Integration Formulas For JEE 2026
Integrals (Indefinite & Definite) is an important topic in Mathematics and carries good weightage, with around 2–3 questions asked every year. This chapter mainly focuses on solving integrals using different methods and understanding how functions behave over a given interval. Students should be clear with the basics of indefinite and definite integrals, as these concepts are used in a wide range of problems.
To perform well, it’s important to practice techniques like substitution, integration by parts, partial fractions, and the properties of definite integrals. Since many questions depend on applying formulas correctly, regular practice and revision are very helpful. For quick revision before exams, students can also use a well-organized JEE Mains Mathematics Formula PDF to go through important formulas and key concepts easily.
What Is Integration
Antiderivative
A function $F(x)$ is called an antiderivative of $f(x)$ if $F'(x) = f(x)$. For example, $F(x) = x^3$ is an antiderivative of $f(x) = 3x^2$.
Indefinite Integral
The family of all antiderivatives of $f(x)$:
$$\int f(x)\,dx = F(x) + C$$where $C$ is the constant of integration (an arbitrary constant). We include $C$ because many functions share the same derivative.
Integration is the reverse of differentiation. If differentiation answers "what is the rate of change?", integration answers "given the rate of change, what is the original quantity?" Integration also computes the area under a curve.
Important: Never forget the $+C$ in indefinite integrals. It is a common mistake in exams and costs marks.
Standard Integral Formulas
Basic Integrals
| Function | Integral |
|---|---|
| $\int x^n\,dx$ $(n \neq -1)$ | $\dfrac{x^{n+1}}{n+1} + C$ |
| $\int \frac{1}{x}\,dx$ | $\ln|x| + C$ |
| $\int e^x\,dx$ | $e^x + C$ |
| $\int a^x\,dx$ $(a > 0, a \neq 1)$ | $\dfrac{a^x}{\ln a} + C$ |
Trigonometric Integrals
| Function | Integral |
|---|---|
| $\int \sin x\,dx$ | $-\cos x + C$ |
| $\int \cos x\,dx$ | $\sin x + C$ |
| $\int \sec^2 x\,dx$ | $\tan x + C$ |
| $\int \csc^2 x\,dx$ | $-\cot x + C$ |
| $\int \sec x \tan x\,dx$ | $\sec x + C$ |
| $\int \csc x \cot x\,dx$ | $-\csc x + C$ |
| $\int \tan x\,dx$ | $-\ln|\cos x| + C = \ln|\sec x| + C$ |
| $\int \cot x\,dx$ | $\ln|\sin x| + C$ |
| $\int \sec x\,dx$ | $\ln|\sec x + \tan x| + C$ |
| $\int \csc x\,dx$ | $\ln|\csc x - \cot x| + C$ |
Inverse Trigonometric Integrals
| Function | Integral |
|---|---|
| $\int \frac{1}{\sqrt{1 - x^2}}\,dx$ | $\sin^{-1} x + C$ |
| $\int \frac{1}{1 + x^2}\,dx$ | $\tan^{-1} x + C$ |
| $\int \frac{1}{x\sqrt{x^2 - 1}}\,dx$ | $\sec^{-1}|x| + C$ |
Worked Example
Evaluate $\int (3x^2 + 5x - 2)\,dx$.
$= 3 \cdot \dfrac{x^3}{3} + 5 \cdot \dfrac{x^2}{2} - 2x + C = \boldsymbol{x^3 + \dfrac{5x^2}{2} - 2x + C}$
Integration by Substitution
Substitution Method
If $\int f(g(x)) \cdot g'(x)\,dx$, let $u = g(x)$, so $du = g'(x)\,dx$. Then:
$$\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du$$After integrating in terms of $u$, substitute back $u = g(x)$.
Worked Example
Evaluate $\int 2x \cos(x^2)\,dx$.
Let $u = x^2 \Rightarrow du = 2x\,dx$.
$\int \cos(u)\,du = \sin u + C = \boldsymbol{\sin(x^2) + C}$
Worked Example
Evaluate $\int \frac{e^{\tan^{-1}x}}{1 + x^2}\,dx$.
Let $u = \tan^{-1} x \Rightarrow du = \dfrac{1}{1+x^2}\,dx$.
