1}{x}$ )
$x^{1/ p}$ and $x^{1/ q}$ together	$x = t^{m}$ with $m = lcm (p, q)$

3.2 Linear over quadratic

Split the numerator

\int \frac{p x + q}{a x ^{2} + b x + c} d x : p x + q = A (2 a x + b) + B

Compare coefficients: $A = \frac{p}{2 a}$ , $B = q - \frac{p b}{2 a}$ . The $A$ part gives $A ln ∣ a x^{2} + b x + c ∣$ , the $B$ part a form from section 2 after completing the square. The same split works for $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ : the $A$ part gives $2 A a x^{2} + b x + c$ , the $B$ part a square-root form from section 2.

3.3 Integration by parts

By parts

\int u v d x = u \int v d x - \int (\frac{d u}{d x} \int v d x) d x

Take $u$ in the order ILATE: Inverse trig, Logarithm, Algebraic, Trigonometric, Exponential. So $\int tan^{- 1} x d x = x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2}) + C$ (with $v = 1$ ).

The

e^{x}

pattern

\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C

Spot $f$ and $f^{'}$ side by side inside the bracket. For $e^{k x}$ : $\int e^{k x} [k f (x) + f^{'} (x)] d x = e^{k x} f (x) + C$ .

e^{a x}

with sine and cosine

\int e^{a x} sin b x d x = \frac{e ^{a x} ( a sin b x - b cos b x )}{a ^{2} + b ^{2}} + C

and $\int e^{a x} cos b x d x = \frac{e ^{a x} ( a cos b x + b sin b x )}{a ^{2} + b ^{2}} + C$ .

3.4 Partial fractions

For a proper fraction (degree of the numerator less than that of the denominator; divide first otherwise):

Fraction	Write it as	When
$\frac{p x + q}{( x - a ) ( x - b )}$	$\frac{A}{x - a} + \frac{B}{x - b}$	$a \neq = b$
$\frac{p x + q}{( x - a ) ^{2}}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$
$\frac{p x ^{2} + q x + r}{( x - a ) ( x - b ) ( x - c )}$	$\frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}$	$a$ , $b$ , $c$ all different
$\frac{p x ^{2} + q x + r}{( x - a ) ^{2} ( x - b )}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$	$a \neq = b$
$\frac{p x ^{2} + q x + r}{( x - a ) ( x ^{2} + b x + c )}$	$\frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$	$x^{2} + b x + c$ has no real roots

4. Definite integrals

Newton-Leibniz formula

\int_{a}^{b} f (x) d x = F (b) - F (a)

$F$ is any antiderivative of $f$ , and $f$ is continuous on $[a, b]$ . No $+ C$ in a definite integral.

Property	Formula	Use it for
Reverse the limits	$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$	Swapping the limits
Split the range	$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$	$[x]$ , $∣ x ∣$ , ${x}$ , max/min and piecewise functions; break where the formula changes
King's property	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$	Write the integral twice, once with $x$ replaced by $a + b - x$ , and add
Zero lower limit	$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$	The same idea on $[0, a]$
Symmetric limits	$\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x$	$= 2 \int_{0}^{a} f (x) d x$ if $f$ is even, $0$ if $f$ is odd
Doubled range	$\int_{0}^{2 a} f (x) d x = \int_{0}^{a} [f (x) + f (2 a - x)] d x$	$= 2 \int_{0}^{a} f (x) d x$ if $f (2 a - x) = f (x)$ , $0$ if $f (2 a - x) = - f (x)$
Periodic functions	$\int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x$ and $\int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x$	$T$ is a period of $f$ and $n$ a positive integer

