Integration Formulas For JEE 2027, Check & Download PDF

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Oct 05, 2026

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Integration Formulas For JEE 2027, Check & Download PDF

Integration Formulas For JEE 2027

Integrals, including Indefinite and Definite Integrals, are important topics for JEE Mathematics preparation. They involve finding antiderivatives and evaluating integrals over a given interval. For JEE 2027, students should build a clear understanding of standard integral formulas, their conditions, and the differences between indefinite and definite integrals.

To prepare well, practise substitution, integration by parts, partial fractions, and properties of definite integrals. A JEE mains Formula PDF can help you revise important results quickly. Taking a Free JEE Mains mock test also helps you practise choosing the right method under timed conditions, identify mistakes, and improve your speed and accuracy.

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

FunctionIntegral
$$\int x^n\,dx$$   $$(n \neq -1)$$$$\dfrac{x^{n+1}}{n+1} + C$$
$$\int \frac{1}{x}\,dx$$$$\ln|x| + C$$
$$\int e^x\,dx$$$$e^x + C$$
$$\int a^x\,dx$$   $$(a > 0, a \neq 1)$$$$\dfrac{a^x}{\ln a} + C$$

Trigonometric Integrals

FunctionIntegral
$$\int \sin x\,dx$$$$-\cos x + C$$
$$\int \cos x\,dx$$$$\sin x + C$$
$$\int \sec^2 x\,dx$$$$\tan x + C$$
$$\int \csc^2 x\,dx$$$$-\cot x + C$$
$$\int \sec x \tan x\,dx$$$$\sec x + C$$
$$\int \csc x \cot x\,dx$$$$-\csc x + C$$
$$\int \tan x\,dx$$$$-\ln|\cos x| + C = \ln|\sec x| + C$$
$$\int \cot x\,dx$$$$\ln|\sin x| + C$$
$$\int \sec x\,dx$$$$\ln|\sec x + \tan x| + C$$
$$\int \csc x\,dx$$$$\ln|\csc x - \cot x| + C$$

Inverse Trigonometric Integrals

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

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 IntegrandSubstitution
$$\sqrt{a^2 - x^2}$$$$x = a\sin\theta$$
$$\sqrt{a^2 + x^2}$$$$x = a\tan\theta$$
$$\sqrt{x^2 - a^2}$$$$x = a\sec\theta$$
$$a + b\cos x$$ 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 DenominatorPartial 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

IntegralResult
$$\int \frac{dx}{x^2 + a^2}$$$$\dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + C$$
$$\int \frac{dx}{x^2 - a^2}$$$$\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C$$
$$\int \frac{dx}{a^2 - x^2}$$$$\dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C$$
$$\int \frac{dx}{\sqrt{a^2 - x^2}}$$$$\sin^{-1}\!\left(\dfrac{x}{a}\right) + C$$
$$\int \frac{dx}{\sqrt{x^2 + a^2}}$$$$\ln\left|x + \sqrt{x^2 + a^2}\right| + C$$
$$\int \frac{dx}{\sqrt{x^2 - a^2}}$$$$\ln\left|x + \sqrt{x^2 - a^2}\right| + C$$

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

$$\int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx = \begin{cases} \dfrac{(n-1)(n-3)\cdots 3 \cdot 1}{n(n-2)\cdots 4 \cdot 2} \cdot \dfrac{\pi}{2} & \text{if } n \text{ is even} \\[12pt] \dfrac{(n-1)(n-3)\cdots 4 \cdot 2}{n(n-2)\cdots 3 \cdot 1} & \text{if } n \text{ is odd} \end{cases}$$

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

PropertyFormula
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 2027: Conclusion

Integration is an important part of JEE Main and JEE Advanced preparation. Students should understand standard integral results, substitution, integration by parts, partial fractions, and properties of definite integrals. Using a JEE Maths Formula Sheet for regular revision helps them recall formulas and recognise when to apply each method.

Solving JEE Integration PYQ problems helps students understand how these concepts have been tested. Practising JEE Indefinite Integration Questions also strengthens their ability to choose suitable methods and avoid calculation errors. Consistent practice, reviewing mistakes, and revising formulas can improve accuracy and build confidence for JEE 2027.

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