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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.

this topic we are given a differential result and then that is evaluated as anti derivative result which is nothing but an integral result.

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.

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 $[\frac {1}{(n+1)}]$ $\times x^{n+1}$

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

$(\frac {1} {x})$ 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

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

We know that the derivative of $\cos (x)$ is $\sin (x)$

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

We know that the derivative of $\sin (x)$ is $\cos (x)$

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

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

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,

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

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,

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

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,