## Basic concepts and Standard Formulas of Indefinite Integration

Contents

- It is a process of finding the limit of a sum of a certain number of elements, as the number of elements tends to infinity when each of the elements tends to zero.
- Integration is a reverse process of differentiation.

If {d (F(x) + c)/dx} = f(x), then

Integration of f(x).dx = F(x) + C ; Here ‘C’ is constant of integration

F(x) is Integral of f(x) or Anti-Derivative of f(x) or Primitive of f(x)

- In differentiation dy/dx is used as a composite symbol or as an operator. In ‘Methods of Differentiation’ these do not have separate meanings.
- But in Integration, it is treated as a fraction where ‘dy’ is called differential of ‘y’ and ‘dx’ is called differential of ‘x’.
- For a single variable function y=f(x), differentiation of ‘y’ is denoted as ‘dy’

i.e., dy = Product of derivative of F(x) w.r.t ‘x’ and differential of x is denoted by ‘dx’.

Ex- y = cos x then dy = -sin x dx

### GEOMETRICAL INTERPRETATION OF INDEFINITE INTEGRATION

y = ∫ f(x) dx denotes a family of curves such that derivative at x= x_{1} of every member of the family is same i.e. tangent

drawn to each member of the family at x = x_{1 }is parallel.

Ex- y = ∫ 2x dx = x^{2} + C

**NOTE-**

- Anti-derivative of a continuous function is always differentiable.
- Anti-derivative of a periodic function needn’t be a periodic function.

### Integral of Standard Functions

∫ (x^{n})dx = (x^{n+1})/(n+1) + C ; n ≠ -1

∫ (ax+b)^{n} dx = (ax+b)^{n+1})/a(n+1) + C ; n ≠ -1

∫1/x dx = ln |x| + C

∫ 1/(ax+b) dx = 1/a ln |ax+b| +C

∫ a^{x} dx = a^{x}/ln a + C ; a>0

∫ a^{px+q} dx = a^{px+q}/p ln a + C ; a>0

∫ e^{x} dx = e^{x} + C

∫ e^{px+q} dx = (e^{px+q})/a + C

**TRIGONOMETRIC INTEGRALS**

∫ sin x dx = – cos x +C

∫ sin (ax+b )dx = – 1/a cos (ax+b) + C

∫ cos x dx = sin x +C

∫ cos (ax+b) dx = 1/a sin (ax+b) + C

∫ sec^{2} x dx = tan x +C

∫ sec^{2} (ax+b) dx = 1/a tan (ax+b) + C

∫ cosec^{2} x dx = – cot x + C

∫ cosec^{2} (ax+b) dx = – 1/a cot (ax+b) + C

∫sec x tan x dx = sec x + C

∫sec (ax+b) tan (ax+b) dx = 1/a sec (ax+b) + C

∫ cosec x cot x dx = – cosec x + C

∫ cosec (ax+b) cot (ax+b) dx = – 1/a cosec (ax+b) + C

#### INVERSE TRIGONOMETRIC INTEGRALS

∫dx/ (1+x^{2}) dx = tan^{-1} x + C

∫dx/ (a^{2 }+ x^{2}) dx = 1/a tan^{-1} x + C

∫dx/ (1- x^{2})^{1/2} dx = sin^{-1} x + C

∫dx/ (a^{2} – x^{2})^{1/2} dx = 1/a sin^{-1} (x/a) + C

∫dx/ x( x^{2} – 1)^{1/2} dx = sec^{-1} x + C

∫dx/ x( x^{2} – a^{2})^{1/2} dx = 1/a sec^{-1} (x/a) + C

If there is any typo error, please do let us know.