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Integration as an Inverse Process of Differentiation

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Integration is the inverse process of differentiation. The derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation. 
Let us consider the following examples:
`d/(dx) sin x = cos x`
we observ that ,the function cos x is the derived function of sin x and also we say that sin x is an anti derivative of cos x .
There is a function F such that
`d/(dx) F(x)` = f(x) , ∀ x ∈ I (interval), then for any arbitrary real number C, (also called constant of integration) 
`d/(dx) [F(x) + C] = = f(x), x ∈ I`    
Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f.
We already know the formulae for the integrals of these functions :

Derivatives Integrals
(Anti derivatives)
`d/(dx) (x^(n+1)/(n+1)) = x^n`   `int x^n dx = x^(n+1)/(n+1) `+C`, n ≠ –1     
`d/(dx)`(x) = 1                                          `int dx` = x + C
`d/(dx)`(sin x) = cos x `int` cos x dx = sin x +C
`d/(dx)` (-cos x) = sin x `int`sin x dx = -cos x +C
`d/(dx)` (tan x) = `sec^2x` `int sec^2 x` dx = tanx +C
`d/(dx)`(-cot x) = `cosec^2x ` `int cosec^2x` dx = -cot x +C
`d/(dx)` (sec x) = sec x tan x  `int` sec x tan x dx = sec x +C
`d/(dx)` (-cosecx) = cosec x cot x `int` cosec x cot x dx = -cosec x +C
`d/(dx) (sin^-1) = 1/(sqrt(1-x^2))` `int (dx)/(sqrt(1-x^2))= sin^(-1) x +C `
`d/(dx) (-cos^(-1)) = 1/(sqrt (1-x^2))` `int (dx)/(sqrt (1-x^2))= -cos^(-1) x + C `
`d/(dx) (tan^(-1) x) = 1/(1+x^2)` `int (dx)/(1+x^2)= tan^(-1) x + C `
 `d/(dx) (-cot^(-1) x) = 1/(1+x^2)` `int (dx)/(1+x^2)= -cot^(-1) x + C `
`d/(dx) (sec^(-1) x) = 1/(x sqrt (x^2 - 1))` `int (dx)/(x sqrt (x^2 - 1))`= `sec^(-1)` x + C
`d/(dx) (-cosec^(-1) x) = 1/(x sqrt (x^2 - 1))` `int (dx)/(x sqrt (x^2 - 1))=-cosec^(-1) x + C `
`d/(dx)(e^x) = e^x` `int e^x dx = e^x + C`
`d/(dx) (log|x|) = 1/x` `int 1/x dx = log|x| +C`
`d/(dx) ((a^x)/(log a)) = a^x` `int a^x dx = a^x/log a` +C
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