In this page, the standard results of derivatives are revised and the same is given in anti-derivative form.

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From the Fundamental theorem of calculus, the indefinite integral of a function is understood to be anti-derivative of a function.

For a given function `f(x)`, there are two possibilities to work out the result of integration `int f(x)dx = g(x)`

• use first principle of `g(x) ``= lim_(n->oo) sum_(i=1)^n ``f((i x)/n) xx x/n`

• use the anti-derivative property to find `g(x)` such that `d/(dx) g(x) = f(x)`.

Finding the anti-derivative is easier than solving limit of summation from the first principles. *And that is the reason, students learn differentiation ahead of integration. *

Let us review the results of anti-derivatives.

Given the result `d/(dx) x^k = k x^(k-1)`

What is the anti-derivative of `x^n`?

- `(x^(n+1))/(n+1) + c`
- `(x^(n+1))/(n+1) + c`
- `(x^(n-1))/(n) + c`

The answer is "`(x^(n+1))/(n+1)`"

Given the result `d/(dx) ax = a`

What is the anti-derivative of `a`?

- `0+c`
- `ax+c`
- `ax+c`

The answer is "`ax+c`"

Given the result `d/(dx) sinx = cos x`

What is the anti-derivative of `cos x`?

- `sin x+c`
- `sin x+c`
- `-sin x+c`

The answer is "`sin x+c`"

Given the result `d/(dx) cosx = -sin x`

What is the anti-derivative of `sin x`?

- `cos x + c`
- `-cos x + c`
- `-cos x + c`

The answer is "`-cos x + c`"

Given the result `d/(dx) tan x = sec^2 x`

What is the anti-derivative of `sec^2 x `?

- `tan x + c`
- `tan x + c`
- `-tan x + c`

The answer is "`tan x + c`"

Given the result `d/(dx) cot x = -csc^2 x`

What is the anti-derivative of `csc^2x`?

- `cot x + c`
- `-cot x + c`
- `-cot x + c`

The answer is "`-cot x + c`"

Given the result `d/(dx) sec x = sec x tan x`

What is the anti-derivative of `sec x tan x` ?

- `sec x + c`
- `sec x + c`
- `(sec x )/(tan x) + c`

The answer is "`sec x + c`"

Given the result `d/(dx) csc x = -csc x cot x`

What is the anti-derivative of `cscx cot x`?

- `csc x + c`
- `-csc x + c`
- `-csc x + c`

The answer is "`-csc x + c`"

Given the result `d/(dx) arcsinx = 1/(sqrt(1-x^2))`

What is the anti-derivative of `1/(sqrt(1-x^2))`?

- `arcsin x + c`
- `arcsin x + c`
- `-arcsin x + c`

The answer is "`arcsin x + c`"

Given the result `d/(dx) arccos x = (-1)/(sqrt(1-x^2))`

What is the anti-derivative of `1/(sqrt(1-x^2))`?

- `arccos x + c`
- `-arccos x+c`
- `-arccos x+c`

The answer is "`-arccos x + c`"

Given the result `d/(dx) arctan x = 1/(1+x^2)`

What is the anti-derivative of `1/(1+x^2)` ?

- `arctan x + c`
- `arctan x + c`
- `-arctan x + c`

The answer is "`arctan x + c`"

Given the result `d/(dx) arcsec x = 1/(|x|sqrt(x^2-1))`

What is the anti-derivative of `1/(xsqrt(x^2-1))`?

- `arcsec x + c`
- `arcsec x + c`
- `sec x + c`

The answer is "`arcsec x + c`"

Given the result `d/(dx) arc csc x = (-1)/(|x|sqrt(x^2-1))`

What is the anti-derivative of `1/(x sqrt(x^2-1))`?

- `arc csc x + c`
- `arc csc x + c`
- `- arc csc x + c`

The answer is "`- arc csc x + c`"

Given the result `d/(dx) arc cot x = (-1)/(1+x^2)`

What is the anti-derivative of `1/(1+x^2)`?

- `arc cot x +c`
- `arc cot x +c`
- `-arc cot x +c`

The answer is "`- arc cot x + c`"

Given the result `d/(dx) e^x =e^x`

What is the anti-derivative of `e^x`?

- `e^x + c`
- `e^x + c`
- `e^(x+1)+c`

The answer is "`e^x + c`"

Given the result `d/(dx) a^x = a^x ln a`

What is the anti-derivative of `a^x`?

- `a^x + c`
- `(a^x)/(ln a) + c`
- `(a^x)/(ln a) + c`

The answer is "`(a^x)/(ln a) + c`"

Given the result `d/(dx) ln x = 1/x `

What is the anti-derivative of `x^(-1)`?

- `ln |x| + c`
- `ln |x| + c`
- `1/(x^0) + c`

The answer is "`ln x + c`"

The standard results of derivatives can be inversed to standard results of anti-derivatives.

**Standard Results of Anti-Derivatives**

`int x^n dx = (x^(n+1))/(n+1) + c`

`int a dx = ax + c`

`int x^(-1) dx = ln x + c`

`int sin x dx = -cos x + c`

`int cos x dx = sinx +c`

`int sec^2 x dx = tan x + c`

`int csc^2 x dx = -cot x + c`

`int sec x tan x dx = sec x + c`

`int csc x cot x dx = -csc x + c`

`int e^x dx = e^x + c`

`int a^x dx = a^x ln a + c`

`int 1/x dx = ln x + c`

`int 1/(sqrt(1-x^2)) dx = arcsin x + c`

`int 1/(sqrt(1-x^2)) dx = -arccos x + c`

`int 1/(xsqrt(x^2-1)) dx = arcsec x + c`

`int 1/(xsqrt(x^2-1)) dx = -arc csc x + c`

`int 1/(1+x^2) dx = arctan x + c`

`int 1/(1+x^2) dx = -arc cot x + c`

*Solved Exercise Problem: *

Integrate `int 1/(sqrt(1-x^2)) dx`

- arcsin x + c
- -arccos x + c
- both the above
- both the above

The answer is "both the above". The difference is the constant of integration `c`. Note `sin(pi/2 + theta) = cos theta`. The two expressions in the choices are equal, but for the constant.

*slide-show version coming soon*