$\int e^u\,du = e^u + C = \boldsymbol{e^{\tan^{-1}x} + C}$
Tip: To spot a substitution, look for a function and its derivative appearing together in the integrand. The "inner function" whose derivative is present is your $u$.
Useful Substitution Patterns
| Expression in Integrand | Substitution |
|---|---|
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ |
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ |
| $a + b\cos x$ or $a + b\sin x$ | $t = \tan(x/2)$ (Weierstrass) |
Integration by Parts
Integration by Parts (IBP)
$$\int u\,dv = uv - \int v\,du$$Or equivalently, if $u$ and $v$ are functions of $x$:
$$\int u \cdot v'\,dx = u \cdot v - \int v \cdot u'\,dx$$ILATE Rule
A guideline for choosing $u$ (the function to differentiate). Pick the function that comes first in the list:
- I — Inverse trigonometric ($\sin^{-1}x$, $\tan^{-1}x$, etc.)
- L — Logarithmic ($\ln x$, $\log x$)
- A — Algebraic ($x$, $x^2$, polynomials)
- T — Trigonometric ($\sin x$, $\cos x$)
- E — Exponential ($e^x$, $a^x$)
The function appearing earlier in ILATE is chosen as $u$.
Worked Example
Evaluate $\int x \, e^x\,dx$.
By ILATE: $u = x$ (Algebraic), $dv = e^x\,dx$ (Exponential).
$du = dx$, $v = e^x$.
$\int x\,e^x\,dx = x \cdot e^x - \int e^x\,dx = xe^x - e^x + C = \boldsymbol{e^x(x - 1) + C}$
Worked Example
Evaluate $\int \ln x\,dx$.
By ILATE: $u = \ln x$ (Logarithmic), $dv = dx$.
$du = \dfrac{1}{x}\,dx$, $v = x$.
$\int \ln x\,dx = x \ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C = \boldsymbol{x\ln x - x + C}$
Worked Example
Evaluate $\int x^2 \sin x\,dx$.
$u = x^2$, $dv = \sin x\,dx$ $\Rightarrow$ $du = 2x\,dx$, $v = -\cos x$.
$= -x^2\cos x + \int 2x\cos x\,dx$
Apply IBP again: $u = 2x$, $dv = \cos x\,dx$ $\Rightarrow$ $du = 2\,dx$, $v = \sin x$.
$= -x^2\cos x + 2x\sin x - \int 2\sin x\,dx$
$= \boldsymbol{(2 - x^2)\cos x + 2x\sin x + C}$
Special IBP Formula
$$\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C$$This shortcut works when the integrand is $e^x$ times a function plus its derivative.
Worked Example
Evaluate $\int e^x(\sin x + \cos x)\,dx$.
Here $f(x) = \sin x$ and $f'(x) = \cos x$. Using the special formula:
$= \boldsymbol{e^x \sin x + C}$
Integration by Partial Fractions
Partial Fraction Decomposition
| Factor in Denominator | Partial Fraction Form |
|---|---|
| $(x - a)$ | $\dfrac{A}{x - a}$ |
| $(x - a)^2$ | $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$ |
| $(x^2 + bx + c)$ (irreducible) | $\dfrac{Ax + B}{x^2 + bx + c}$ |
The degree of the numerator must be less than the degree of the denominator. If not, perform polynomial long division first.
Worked Example
Evaluate $\int \frac{1}{(x-1)(x+2)}\,dx$.
Decompose: $\dfrac{1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$
Multiply both sides by $(x-1)(x+2)$: $1 = A(x+2) + B(x-1)$
Put $x = 1$: $1 = 3A \Rightarrow A = \frac{1}{3}$. Put $x = -2$: $1 = -3B \Rightarrow B = -\frac{1}{3}$.
$\int \frac{1}{(x-1)(x+2)}\,dx = \frac{1}{3}\ln|x-1| - \frac{1}{3}\ln|x+2| + C = \boldsymbol{\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C}$
Standard Quadratic Integral Results
| Integral | Result |
|---|---|
| $\int \frac{dx}{x^2 + a^2}$ | $\dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + C$ |
| $\int \frac{dx}{x^2 - a^2}$ | $\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C$ |
| $\int \frac{dx}{a^2 - x^2}$ | $\dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C$ |
| $\int \frac{dx}{\sqrt{a^2 - x^2}}$ | $\sin^{-1}\!\left(\dfrac{x}{a}\right) + C$ |
| $\int \frac{dx}{\sqrt{x^2 + a^2}}$ | $\ln\left|x + \sqrt{x^2 + a^2}\right| + C$ |
| $\int \frac{dx}{\sqrt{x^2 - a^2}}$ | $\ln\left|x + \sqrt{x^2 - a^2}\right| + C$ |
Worked Example
Evaluate $\int \frac{dx}{x^2 + 4x + 13}$.
