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Standard Results of Anti-Derivatives

what you'll learn...

Overview

»  AntiDerivative Standard Results
→  Inverse of standard results of derivatives

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

$\int adx=ax+c$$\int a \mathrm{dx} = a x + c$

$\int {x}^{-1}dx=\mathrm{ln}x+c$$\int {x}^{- 1} \mathrm{dx} = \ln x + c$

$\int \mathrm{sin}xdx=-\mathrm{cos}x+c$$\int \sin x \mathrm{dx} = - \cos x + c$

$\int \mathrm{cos}xdx=\mathrm{sin}x+c$$\int \cos x \mathrm{dx} = \sin x + c$

$\int {\mathrm{sec}}^{2}xdx=\mathrm{tan}x+c$$\int {\sec}^{2} x \mathrm{dx} = \tan x + c$

$\int {\mathrm{csc}}^{2}xdx=-\mathrm{cot}x+c$$\int {\csc}^{2} x \mathrm{dx} = - \cot x + c$

$\int \mathrm{sec}x\mathrm{tan}xdx=\mathrm{sec}x+c$$\int \sec x \tan x \mathrm{dx} = \sec x + c$

$\int \mathrm{csc}x\mathrm{cot}xdx=-\mathrm{csc}x+c$$\int \csc x \cot x \mathrm{dx} = - \csc x + c$

$\int {e}^{x}dx={e}^{x}+c$$\int {e}^{x} \mathrm{dx} = {e}^{x} + c$

$\int {a}^{x}dx={a}^{x}\mathrm{ln}a+c$$\int {a}^{x} \mathrm{dx} = {a}^{x} \ln a + c$

$\int \frac{1}{x}dx=\mathrm{ln}x+c$$\int \frac{1}{x} \mathrm{dx} = \ln x + c$

$\int \frac{1}{\sqrt{1-{x}^{2}}}dx=\mathrm{arcsin}x+c$$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \arcsin x + c$

$\int \frac{1}{\sqrt{1-{x}^{2}}}dx=-\mathrm{arccos}x+c$$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = - \arccos x + c$

$\int \frac{1}{x\sqrt{{x}^{2}-1}}dx=arc\mathrm{sec}x+c$$\int \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx} = a r c \sec x + c$

$\int \frac{1}{x\sqrt{{x}^{2}-1}}dx=-arc\mathrm{csc}x+c$$\int \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx} = - a r c \csc x + c$

$\int \frac{1}{1+{x}^{2}}dx=\mathrm{arctan}x+c$$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan x + c$

$\int \frac{1}{1+{x}^{2}}dx=-arc\mathrm{cot}x+c$$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = - a r c \cot x + c$

simplify

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\left(x\right)$$f \left(x\right)$, there are two possibilities to work out the result of integration $\int f\left(x\right)dx=g\left(x\right)$$\int f \left(x\right) \mathrm{dx} = g \left(x\right)$

• use first principle of $g\left(x\right)$$g \left(x\right)$$=\underset{n\to \infty }{lim}\sum _{i=1}^{n}$$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$f\left(\frac{ix}{n}\right)×\frac{x}{n}$$f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

• use the anti-derivative property to find $g\left(x\right)$$g \left(x\right)$ such that $\frac{d}{dx}g\left(x\right)=f\left(x\right)$$\frac{d}{\mathrm{dx}} g \left(x\right) = f \left(x\right)$.

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.

exponent

Given the result $\frac{d}{dx}{x}^{k}=k{x}^{k-1}$$\frac{d}{\mathrm{dx}} {x}^{k} = k {x}^{k - 1}$
The anti-derivative of ${x}^{n}$${x}^{n}$ is "$\frac{{x}^{n+1}}{n+1}$$\frac{{x}^{n + 1}}{n + 1}$"

constant

Given the result $\frac{d}{dx}ax=a$$\frac{d}{\mathrm{dx}} a x = a$
The anti-derivative of $a$$a$ "$ax+c$$a x + c$"

cosine

Given the result $\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$
The anti-derivative of $\mathrm{cos}x$$\cos x$ is "$\mathrm{sin}x+c$$\sin x + c$"

