Integral Calculus : Standard Results of Anti-Derivatives

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Standard Results of Anti-Derivatives

what you'll learn...

Overview

» AntiDerivative Standard Results
→ Inverse of standard results of derivatives

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ $int x^n dx = (x^(n+1))/(n+1) + c$

$\int a d x = a x + c$ $int a dx = ax + c$

$\int x^{- 1} d x = \ln x + c$ $int x^(-1) dx = ln x + c$

$\int \sin x d x = - \cos x + c$ $int sin x dx = -cos x + c$

$\int \cos x d x = \sin x + c$ $int cos x dx = sinx +c$

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

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

$\int \sec x \tan x d x = \sec x + c$ $int sec x tan x dx = sec x + c$

$\int \csc x \cot x d x = - \csc x + c$ $int csc x cot x dx = -csc x + c$

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

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

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

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + c$ $int 1/(sqrt(1-x^2)) dx = arcsin x + c$

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = - \arccos x + c$ $int 1/(sqrt(1-x^2)) dx = -arccos x + c$

$\int \frac{1}{x \sqrt{x^{2} - 1}} d x = a r c \sec x + c$ $int 1/(xsqrt(x^2-1)) dx = arcsec x + c$

$\int \frac{1}{x \sqrt{x^{2} - 1}} d x = - a r c \csc x + c$ $int 1/(xsqrt(x^2-1)) dx = -arc csc x + c$

$\int \frac{1}{1 + x^{2}} d x = \arctan x + c$ $int 1/(1+x^2) dx = arctan x + c$

$\int \frac{1}{1 + x^{2}} d x = - a r c \cot x + c$ $int 1/(1+x^2) dx = -arc cot x + c$

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 $f (x)$ $f(x)$ , there are two possibilities to work out the result of integration $\int f (x) d x = g (x)$ $int f(x)dx = g(x)$

• use first principle of $g (x)$ $g(x)$ $= lim_{n \to \infty} \sum_{i = 1}^{n}$ $= lim_(n->oo) sum_(i=1)^n$ $f (\frac{i x}{n}) \times \frac{x}{n}$ $f((i x)/n) xx x/n$

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

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.

exponent

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

constant

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

cosine

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

sine

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

sec squared

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

cosecant squared

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

sec tan

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

cosecant cot

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

arcsine

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

arccos

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

arctan

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

arcsec

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

arccosec

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

arccot

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

e power

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

a power

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

log base e

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

tricky example

Integrate $\int \frac{1}{\sqrt{1 - x^{2}}} d x$ $int 1/(sqrt(1-x^2)) dx$

Both the following are answers.

$\arcsin x + c$ $arcsin x + c$

$- \arccos x + c$ $-arccos x + c$

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

summary

Standard Results of Anti-Derivatives

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ $int x^n dx = (x^(n+1))/(n+1) + c$

$\int a d x = a x + c$ $int a dx = ax + c$

$\int x^{- 1} d x = \ln x + c$ $int x^(-1) dx = ln x + c$

$\int \sin x d x = - \cos x + c$ $int sin x dx = -cos x + c$

$\int \cos x d x = \sin x + c$ $int cos x dx = sinx +c$

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

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

$\int \sec x \tan x d x = \sec x + c$ $int sec x tan x dx = sec x + c$

$\int \csc x \cot x d x = - \csc x + c$ $int csc x cot x dx = -csc x + c$

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

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

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

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + c$ $int 1/(sqrt(1-x^2)) dx = arcsin x + c$

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = - \arccos x + c$ $int 1/(sqrt(1-x^2)) dx = -arccos x + c$

$\int \frac{1}{x \sqrt{x^{2} - 1}} d x = a r c \sec x + c$ $int 1/(xsqrt(x^2-1)) dx = arcsec x + c$

$\int \frac{1}{x \sqrt{x^{2} - 1}} d x = - a r c \csc x + c$ $int 1/(xsqrt(x^2-1)) dx = -arc csc x + c$

$\int \frac{1}{1 + x^{2}} d x = \arctan x + c$ $int 1/(1+x^2) dx = arctan x + c$

$\int \frac{1}{1 + x^{2}} d x = - a r c \cot x + c$ $int 1/(1+x^2) dx = -arc cot x + c$

Outline

The outline of material to learn "Integral Calculus" is as follows.

• Detailed outline of Integral Calculus

→ Application Scenario

→ Integration First Principles

→ Graphical Meaning of Integration

→ Definition of Integrals

→ Fundamental Theorem of Calculus

→ Algebra of Integrals

→ Antiderivatives: Standard results

→ Integration of Expressions

→ Integration by Substitution

→ Integration using Identities

→ Integration by Parts

→ Integration by Partial Fraction

→ Integration: Combination of Methods