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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