sine

Given the result $\frac{d}{dx}\mathrm{cos}x=-\mathrm{sin}x$$\frac{d}{\mathrm{dx}} \cos x = - \sin x$
The anti-derivative of $\mathrm{sin}x$$\sin x$ is "$-\mathrm{cos}x+c$$- \cos x + c$"

sec squared

Given the result $\frac{d}{dx}\mathrm{tan}x={\mathrm{sec}}^{2}x$$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$
The anti-derivative of ${\mathrm{sec}}^{2}x$${\sec}^{2} x$ is "$\mathrm{tan}x+c$$\tan x + c$"

cosecant squared

Given the result $\frac{d}{dx}\mathrm{cot}x=-{\mathrm{csc}}^{2}x$$\frac{d}{\mathrm{dx}} \cot x = - {\csc}^{2} x$
The anti-derivative of ${\mathrm{csc}}^{2}x$${\csc}^{2} x$ is "$-\mathrm{cot}x+c$$- \cot x + c$"

sec tan

Given the result $\frac{d}{dx}\mathrm{sec}x=\mathrm{sec}x\mathrm{tan}x$$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$
The anti-derivative of $\mathrm{sec}x\mathrm{tan}x$$\sec x \tan x$ is "$\mathrm{sec}x+c$$\sec x + c$"

cosecant cot

Given the result $\frac{d}{dx}\mathrm{csc}x=-\mathrm{csc}x\mathrm{cot}x$$\frac{d}{\mathrm{dx}} \csc x = - \csc x \cot x$
The anti-derivative of $\mathrm{csc}x\mathrm{cot}x$$\csc x \cot x$ is "$-\mathrm{csc}x+c$$- \csc x + c$"

arcsine

Given the result $\frac{d}{dx}\mathrm{arcsin}x=\frac{1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arcsin x = \frac{1}{\sqrt{1 - {x}^{2}}}$
The anti-derivative of $\frac{1}{\sqrt{1-{x}^{2}}}$$\frac{1}{\sqrt{1 - {x}^{2}}}$ is "$\mathrm{arcsin}x+c$$\arcsin x + c$"

arccos

Given the result $\frac{d}{dx}\mathrm{arccos}x=\frac{-1}{\sqrt{1-{x}^{2}}}$$\frac{d}{\mathrm{dx}} \arccos x = \frac{- 1}{\sqrt{1 - {x}^{2}}}$
The anti-derivative of $\frac{1}{\sqrt{1-{x}^{2}}}$$\frac{1}{\sqrt{1 - {x}^{2}}}$ is "$-\mathrm{arccos}x+c$$- \arccos x + c$"

arctan

Given the result $\frac{d}{dx}\mathrm{arctan}x=\frac{1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$
The anti-derivative of $\frac{1}{1+{x}^{2}}$$\frac{1}{1 + {x}^{2}}$ is "$\mathrm{arctan}x+c$$\arctan x + c$"

arcsec

Given the result $\frac{d}{dx}arc\mathrm{sec}x=\frac{1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \sec x = \frac{1}{| x | \sqrt{{x}^{2} - 1}}$
The anti-derivative of $\frac{1}{x\sqrt{{x}^{2}-1}}$$\frac{1}{x \sqrt{{x}^{2} - 1}}$ is "$arc\mathrm{sec}x+c$$a r c \sec x + c$"

arccosec

Given the result $\frac{d}{dx}arc\mathrm{csc}x=\frac{-1}{|x|\sqrt{{x}^{2}-1}}$$\frac{d}{\mathrm{dx}} a r c \csc x = \frac{- 1}{| x | \sqrt{{x}^{2} - 1}}$
The anti-derivative of $\frac{1}{x\sqrt{{x}^{2}-1}}$$\frac{1}{x \sqrt{{x}^{2} - 1}}$ is "$-arc\mathrm{csc}x+c$$- a r c \csc x + c$"