4.1 Patterns that the properties produce

Write it twice: $\int_{a}^{b} \frac{g ( x )}{g ( x ) + g ( a + b - x )} d x = \frac{b - a}{2}$ when the denominator is never zero. Favourite JEE cases: $g = 3 cos 2 x$ on $[\frac{π}{24}, \frac{5 π}{24}]$ , which gives the integrand $\frac{1}{1 + 3 tan 2 x}$ (PYQ 1 below), and $g = sin^{3/2} x$ on $[0, \frac{π}{2}]$ .
$x$ times a function of $sin x$ : $\int_{0}^{π} x f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$ .
Symmetric limits with $1 + b^{x}$ : $\int_{- a}^{a} \frac{g ( x )}{1 + b ^{x}} d x = \int_{0}^{a} g (x) d x$ for even $g$ and $b > 0$ , because $\frac{1}{1 + b ^{x}} + \frac{1}{1 + b ^{- x}} = 1$ .
Greatest integer and fractional part: $\int_{0}^{n} [x] d x = \frac{n ( n - 1 )}{2}$ and $\int_{0}^{n} {x} d x = \frac{n}{2}$ for a positive integer $n$ . For $[x^{2}]$ or $[2 sin x]$ , break where the inside crosses an integer.
Wallis: $\int_{0}^{π /2} sin^{n} x d x = \int_{0}^{π /2} cos^{n} x d x = \frac{( n - 1 ) ( n - 3 ) \dots}{n ( n - 2 ) \dots} \times k$ for an integer $n \geq 2$ , with $k = \frac{π}{2}$ for even $n$ and $k = 1$ for odd $n$ ; each product stops at $2$ or $1$ .
A log result: $\int_{0}^{π /2} ln sin x d x = \int_{0}^{π /2} ln cos x d x = - \frac{π}{2} ln 2$ .
A function and its inverse: for continuous $f$ , strictly increasing or strictly decreasing, $\int_{a}^{b} f (x) d x + \int_{f (a)}^{f (b)} f^{- 1} (y) d y = b f (b) - a f (a)$ . JEE pairs $lo g_{2} (x^{2} + 4)$ with $2^{x} - 4$ , or $lo g_{2} (x^{3} + 1)$ with $(2^{x} - 1)^{1/3}$ .

5. Differentiating an integral (Leibniz rule)

Leibniz rule

\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)

$f$ continuous, $g$ and $h$ differentiable. With a constant lower limit: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ .

An integral equation: when $\int_{0}^{x}$ of the unknown function appears, differentiate both sides; you get a differential equation. Put $x$ equal to the lower limit to find the constant.
A definite integral is a number: if $f (x) = g (x) + \int_{0}^{1} k (t) f (t) d t$ , the integral is a constant. Call it $λ$ , write $f (x) = g (x) + λ$ and substitute back.
Maxima and minima: for $F (x) = \int_{0}^{h (x)} f (t) d t$ , the critical points are the zeros of $f (h (x)) h^{'} (x)$ .

6. Definite integral as the limit of a sum

Limit of a sum

n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x

The same with $r$ from $0$ to $n - 1$ . If $r$ runs from $α n$ to $β n$ , the limits become $α$ and $β$ .

Take out $\frac{1}{n}$ so that the rest is a function of $\frac{r}{n}$ only.
Replace $\frac{r}{n}$ by $x$ and $\frac{1}{n}$ by $d x$ ; the sum becomes $\int$ .
Limits: the smallest and largest values of $\frac{r}{n}$ as $n \to \infty$ .

7. Exam traps

Watch out

$\int \frac{d x}{x} = ln ∣ x ∣$ , not $ln x$ : keep the modulus in $ln ∣ sec x ∣$ , $ln ∣ f (x) ∣$ and the other log results.
$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ fails for $n = - 1$ .
King's property changes $x$ to $a + b - x$ in the integrand only; the limits stay. It is for definite integrals.
Test even or odd on the whole integrand. $\frac{x ^{2}}{1 + e ^{x}}$ is neither; use $f (x) + f (- x)$ .
For $[x^{2}]$ , $[2 sin x]$ or $∣ x^{2} + 2 x - 3∣$ , split at every point where the inside crosses an integer or zero, not only at whole-number $x$ .
Leibniz rule: multiply by the derivative of the limit. $\frac{d}{d x} \int_{0}^{x^{2}} f (t) d t = 2 x f (x^{2})$ .
After a substitution in a definite integral, change the limits, or go back to $x$ before putting them in.
An indefinite integral with a condition such as $f (0) = 0$ : find $C$ first, then evaluate.

8. Past-paper questions, solved

Ten questions from our past papers, chosen to cover the formulas used most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: King's property

JEE Main 2026 Jan 23 Shift 1, Mathematics Q10

The value of the integral $\int_{π /24}^{5 π /24} \frac{d x}{1 + 3 tan 2 x}$ is :

(A) $\frac{π}{12}$
(B) $\frac{π}{18}$
(C) $\frac{π}{6}$
(D) $\frac{π}{3}$

Solution:

Here $a + b = \frac{π}{24} + \frac{5 π}{24} = \frac{π}{4}$ . Replace $x$ by $\frac{π}{4} - x$ ; since $tan (\frac{π}{2} - 2 x) = cot 2 x$ ,
$I = \int_{π /24}^{5 π /24} \frac{3 tan 2 x}{3 tan 2 x + 1} d x$
Add this to the first form: the two integrands add to $1$ , so
$2 I = \frac{5 π}{24} - \frac{π}{24} = \frac{π}{6}$
Hence $I = \frac{π}{12}$ .

