Indefinite Integrals
Definition of Integral
Differentiation operation is the process of determining the derivative of a function $F'\left( x \right)$ if the function $F\left( x \right)$ is known. The process of determining the function $F\left( x \right)$ if $F'\left( x \right) = f\left( x \right)$ is known to be the inverse of the Differentiation operation and is commonly referred to as the Integration Operation.
To understand the relationship between the integration operation and the differentiation operation, see the example in the following table:
$F\left( x \right)$ | $F'\left( x \right) = f\left( x \right)$ |
---|---|
$F\left( x \right) = x$ | $F'\left( x \right) = 1$ |
$F\left( x \right) = x + 1$ | $F'\left( x \right) = 1$ |
$F\left( x \right) = x + 5$ | $F'\left( x \right) = 1$ |
$F\left( x \right) = x - 3$ | $F'\left( x \right) = 1$ |
Note that each of the functions $F\left( x \right)$ from the table above differs only in constants, while the other terms remain the same. Therefore, the set of all functions resulting from the integration operation $F'\left( x \right) = f\left( x \right) = 1$ can be written as $F\left( x \right) = x + C$ where $C$ is a constant and $C \in \Re $.
Based on the description above, the integration operation can be defined as follows:
Suppose $F\left( x \right)$ is a general function that can be differentiated so that $F'\left( x \right) = f\left( x \right)$. In this case, $F\left( x \right)$ is called the Anti-differential (Anti-derivative) set or the integral set of the function $F'\left( x \right) = f\left( x \right)$ .
Integral Notation and Definition of Indefinite Integral
The integration of the function $f\left( x \right)$ to the variable $x$ written in the form $\int {f\left( x \right)} dx$ is called the Indefinite Integral of the function $f\left ( x \right)$ against $x$. The indefinite integral of the function $f\left( x \right)$ against $x$ is a general function defined by the following relationship:
\[\boxed{\int {f\left( x \right)} dx = F\left( x \right) + C}\]with:
- $F\left( x \right)$ is called a general integral function
- $f\left( x \right)$ is called the integral function
- $C$ a real constant called the constant of integration
Indeterminate Integrals of Algebraic Functions
Indefinite Integral Formulas for Algebraic Functions
Suppose $a$ any real constant, $f\left( x \right)$ and $g\left( x \right)$ each are integral functions that can be determined by their general integral functions, then:
- $\int {dx = x + C} $
- $\int {adx = ax + C} $
- $\int {\left\{ {f\left( x \right) \pm g\left( x \right)} \right\}} dx = \int {f\left( x \right)} dx \pm \int {g\left( x \right)} dx$
- $\int {{x^n}} dx = \frac{1}{{n + 1}}{x^{n + 1}} + C$
- $\int {a{x^n}} dx = \frac{a}{{n + 1}}{x^{n + 1}} + C$
Example
Find the following indefinite integrals:
- $\int {4xdx} $
- $\int {2{x^3}dx} $
- $\int {\frac{7}{{2{x^2}}}} dx$
- $\int {\left( {{x^2} - 3x + 4} \right)} dx$
- $\int {\left( {\sqrt[3]{{{x^2}}} - \frac{2}{{\sqrt[4]{{{x^3}}}}}} \right)} dx$
Answer
- $\int {4xdx} = \frac{4}{{1 + 1}}{x^{1 + 1}} + C = 2{x^2} + C$
- $\int {2{x^3}dx} = \frac{2}{{3 + 1}}{x^{3 + 1}} + C = \frac{1}{2}{x^4} + C$
- $\int {\frac{7}{{2{x^2}}}} dx = \int {\frac{7}{2}} {x^{ - 2}}dx = \frac{7}{{2\left( { - 2 + 1} \right)}}{x^{ - 2 + 1}} + C = \frac{7}{{ - 2}}{x^{ - 1}} + C = - \frac{7}{{2x}} + C$