arccot

Given the result $\frac{d}{dx}arc\mathrm{cot}x=\frac{-1}{1+{x}^{2}}$$\frac{d}{\mathrm{dx}} a r c \cot x = \frac{- 1}{1 + {x}^{2}}$
The anti-derivative of $\frac{1}{1+{x}^{2}}$$\frac{1}{1 + {x}^{2}}$ is "$-arc\mathrm{cot}x+c$$- a r c \cot x + c$"

e power

Given the result $\frac{d}{dx}{e}^{x}={e}^{x}$$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$
The anti-derivative of ${e}^{x}$${e}^{x}$? is "${e}^{x}+c$${e}^{x} + c$"

a power

Given the result $\frac{d}{dx}{a}^{x}={a}^{x}\mathrm{ln}a$$\frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \ln a$
The anti-derivative of ${a}^{x}$${a}^{x}$?

log base e

Given the result $\frac{d}{dx}\mathrm{ln}x=\frac{1}{x}$$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$
The anti-derivative of ${x}^{-1}$${x}^{- 1}$?

tricky example

Integrate $\int \frac{1}{\sqrt{1-{x}^{2}}}dx$$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$\mathrm{arcsin}x+c$$\arcsin x + c$

$-\mathrm{arccos}x+c$$- \arccos x + c$

The difference between the two expressions is the constant of integration $c$$c$. Note $\mathrm{sin}\left(\frac{\pi }{2}+\theta \right)=\mathrm{cos}\theta$$\sin \left(\frac{\pi}{2} + \theta\right) = \cos \theta$. The two expressions in the choices are equal, but for the constant.

summary

Standard Results of Anti-Derivatives

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

$\int adx=ax+c$$\int a \mathrm{dx} = a x + c$

$\int {x}^{-1}dx=\mathrm{ln}x+c$$\int {x}^{- 1} \mathrm{dx} = \ln x + c$

$\int \mathrm{sin}xdx=-\mathrm{cos}x+c$$\int \sin x \mathrm{dx} = - \cos x + c$

$\int \mathrm{cos}xdx=\mathrm{sin}x+c$$\int \cos x \mathrm{dx} = \sin x + c$

$\int {\mathrm{sec}}^{2}xdx=\mathrm{tan}x+c$$\int {\sec}^{2} x \mathrm{dx} = \tan x + c$

$\int {\mathrm{csc}}^{2}xdx=-\mathrm{cot}x+c$$\int {\csc}^{2} x \mathrm{dx} = - \cot x + c$

$\int \mathrm{sec}x\mathrm{tan}xdx=\mathrm{sec}x+c$$\int \sec x \tan x \mathrm{dx} = \sec x + c$

$\int \mathrm{csc}x\mathrm{cot}xdx=-\mathrm{csc}x+c$$\int \csc x \cot x \mathrm{dx} = - \csc x + c$

$\int {e}^{x}dx={e}^{x}+c$$\int {e}^{x} \mathrm{dx} = {e}^{x} + c$

$\int {a}^{x}dx={a}^{x}\mathrm{ln}a+c$$\int {a}^{x} \mathrm{dx} = {a}^{x} \ln a + c$

$\int \frac{1}{x}dx=\mathrm{ln}x+c$$\int \frac{1}{x} \mathrm{dx} = \ln x + c$

$\int \frac{1}{\sqrt{1-{x}^{2}}}dx=\mathrm{arcsin}x+c$$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \arcsin x + c$

$\int \frac{1}{\sqrt{1-{x}^{2}}}dx=-\mathrm{arccos}x+c$$\int \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = - \arccos x + c$

$\int \frac{1}{x\sqrt{{x}^{2}-1}}dx=arc\mathrm{sec}x+c$$\int \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx} = a r c \sec x + c$

$\int \frac{1}{x\sqrt{{x}^{2}-1}}dx=-arc\mathrm{csc}x+c$$\int \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx} = - a r c \csc x + c$

$\int \frac{1}{1+{x}^{2}}dx=\mathrm{arctan}x+c$$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan x + c$

$\int \frac{1}{1+{x}^{2}}dx=-arc\mathrm{cot}x+c$$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = - a r c \cot x + c$

Outline