Answer key (A) $\frac{π}{12}$

PYQ 2: Substitution

x \to \frac{1}{x}

, then an inverse-trig form

JEE Advanced 2025 Paper 2, Mathematics Section 3 Q8

$α = \int_{1/2}^{2} \frac{tan ^{- 1} x}{2 x ^{2} - 3 x + 2} d x,$

then the value of $7 tan (\frac{2 α 7}{π})$ is ________.

(Here, the inverse trigonometric function $tan^{- 1} x$ assumes values in $(- \frac{π}{2}, \frac{π}{2})$ .)

Solution:

Put $x = \frac{1}{t}$ . Then $d x = - \frac{d t}{t ^{2}}$ and $2 x^{2} - 3 x + 2 = \frac{2 t ^{2} - 3 t + 2}{t ^{2}}$ , the limits $\frac{1}{2}$ and $2$ swap, and $tan^{- 1} \frac{1}{t} = cot^{- 1} t$ for $t > 0$ :
$α = \int_{1/2}^{2} \frac{cot ^{- 1} t}{2 t ^{2} - 3 t + 2} d t$
Add the two forms and use $tan^{- 1} x + cot^{- 1} x = \frac{π}{2}$ :
$2 α = \frac{π}{2} \int_{1/2}^{2} \frac{d x}{2 x ^{2} - 3 x + 2}$
Complete the square, $2 x^{2} - 3 x + 2 = 2 [(x - \frac{3}{4})^{2} + \frac{7}{16}]$ :
$\int_{1/2}^{2} \frac{d x}{2 x ^{2} - 3 x + 2} = \frac{2}{7} [tan^{- 1} \frac{4 x - 3}{7}]_{1/2}^{2}$
That is $\frac{2}{7} (tan^{- 1} \frac{5}{7} + tan^{- 1} \frac{1}{7})$ , and as $\frac{5}{7} \cdot \frac{1}{7} < 1$ the sum is $tan^{- 1} (37)$ . So
$α = \frac{π}{2 7} tan^{- 1} (37)$
Then $\frac{2 α 7}{π} = tan^{- 1} (37)$ and the required value is $7 \cdot 37 = 21$ .

Answer key 21

PYQ 3: Substitution

t = sin θ - cos θ

JEE Main 2025 Jan 29 Shift 1, Mathematics Q16

The integral $80 \int_{0}^{π /4} \frac{sin θ + cos θ}{9 + 16 sin 2 θ} d θ$ is equal to :

(A) $3 lo g_{e} 4$
(B) $6 lo g_{e} 4$
(C) $4 lo g_{e} 3$
(D) $2 lo g_{e} 3$

Solution:

Use $sin 2 θ = 1 - (sin θ - cos θ)^{2}$ and put $t = sin θ - cos θ$ : then $d t = (cos θ + sin θ) d θ$ , and $θ$ from $0$ to $\frac{π}{4}$ takes $t$ from $- 1$ to $0$ .
The denominator becomes $9 + 16 (1 - t^{2}) = 25 - 16 t^{2}$ :
$I = 80 \int_{- 1}^{0} \frac{d t}{25 - 16 t ^{2}} = 5 \int_{- 1}^{0} \frac{d t}{( 5/4 ) ^{2} - t ^{2}}$
With $\int \frac{d t}{a ^{2} - t ^{2}} = \frac{1}{2 a} ln \frac{a + t}{a - t}$ and $a = \frac{5}{4}$ :
$I = 2 [ln \frac{5 + 4 t}{5 - 4 t}]_{- 1}^{0} = 2 ln 9$
So $I = 4 ln 3$ .

