#### 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 | `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 |

#### Shaalaa.com | Integrals part 3 (Integration as process of differentiation)

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