Antiderivative

Anti derivatives  are an integral part of calculus, its a reverse process of differentiation. Here in
this topic we are given a differential result and then that is evaluated as anti derivative result which is nothing but an integral result.

Antiderivative of a Function

A function $p(x)$ is called a primitive or an anti derivative of a
function $f(x)$ if $p’(x) = f(x)$.

Let p(x) be an anti derivative of a function $f(x)$ and Let C be any constant. Then
Derivative of $[p(x) + C] = p’(x) = f(x)$.
 
This means $p(x) + C$ is also an anti derivative of $f(x)$.

Thus, if a function $f(x)$ possesses an anti derivative, then it possess infinitely many anti derivatives which are contained in the expression $p(x) + C$, where C is a constant.

Antiderivative Rules


1. Anti derivative of $(x^{n})$

Anti derivative of $(x^{n})$ is a function whose derivative is $f(x) = x^{n}$. Since the differentiation of a power reduces the power by one, finding the opposite means the exponent increases by 1.

Therefore let us guess the anti derivative would be $f(x) = x^{n + 1}$ - - - - - - - - - - - - -  (i)

Let us see the derivative of $( x^{n + 1} )$ is equal to $[\frac {1} {(n+1)} ]$ * $x^{n}$ This has the same power but a wrong coefficient

Therefore, we conclude that

$\int x^{n} dx$ = $\frac {1}{n+1}$ $x^{n+1}$

Anti derivative of $(x^{n})$ is equal to $[\frac {1}{(n+1)}]$ $\times x^{n+1}$

2. Anti derivative of $\log (x)$

Integration of $\log (x)$ is derived using the Integration by parts method

$f(x)$ = $\int \log (x) dx$

Let us assume $\log (x)$ = u $\Rightarrow$ du = $(\frac {1}{x})$ dx

         $\int  \log ( x ) dx$ = $\int   \log (x) * 1\ dx$

      = $ \log (x) * (x)$  -  $\int (\frac {1}{x}). x\ dx$

      = $ \log (x) * (x) - \int 1\ dx$

      = $ \log(x).x - x + C$

      = $x \log (x) - x + C$.
Therefore the Anti derivative of log (x) is equal to x log (x) – x + C  

3. Anti derivative of  $\int$ $(\frac {1} {x})$

$(\frac {1} {x})$ is equal to x to the power -1,

We know that the derivative of $\log (x)$ is equal to $(\frac {1} {x})$, this function is defined only for positive values of ‘x’. so, if ‘x’ is negative then ‘- x’ is positive.
Therefore Anti derivative of $(\frac {1} {x})$ is equal to $\log | x | + C$

4. Anti derivative of  $e^{x}$

The distinguishing feature of the exponential function is that it is its own derivative. But then it is also its own Anti derivative.
Therefore the Anti derivative of  $\int e^{x} = e^{x}$

Antiderivative of Sin x


We know that the derivative of $\cos (x)$ is $\sin (x)$
Therefore, the Anti Derivative of $\sin (x)$ is equal to $– \cos (x) + C$

Antiderivative of Cos x


We know that the derivative of $\sin (x)$ is $\cos (x)$
Therefore, the Anti derivative of $\cos (x)$ is equal to $\sin (x) + C$

Antiderivative of Tan x

Let us assume $f( x ) =  \tan (x) dx$. Then,

                     $f (x)$ =  $\frac {(\sin x)} {(\cos x)} dx$.

Let $\cos x = t$. Then

    $\Rightarrow$     $d (\cos x) = dt$

    $\Rightarrow$   $ - \sin x = dt $

    $\Rightarrow$   $dx$ = -$\frac {dt} {\sin x}$

Putting $\cos x = t$, and dx = $\frac {- dt} {\sin x}$, we get

$f(x)$ = $\int$ $\frac {\sin x}{\cos x}$ $\times$ $\frac {-dt}{\sin x}$

$f( x )$ = -  $\int$ $(\frac {1} { t})$ $dt$

$f( x )$ = -$\log | t | + C$

$f( x )$ = - $\log | \cos x | + C$
Therefore, the Anti derivative of $\tan ( x )$ is equal to $- \log | \cos x | + C$

Antiderivative of Cot x

Let us assume that $f(x)$ =  $\cot (x) dx$. Then,

                     $f(x)$ =  $\frac {(\cos x)} {(sin x)}$ $dx$.

Let $\sin x = t$. Then

$\Rightarrow$ $d (\sin x) = dt$

$\Rightarrow$ $\cos x = dt $

$\Rightarrow$ dx =  $\frac {dt} {\cos x}$

Putting $\sin x = t$, and dx = $\frac {dt} {\cos x}$, we get

$f(x)$ = $\int$ $\frac {\cos x} {\sin x}$ $\times$ $\frac {dt} {\cos x}$

$f(x)$ = $\int$ $(\frac {1} {t})$ $dt$

$f(x) = \log | t | + C$

$f(x) = \log | \sin x | + C$
Therefore, the Anti derivative of $\cot (x)$ is equal to $\log | \sin x | + C$

Antiderivative of Sec x

Let us assume that f(x) =  $\sec$ (x) dx. Then,

Multiplying and dividing with $\sec (x) + \tan (x)$ we get

      f(x) =$ \int \sec (x) \times$ $\frac {\sec (x) + \tan (x) } {\sec (x) + \tan (x) }$ $dx$

Let $\sec (x) + \tan (x) = t$. Then

$\Rightarrow$      $d (\sec x + \tan x) = dt$

$\Rightarrow$      $(\sec x. \tan x + \sec^{2}x ) dx$ = dt

$\Rightarrow$     $dx$ =  $\frac {dt} {(\sec x. \tan x + \sec^{2}x )}$

Putting $\sec x + \tan x = t$, and dx =  $\frac {dt} {(\sec x. \tan x + \sec^{2}x )}$, we get

f(x) = $\int$ $\frac {\sec x (\sec (x) + \tan (x))} {t}$ $\times$ $\frac {dt} {\sec x (\sec (x) + \tan (x))}$

f(x) =  $ (\frac {1} {t})$ $dt$

f(x) = $\log | t | + C$

f(x) =$ \log | \sec x + \tan x | + C$
Therefore, the Anti derivative of $\sec (x)$ is equal to $\log | \sec x + \tan x | + C$

Anti Derivative of Csc x

Let us assume f( x ) =  $\csc (x) dx$. Then,

Multiplying and dividing with $\csc (x) - \cot (x)$ we get

                          f( x ) = dx

Let $\csc (x) - \cot (x)$ = t. Then

     $\Rightarrow$  $d (\csc (x) - \cot (x))$ = dt

     $\Rightarrow$  $(- \csc x. \cot x + \csc^{2}x ) dx$ = dt
 
     $\Rightarrow$  dx =  $\frac {dt} {(- \csc x. \cot x + \csc^{2}x)}$

Putting $\csc (x) - \cot (x)$ = t, and dx = $\frac {dt} {(-\csc x. \cot x + \csc^{2}x)}$, we get

f(x) = $\int$ $\frac {\csc x (\csc (x) - \cot (x))} {t}$ $\times$ $\frac {dt} {\csc x (\csc (x) - \cot (x))}$

f(x) =   $(\frac {1} {t})$ dt

f(x) = $\log | t | + C$

f(x) = $\log | \csc x - \cot x | + C$
Therefore, the Anti derivative of $\csc (x)$ is equal to $\log | \csc x - \cot x | + C$