Answer key (C) $4 lo g_{e} 3$

PYQ 4: Leibniz rule

JEE Main 2025 Jan 22 Shift 2, Mathematics Q8

Let $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}} d t$ , $x \in R$ . Then the numbers of local maximum and local minimum points of $f$ , respectively, are :

(A) 2 and 3
(B) 3 and 2
(C) 1 and 3
(D) 2 and 2

Solution:

By the Leibniz rule, with upper limit $x^{2}$ :
$f^{'} (x) = \frac{x ^{4} - 8 x ^{2} + 15}{e ^{x^{2}}} \cdot 2 x = \frac{2 x ( x ^{2} - 3 ) ( x ^{2} - 5 )}{e ^{x^{2}}}$
$f^{'} (x) = 0$ at $x = - 5, - 3, 0, 3, 5$ . As $e^{x^{2}} > 0$ , the sign of $f^{'}$ is that of $2 x (x^{2} - 3) (x^{2} - 5)$ : from left to right $-, +, -, +, -, +$ .
A change from $+$ to $-$ is a local maximum ( $x = \pm 3$ ); from $-$ to $+$ a local minimum ( $x = - 5$ , $0$ , $5$ ). So $2$ maxima and $3$ minima.

Answer key (A) 2 and 3

PYQ 5: Splitting the range

JEE Main 2026 Jan 28 Shift 2, Mathematics Q13

Let $[\cdot]$ denote the greatest integer function. Then $\int_{- π /2}^{π /2} (\frac{12 ( 3 + [ x ])}{3 + [ sin x ] + [ cos x ]}) d x$ is equal to :

(A) $15 π + 4$
(B) $11 π + 2$
(C) $13 π + 1$
(D) $12 π + 5$

Solution:

Break $[- \frac{π}{2}, \frac{π}{2}]$ where $[x]$ , $[sin x]$ or $[cos x]$ changes: at $x = - 1, 0, 1$ . (At single points such as $x = 0$ , where $cos x = 1$ , the value does not change the integral.)
On $[- \frac{π}{2}, - 1)$ : $[x] = - 2$ , $[sin x] = - 1$ , $[cos x] = 0$ , so the integrand is $\frac{12 \cdot 1}{2} = 6$ . On $[- 1, 0)$ : $[x] = - 1$ , $[sin x] = - 1$ , integrand $\frac{12 \cdot 2}{2} = 12$ .
On $(0, 1)$ : $[x] = [sin x] = [cos x] = 0$ , integrand $12$ . On $[1, \frac{π}{2})$ : $[x] = 1$ , integrand $\frac{12 \cdot 4}{3} = 16$ .
Add length times value:
$6 (\frac{π}{2} - 1) + 24 + 16 (\frac{π}{2} - 1) = 11 π + 2$

Answer key (B) $11 π + 2$

PYQ 6: By parts for

tan^{- 1} x

JEE Main 2026 Apr 4 Shift 2, Mathematics Q20

The integral $\int_{0}^{1} cot^{- 1} (1 + x + x^{2}) d x$ is equal to:

(A) $2 tan^{- 1} 2 + \frac{1}{2} lo g_{e} (\frac{5}{4}) + \frac{π}{2}$
(B) $2 tan^{- 1} 2 + \frac{1}{2} lo g_{e} (\frac{5}{4}) - \frac{π}{2}$
(C) $2 tan^{- 1} 2 - \frac{1}{2} lo g_{e} (\frac{5}{4}) + \frac{π}{2}$
(D) $2 tan^{- 1} 2 - \frac{1}{2} lo g_{e} (\frac{5}{4}) - \frac{π}{2}$

Solution:

For $x \geq 0$ :
$cot^{- 1} (1 + x + x^{2}) = tan^{- 1} \frac{( x + 1 ) - x}{1 + x ( x + 1 )} = tan^{- 1} (x + 1) - tan^{- 1} x$
By parts with $v = 1$ : $\int tan^{- 1} u d u = u tan^{- 1} u - \frac{1}{2} ln (1 + u^{2})$ . Hence
$\int_{0}^{1} tan^{- 1} (x + 1) d x = \int_{1}^{2} tan^{- 1} u d u = 2 tan^{- 1} 2 - \frac{π}{4} - \frac{1}{2} ln \frac{5}{2}$
and $\int_{0}^{1} tan^{- 1} x d x = \frac{π}{4} - \frac{1}{2} ln 2$ .
Subtract:
$I = 2 tan^{- 1} 2 - \frac{π}{2} - \frac{1}{2} ln \frac{5}{4}$

Answer key (D) $2 tan^{- 1} 2 - \frac{1}{2} lo g_{e} (\frac{5}{4}) - \frac{π}{2}$

PYQ 7: Symmetric limits

JEE Main 2025 Apr 2 Shift 2, Mathematics Q7

Let $(a, b)$ be the point of intersection of the curve $x^{2} = 2 y$ and the straight line $y - 2 x - 6 = 0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x ^{2}}{1 + 5 ^{x}} d x$ is equal to :

