# Integration as an Inverse Process of Differentiation

#### notes

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 intsin 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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Integrals part 3 (Integration as process of differentiation) [00:04:16]
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