- $\int {\left( {{x^2} - 3x + 4} \right)} dx = \int {{x^2}} dx - \int {3x} dx + \int {4dx = \frac{1}{{2 + 1}}} {x^{2 + 1}} - \frac{3}{{1 + 1}}{x^{1 + 1}} + 4x + C = \frac{1}{3}{x^3} - \frac{3}{2}{x^2} + 4x + C$
- $\int {\left( {\sqrt[3]{{{x^2}}} - \frac{2}{{\sqrt[4]{{{x^3}}}}}} \right)} dx = \int {\left( {{x^{\frac{2}{3}}} - 2{x^{ - \frac{3}{4}}}} \right)} dx = \frac{1}{{\frac{2}{3} + 1}}{x^{\frac{2}{3} + 1}} - \frac{2}{{ - \frac{3}{4} + 1}}{x^{ - \frac{3}{4} + 1}} + C = \frac{1}{{\frac{5}{3}}}{x^{\frac{5}{3}}} - \frac{2}{{\frac{1}{4}}}{x^{\frac{1}{4}}} + C = \frac{3}{5}\sqrt[3]{{{x^5}}} - 8\sqrt[4]{x} + C$
Indeterminate Integrals of Trigonometric Functions
Formulas of Indefinite Integrals of Trigonometric Functions
To determine the rules for indefinite integrals for trigonometric functions, it is necessary to recall the derivatives of trigonometric functions as shown in the following table:
$F\left( x \right)$ | $F'\left( x \right) = f\left( x \right)$ |
---|---|
$\sin x$ | $\cos x$ |
$\cos x$ | $ - \sin x$ |
$\tan x$ | ${\sec ^2}x$ |
$\cot x$ | $ - \cos e{c^2}x$ |
$\sec x$ | $\tan x.\sec x$ |
$\cos ecx$ | $ - \cot x.\cos ecx$ |
Using the indefinite integral rule $\int {f\left( x \right)} dx = F\left( x \right) + C$ which has the property that $F'\left( x \right) = f\left( x \right)$, then the indefinite integral of trigonometric functions can be formulated as follows:
- $\int {\sin xdx = - \cos x + C} $
- $\int {\cos xdx = \sin x + C} $
- $\int {{{\sec }^2}xdx = \tan x + C} $
- $\int {\cos e{c^2}xdx = - \cot x + C} $
- $\int {\tan x.\sec xdx = \sec x + C} $
- $\int {\cot x.\cos ecxdx = - \cos ecx + C} $
- $\int {\sin \left( {ax + b} \right)xdx = - \frac{1}{a}\cos \left( {ax + b} \right) + C} $
- $\int {\cos \left( {ax + b} \right)dx = \frac{1}{a}\sin \left( {ax + b} \right) + C} $
- $\int {{{\sec }^2}\left( {ax + b} \right)dx = \frac{1}{a}\tan \left( {ax + b} \right) + C} $
- $\int {\cos e{c^2}\left( {ax + b} \right)dx = - \frac{1}{a}\cot \left( {ax + b} \right) + C} $
- $\int {\tan \left( {ax + b} \right).\sec \left( {ax + b} \right)dx = \frac{1}{a}\sec \left( {ax + b} \right) + C} $
- $\int {\cot \left( {ax + b} \right).\cos ec\left( {ax + b} \right)dx = - \frac{1}{a}\cos ec\left( {ax + b} \right) + C} $
where $a$ and $b$ are real numbers and $a \ne 0$ respectively.
Example
Find the following indefinite integrals:
- $\int {\left( {2x - \cos x} \right)} dx$
- $\int {\left( {\sin 2x + \cos x} \right)} dx$
- $\int {3{{\sec }^2}} xdx$
- $\int {\left( {{{\tan }^2}x - 3} \right)} dx$
- ${\int {\left( {\sin x + \cos x} \right)} ^2}dx$
Answer
- $\int {\left( {2x - \cos x} \right)} dx = \int {2xdx - \int {\cos xdx = \frac{2}{{1 + 1}}} } {x^2} - \sin x + C = {x^2} - \sin x + C$
- $\int {\left( {\sin 2x + \cos x} \right)} dx = \int {\sin 2xdx + \int {\cos xdx = - \frac{1}{2}} } \cos 2x + \sin x + C$
- $\int {5{{\sec }^2}} xdx = 5\int {{{\sec }^2}} xdx = 5.\tan x + C = 5\tan x + C$
- $\int {\left( {{{\tan }^2}x - 3} \right)} dx = \int {\left( {\left( {{{\tan }^2}x + 1} \right) - 4} \right)} dx = \int {\left( {{{\sec }^2}x - 4} \right)} dx = \int {{{\sec }^2}} xdx - \int {4dx = \tan x - 4x + C} $
- ${\int {\left( {\sin x + \cos x} \right)} ^2}dx = \int {\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)} dx = \int {\left( {1 + \sin 2x} \right)} dx = \int {dx + \int {\sin 2xdx = x - \frac{1}{2}\cos 2x + C} } $
This is a discussion of indefinite integrals in algebraic functions and trigonometric functions. Hopefully useful.
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