Please list the formulae required in Integration in Physics for Class-11 (IIT-JEE)

Hint: We know that integration is the reverse operation to differentiation i.e. it is the process of getting from the derivative start fraction, d, g, left bracket, x, right bracket, divided by, d, x, end fraction, equals, g, prime, left bracket, x, right bracket, dxdg(x)=g′(x) to the function g, left bracket, x, right bracket,g(x). So, one possible use of integration is to find distance using velocity, or finding velocity using acceleration. If a function of one of these components over time is known, then integration is the fastest method to apply.

Complete step by step answer
The main indefinite and definite integration formulas as well as some main properties of integration. In general, integration is the reverse operation of differentiation. It is also called antiderivative. The formulas provided here will help students to easily remember them for the exam and score higher marks in the exams.
Indefinite Integration:
1. If $f$ and $g$ are functions of $x$ such that $g^{\prime}(x)=f(x)$ then
$\int f(x) d x=g(x)+c \Leftrightarrow \dfrac{d}{d x}\{g(x)+c\}=f(x),$ where $c$ is called the
constant of integration.
2. Standard formula:
a) $\int x n d x=\dfrac{x^{n+1}}{n+1}+c, n \neq-1$
b) $\int \dfrac{1}{x} d x=\log _{e}|x|+c$
(c) $\int e^{x} d x=e^{x}+c$
(d) $\int \mathrm{a}^{\mathrm{x}} \mathrm{dx}=\dfrac{a^{x}}{\log _{e} a}+c$
(e) $\int \sin x d x=-\cos x+c$
(f) $\int \cos x d x=\sin x+c$
(g) $\int \sec ^{2} x d x=\tan x+c$
(h) $\int \operatorname{cosec}^{2} x d x=-\cot x+c$
(h) $\int \sec x \tan x d x=\sec x+c$
(i) $\int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+c$
(j) $\int \cot x d x=\log |\sin x|+c$
$(\mathrm{k}) \int \tan \mathrm{x} \mathrm{d} \mathrm{x}=-\log |\cos x|+c$
(I) $\int \sec x d x=\log |\sec x+\tan x|+c$
$(\mathrm{m}) \int \operatorname{cosec} \mathrm{x} \mathrm{dx}=\log |\operatorname{cosec} x-\cot x|+c$
(n) $\int \dfrac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\dfrac{x}{a}\right)+c$
(o) $\int-\dfrac{1}{\sqrt{a^{2}-x^{2}}} d x=\cos ^{-1}\left(\dfrac{x}{a}\right)+c$
(p) $\int \dfrac{1}{a^{2}+x^{2}} d x=\dfrac{1}{a} \tan ^{-1}\left(\dfrac{x}{a}\right)+c$
(q)$\int-\dfrac{1}{a^{2}+x^{2}} d x=\dfrac{1}{a} \cot ^{-1}\left(\dfrac{x}{a}\right)+c$
(r) $\int \dfrac{1}{x \sqrt{x^{2}-a^{2}}} d x=\dfrac{1}{a} \sec ^{-1}\left(\dfrac{x}{a}\right)+c$
(s) $\int-\dfrac{1}{x \sqrt{x^{2}-a^{2}}} d x=\dfrac{1}{a} \operatorname{cosec}^{-1}\left(\dfrac{x}{a}\right)+c$

Note We know that integrals can be used for computing the area of a two-dimensional region that has a curved boundary, as well as computing the volume of a three-dimensional object that has a curved boundary. The area of a two-dimensional region can be calculated using the aforementioned definite integral. There are two forms of the integrals. Indefinite Integrals: It is an integral of a function when there is no limit for integration. It contains an arbitrary constant. Definite Integrals: An integral of a function with limits of integration.