Complete the square: $x^2 + 4x + 13 = (x+2)^2 + 9 = (x+2)^2 + 3^2$.
Let $u = x + 2$, $du = dx$:
$\int \frac{du}{u^2 + 3^2} = \frac{1}{3}\tan^{-1}\!\left(\frac{u}{3}\right) + C = \boldsymbol{\frac{1}{3}\tan^{-1}\!\left(\frac{x+2}{3}\right) + C}$
Worked Example
Evaluate $\int \frac{dx}{\sqrt{5 - 4x - x^2}}$.
Rewrite: $5 - 4x - x^2 = -(x^2 + 4x - 5) = -[(x+2)^2 - 9] = 9 - (x+2)^2 = 3^2 - (x+2)^2$.
Let $u = x+2$:
$\int \frac{du}{\sqrt{3^2 - u^2}} = \sin^{-1}\!\left(\frac{u}{3}\right) + C = \boldsymbol{\sin^{-1}\!\left(\frac{x+2}{3}\right) + C}$
Tip: For integrals of the form $\int \frac{dx}{ax^2 + bx + c}$ or $\int \frac{dx}{\sqrt{ax^2 + bx + c}}$: always complete the square first, then match with a standard form.
Trigonometric Integrals
Powers of Sine and Cosine
- $\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ (use $\sin^2 x = \frac{1 - \cos 2x}{2}$)
- $\int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$ (use $\cos^2 x = \frac{1 + \cos 2x}{2}$)
- For $\sin^n x$ or $\cos^n x$ with $n$ odd: save one factor for substitution.
- For $\sin^m x \cos^n x$ with $m + n$ even: use half-angle identities.
Worked Example
Evaluate $\int \sin^3 x\,dx$.
$\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x)\sin x$
Let $u = \cos x$, $du = -\sin x\,dx$:
$= -\int (1 - u^2)\,du = -u + \frac{u^3}{3} + C = \boldsymbol{\frac{\cos^3 x}{3} - \cos x + C}$
Definite Integrals
$\int_a^b f(x)\,dx = F(b) - F(a)$, where $F$ is any antiderivative of $f$. Here $a$ is the lower limit and $b$ is the upper limit. A definite integral gives a number representing the signed area between the curve and the $x$-axis.
Definite Integral as Limit of a Sum
Limit of Sum Definition
$$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{r=0}^{n-1} f(a + rh) \cdot h, \quad \text{where } h = \frac{b-a}{n}$$Useful summation formulas:
- $\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}$
- $\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$
- $\sum_{r=1}^{n} r^3 = \left[\dfrac{n(n+1)}{2}\right]^2$
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus (FTC)
Part 1: If $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$.
Part 2 (Evaluation Theorem): If $F'(x) = f(x)$, then:
$$\int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b$$Worked Example
Evaluate $\int_0^{\pi/2} \cos x\,dx$.
$\int_0^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = \textbf{1}$
Worked Example
Evaluate $\int_1^e \frac{1}{x}\,dx$.
$= \Big[\ln x\Big]_1^e = \ln e - \ln 1 = 1 - 0 = \textbf{1}$
Properties of Definite Integrals
Basic Properties
- $\int_a^a f(x)\,dx = 0$
- $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ (reversing limits changes sign)
- $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ (splitting the interval)
- $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ (the variable of integration is a "dummy variable")
King's Rule (Most Important!)
$$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$$This replaces $x$ with $(a + b - x)$. It is the single most useful property for JEE.
Worked Example
Evaluate $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$.
By King's rule (replace $x$ with $\frac{\pi}{2} - x$):
$I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}\,dx$
Add the two expressions:
$2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$
$I = \boldsymbol{\dfrac{\pi}{4}}$
Tip: King's rule is your go-to technique whenever you see $\int_0^a$ with symmetric-looking trigonometric functions. It often creates a companion integral that, when added to the original, simplifies beautifully.
