Indefinite Integral (Antiderivative)

Indefinite Integral (Antiderivative)

¡ì. INDEFINITE INTEGRAL (ANTIDERIVATIVE)

.1. Definitions

The main task of differential calculus is to find the derivative

f ' ( x ) or the differential df ( x ) = f ' ( x ) dx of the function f ( x ) . The

integral calculus solves the inverse problem ¨C finding the function

F ( x ) whose derivative is the given function f ( x ) , F' ( x ) = f ( x ) or

dF ( x ) = F ' ( x ) dx = f ( x ) dx .

The integral calculus are applied in Geometry, Mechanics,

Physics, Techniques and etc.

Definition. The function F ( x ) , x ¡Ê ( a,b ) is an antiderivative of

the function f ( x ) in the interval ( a,b ) if it is differentiable

?x ¡Ê ( a,b ) and F' ( x ) = f ( x ) or dF ( x ) = f ( x ) dx .

Definition. The set of all antiderivative functions of f ( x ) in a

given interval ( a,b ) , { F ( x ) + C} , where C is a constant, is indefinite

integral of f ( x ) for all x in ( a,b ) and it is denoted as

The

¡Ò f ( x ) dx = F ( x ) + C .

symbol ¡Ò is called integral

sign, f ( x ) - integrand, x -

variable of integration, the symbol dx indicates the variable in which

the antiderivative is taken and C - constant of integration.

Rules of Integration.

¡ä

f

x

dx

= f ( x),

(

)

¡Ò

(

d

)

( ¡Ò f ( x ) dx ) = f ( x ) dx ,

¡Ò af ( x ) dx = a ¡Ò f ( x ) dx , a - constant,

¡Ò ( f1 ( x ) ¡À f2 ( x ) ) dx = ¡Ò f1 ( x ) dx ¡À ¡Ò f2 ( x ) dx .

1

f

x

dx

=

f ( x ) dAx , A - constant,

(

)

¡Ò

A¡Ò

1

Indefinite Integral (Antiderivative)

¡Ò f ( x ) dx = ¡Ò f ( x ) d ( x ¡À A) , A - constant,

¡Ò f ( x ) dx = F ( x ) + C ? ¡Ò f ( u ( x ) ) du ( x ) = F ( u ( x ) ) + C ,

where u ( x ) is a differentiable function.

General Rules of Integration.

d ¡Ò f ( u ) du = f ( u ) du ,

(

)

¡Ò dF ( u ) = F ( u ) + C

¡Ò af ( u ) du = a ¡Ò f ( u ) du ,

¡Ò ( f1 ( u ) ¡À f2 ( u ) ) du = ¡Ò f1 ( u ) du ¡À ¡Ò f2 ( u ) du ,

where u is a differentiable function.

Rules for Finding the Antiderivatives.

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

¡Ò

¡Ò

¡Ò

¡Ò

¡Ò

¡Ò

¡Ò

¡Ò

x n +1

x dx =

+ C,n ¡Ù ?1 ,

n +1

n

e x dx = e x + C ,

ax

a dx =

+ C,a ¡Ù 1,

ln a

x

dx

= ln x + C ,

x

sin xdx = ? cos x + C ,

cos xdx = sin x + C ,

d

¦Ð

=

+

¡Ù

2

+

1

,

tgx

C,x

k

(

)

2

cos 2 x

dx

= ?ctgx + C,x ¡Ù k¦Ð ,

2

sin x

2

Indefinite Integral (Antiderivative)

(9)

(10)

(11)

(12)

(13)

(14)

(15)

(16)

(17)

¡Ò

¡Ò

tgxdx = ? ln cos x + C,x ¡Ù ( 2k + 1)

¦Ð

2

,

ctgxdx = ln sin x + C,x ¡Ù k¦Ð ,

¡Ò

?1

a+x

+ C, a ¡Ù 0

ln

?

dx

? 2a a ? x

=?

,

2

2

a ?x

? 1 arctg x + C, x a

?? a

a

¡Ò

dx

¡Ò x ¡À a = ln x + x ¡À a + C, x > a ,

¡Ò chxdx = shx + C ,

¡Ò shxdx = chx + C ,

dx

x

= arcsin + C, x < a ,

¡Ò a ?x

a

x

a

x + a dx =

x + a + ln x + x + a

¡Ò

x

a

x

a ? x dx =

a ? x + arcsin + C .

¡Ò

a

dx

1

x

=

arctg

+ C, §Ñ ¡Ù 0 ,

a

x2 + a2 a

2

2

2

2

2

2

2

2

2

(18)

2

2

2

2

2

2

(19)

2

2

2

2

2

2

3

2

+C,

Indefinite Integral (Antiderivative)

Maple commands.

> int(f,x);

> Int(f,x);

where f is integrand, x - variable of integration.

Checking of the results from J:=int(F,x) is with command:

> diff(J,x);

.2. Integration by Formulae

There exist many integration formulae. We will use (1)¡Â(19) and

also the rules for integrations and the rules for finding of the

antiderivatives.

The integral

¡Ò f ( x ).g' ( x ) dx

is denoted very often as

¡Ò f ( x ).dg ( x ) .

This process is carrying out of the function g' ( x ) under

differential.

Example. Evaluate the integral

J1 = ¡Ò x 4 + 12 x3 ? 3 x + 5 dx .

(

)

Mathematical Solution. From formula (1):

J1 = ¡Ò x 4 dx + 12 ¡Ò x3dx ? 3¡Ò xdx + 5¡Ò dx =

x5

x4

x2

x5

3x 2

4

=

+ 12. ? 3. + 5 + C =

+ 3x ?

+5+C.

5

4

2

5

2

Solution with Maple.

>J[1]:=int(x^4+12*x^3-3*x+5,x);

x5

3x 2

4

J1 :=

+ 3x ?

+5+C.

5

2

4

Indefinite Integral (Antiderivative)

x5

3x 2

4

The results is: J1 + C , i.e.

+ 3x ?

+5+C.

5

2

It is better to use .

>J[1]:=Int(x^4+12*x^3-3*x+5,x)=

int(x^4+12*x^3-3*x+5,x);

x5

3x 2

4

3

4

J1 := ¡Ò x + 12 x ? 3x + 5 dx =

+ 3x ?

+ 5.

5

2

Checking:

>diff(J[1],x);

x 4 + 12 x3 ? 3x + 5 .

(

)

Example. Evaluate the integral

J 2 = ¡Ò 4 sin3 x.cos xdx

Mathematical Solution. From (1) and the rules

J 2 = 4 ¡Ò sin x.( cos x ) dx = 4 ¡Ò sin

3

3

4

sin x )

(

xd sin x = 4.

4

= sin 4 x + C .

Solution with Maple.

>J[2]:=Int(4*sin(x)^3*cos(x),x)=

int(4*sin(x)^3*cos(x),x);;

J 2 := ¡Ò 4 sin3 x.cos xdx = sin ( x )

4

Example. Evaluate the integral

dx

I1 =

.

2

1 ? 8x

¡Ò

Mathematical Solution. From (17) and the rules

1

2

dx 2 2

I1 =

=

arcsin 2 2 x + C

2

4

2 2

1 ? 2 2x

¡Ò

(

)

Solutions with Maple.

>I[1]:=Int(1/sqrt(1-8*x^2),x)=

int(1/sqrt(1-8*x^2),x);

5

(

)

=

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