(A) $24$
(B) $27$
(C) $18$
(D) $21$

Solution:

$x^{2} = 2 y$ and $y = 2 x + 6$ give $x^{2} - 4 x - 12 = 0$ , so $x = 6$ or $x = - 2$ . The point in the second quadrant is $(- 2, 2)$ : $a = - 2$ , $b = 2$ .
With $f (x) = \frac{9 x ^{2}}{1 + 5 ^{x}}$ :
$f (x) + f (- x) = 9 x^{2} (\frac{1}{1 + 5 ^{x}} + \frac{5 ^{x}}{5 ^{x} + 1}) = 9 x^{2}$
So $I = \int_{0}^{2} [f (x) + f (- x)] d x = \int_{0}^{2} 9 x^{2} d x = 24$ .

Answer key (A) $24$

PYQ 8: The

e^{x} [f + f^{'}]

form

JEE Main 2026 Jan 23 Shift 1, Mathematics Q2

Let $f (x) = \int \frac{( 2 - x ^{2} ) e ^{x}}{( 1 + x ) ( 1 - x ) ^{3/2}} d x$ . If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to :

(A) $3 e - 1$
(B) $2 e + 1$
(C) $2 e - 1$
(D) $3 e + 1$

Solution:

Take $g (x) = \frac{1 + x}{1 - x}$ . Writing both over the same denominator:
$g (x) g^{'} (x) = \frac{1 - x ^{2}}{1 + x ( 1 - x ) ^{3/2}} = \frac{1}{1 + x ( 1 - x ) ^{3/2}}$
So $g + g^{'} = \frac{2 - x ^{2}}{1 + x ( 1 - x ) ^{3/2}}$ , and the integrand is $e^{x} [g (x) + g^{'} (x)]$ .
Hence $f (x) = e^{x} g (x) + C$ . From $f (0) = 0$ , $C = - 1$ , and
$f (\frac{1}{2}) = e \cdot 3 - 1 = 3 e - 1$

Answer key (A) $3 e - 1$

PYQ 9: Periodic functions

JEE Main 2026 Apr 2 Shift 2, Mathematics Q17

The value of $\int_{0}^{20 π} (sin^{4} x + cos^{4} x) d x$ is equal to:

(A) $\frac{15 π}{2}$
(B) $25 π$
(C) $15 π$
(D) $\frac{25 π}{2}$

Solution:

Rewrite the integrand:
$sin^{4} x + cos^{4} x = 1 - \frac{sin ^{2} 2 x}{2} = \frac{3}{4} + \frac{cos 4 x}{4}$
It has period $\frac{π}{2}$ .
$[0, 20 π]$ holds $40$ periods:
$I = 40 \int_{0}^{π /2} (\frac{3}{4} + \frac{cos 4 x}{4}) d x = 40 \cdot \frac{3 π}{8} = 15 π$

Answer key (C) $15 π$

PYQ 10: Limit of a sum

JEE Advanced 2022 Paper 2, Mathematics Section 3 Q4

For positive integer $n$ , define

$f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$ .

Then, the value of $n \to \infty lim f (n)$ is equal to

(A) $3 + \frac{4}{3} lo g_{e} 7$
(B) $4 - \frac{3}{4} lo g_{e} (\frac{7}{3})$
(C) $4 - \frac{4}{3} lo g_{e} (\frac{7}{3})$
(D) $3 + \frac{3}{4} lo g_{e} 7$

Solution:

The $r$ th fraction is $\frac{16 r + ( 9 - 4 r ) n - 3 n ^{2}}{4 r n + 3 n ^{2}}$ for $r = 1, 2, \dots, n$ ; the last one, $r = n$ , is $\frac{25 n - 7 n ^{2}}{7 n ^{2}}$ .
Share the leading $n$ out as $1$ per fraction ( $n$ fractions):
$f (n) = r = 1 \sum n \frac{16 r + 9 n}{4 r n + 3 n ^{2}} = \frac{1}{n} r = 1 \sum n \frac{16 ( r / n ) + 9}{4 ( r / n ) + 3}$
So the limit is
$\int_{0}^{1} \frac{16 x + 9}{4 x + 3} d x = \int_{0}^{1} (4 - \frac{3}{4 x + 3}) d x = 4 - \frac{3}{4} ln \frac{7}{3}$

Answer key (B) $4 - \frac{3}{4} lo g_{e} (\frac{7}{3})$

Every formula on this page is checked with a computer algebra system, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026; the questions were read by hand (lists checked 2026-10-15).

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