Even and Odd Function Properties
If $f$ is defined on $[-a, a]$:
- Even function ($f(-x) = f(x)$): $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$
- Odd function ($f(-x) = -f(x)$): $\int_{-a}^{a} f(x)\,dx = 0$
Worked Example
Evaluate $\int_{-1}^{1} x^3\,dx$.
$f(x) = x^3$ is an odd function since $f(-x) = (-x)^3 = -x^3 = -f(x)$.
$\int_{-1}^{1} x^3\,dx = \textbf{0}$
Worked Example
Evaluate $\int_{-2}^{2} (x^4 + x^2)\,dx$.
$f(x) = x^4 + x^2$ is even (only even powers). So:
$= 2\int_0^2 (x^4 + x^2)\,dx = 2\left[\frac{x^5}{5} + \frac{x^3}{3}\right]_0^2 = 2\left(\frac{32}{5} + \frac{8}{3}\right) = 2 \cdot \frac{96 + 40}{15} = \boldsymbol{\dfrac{272}{15}}$
Property for $[0, 2a]$ Integrals
$$\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx$$Special cases:
- If $f(2a - x) = f(x)$: $\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$
- If $f(2a - x) = -f(x)$: $\int_0^{2a} f(x)\,dx = 0$
Periodic Function Property
If $f(x)$ is periodic with period $T$ (i.e., $f(x + T) = f(x)$), then:
$$\int_0^{nT} f(x)\,dx = n \int_0^T f(x)\,dx \quad \text{for any positive integer } n$$More generally: $\int_a^{a+T} f(x)\,dx = \int_0^T f(x)\,dx$ for any $a$.
Worked Example
Evaluate $\int_0^{5\pi} |\sin x|\,dx$.
$|\sin x|$ has period $\pi$.
$= 5\int_0^{\pi} |\sin x|\,dx = 5\int_0^{\pi} \sin x\,dx = 5[-\cos x]_0^{\pi} = 5(1 + 1) = \textbf{10}$
Walli's Formula
Worked Example
Evaluate $\int_0^{\pi/2} \sin^4 x\,dx$.
$n = 4$ (even). $= \dfrac{3 \cdot 1}{4 \cdot 2} \cdot \dfrac{\pi}{2} = \dfrac{3}{8} \cdot \dfrac{\pi}{2} = \boldsymbol{\dfrac{3\pi}{16}}$
Worked Example
Evaluate $\int_0^{\pi/2} \cos^5 x\,dx$.
$n = 5$ (odd). $= \dfrac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \boldsymbol{\dfrac{8}{15}}$
General Walli's Formula
$$\int_0^{\pi/2} \sin^m x \cos^n x\,dx = \frac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots} \times k$$where $k = \frac{\pi}{2}$ if both $m$ and $n$ are even, and $k = 1$ otherwise. Each product reduces by 2 until reaching 1 or 0.
Worked Example
Evaluate $\int_0^{\pi/2} \sin^2 x \cos^4 x\,dx$.
$m = 2$, $n = 4$ (both even, so $k = \pi/2$).
Numerator: $(2-1) \cdot (4-1)(4-3) = 1 \cdot 3 \cdot 1 = 3$
Denominator: $(6)(4)(2) = 48$
$= \dfrac{3}{48} \cdot \dfrac{\pi}{2} = \boldsymbol{\dfrac{\pi}{32}}$
Quick Reference
Definite Integral Properties Summary
| Property | Formula |
|---|---|
| King's rule | $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ |
| Even function | $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ |
| Odd function | $\int_{-a}^a f(x)\,dx = 0$ |
| Periodic ($T$) | $\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx$ |
| $f(2a-x) = f(x)$ | $\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ |
| $f(2a-x) = -f(x)$ | $\int_0^{2a} f(x)\,dx = 0$ |
Integration Formulas For JEE 2026: Conclusion
Integration is a fundamental topic in Mathematics that plays a crucial role in solving a wide range of problems. A clear understanding of concepts like antiderivatives, substitution, and integration by parts helps in tackling both simple and complex questions effectively.
Regular practice of standard formulas and properties of definite integrals improves speed and accuracy. Focusing on key techniques and revising them consistently can significantly enhance performance in the mathematics section of the exam.
