INTEGRAL (KALKULUS 2)
1. Integral as Anti-Derivative
Integration is notated ∫ , introduced by Leibniz (1646-1716) from German. The relationship between Integral andDifferentiation can be written
F’ (x) = dF/dx = f(x) ó ∫f(x) dx = F(x)
F’ (x) = dF/dx = f(x) ó ∫f(x) dx = F(x)
2. Indefinite Integral
Suppose we have :
F(x) = 3x2 + 5x - 7 , then F’(x) = 6x + 5.
F(x) = 3x2 + 5x + 8 , then F’(x) = 6x + 5
F(x) = 3x2 + 5x + c , then F’(x) = 6x + 5
So we have, ∫f(x) dx = F(x) + C, called as an indefinite integral.
F(x) = 3x2 + 5x - 7 , then F’(x) = 6x + 5.
F(x) = 3x2 + 5x + 8 , then F’(x) = 6x + 5
F(x) = 3x2 + 5x + c , then F’(x) = 6x + 5
So we have, ∫f(x) dx = F(x) + C, called as an indefinite integral.
C may value 1,2,3, …(indefinite) = constant of integration
f(x) is integrand and F(x) is common integral function.
a. Indefinite integral of Algebraic Function
Suppose a is any real constant, f(x) and g(x) is each an integral function whose common integral function can be defined, then we have some formulas and properties as follows:
1). ∫ d ( F (x) ) = F (x ) + C
2). ∫ k d x = k x + C
3). ∫ xn dx = x n+1/ n +1 + C, with n ≠ - 1
4). For n = - 1, formula ( 3 ) can be : ∫ 1/x dx = ln x + C
5). ∫ k f ( x ) dx = k ∫ f (x) dx
6). ∫ [ f ( x ) + g (x) ] dx = ∫ f (x) dx + ∫ g ( x ) dx
7). ∫ [ f ( x ) - g (x) ] dx = ∫ f (x) dx - ∫ g ( x ) dx
Example:
1) Find the result of ∫ ( 5x9– 3x5 + 2xÖx + x-1 - 5)dx
= 5x10 – 3 x6 + 2x5/2 + Ln x + C
10 6 5/2
= 5x10 – 3 x6 + 4 x5/2+ Lnx + C
10 6 5
= 1x10 – 1 x6 + 4√x5+ Lnx + C
2 2 5
Exercise:
A. Find the result of these integrals !
1. ∫ F(x) dx, where F(x) = …………………………
2. ∫ F(s) ds, where F(s) = …………………………
3. ∫ G(t) dt , where G(t) = …………………………
B. Find function F if the following are given : F’(x) = 6x2 , F(0) = 0
C. Given that F’(x) = 4x-1 and F(3) = 20. Find the F(x) !
D. The slope of tangent of a curve at each point (x,y) is expressed by dy/dx = 3x2– 6x + 1. If the curve passes point (2,-3), find its equation !
Solution :
b. Indefinite integral of Trigonometric Function
The rule of determining the integral of a trigonometric functions based on the differentiation of each function is as follows:
1). If y = Sin x → then dy/dx = Cos x → dy = Cos x dx
∫ dy = ∫ Cos x dx
y = Sin x + C
2). If y = Cos x → dy/dx = - Sin x → dy = - Sin x dx
∫ dy = ∫ - Sin x dx
y = Cos x + C
∫ dy = ∫ Sin x dx
y = - Cos x + C
3). If Y= tan x = sin x/ cos x = U/V = (U'.V - V'.U) / V2
Then, Y' = (Cos x Cos x – (- Sin x ) Sin x) / Cos2 x
= (Cos2x + Sin2 x) / Cos2x
= 1/ Cos2 x = Sec2x
So, = ∫ Sec 2x dx = tan x + C
4). If Y= Cot x = Cos x/ sin x = U/V = (U'.V - V'.U) / V2
Then Y' = (- Sin x Sin x – Cos x Cos x) / Sin2x
= (- Sin2 x – Cos2 x) / Sin2x
= - ( Sin2 x + Cos2 x) / Sin2x
= -1/ Cos2x = - Cosec2 x
So, = ∫ Cosec2 x dx = Cot x + C
FORMULAS OF INTEGRAL TRIGONOMETRIC
1). ∫ Sin x dx = - Cos x + C
2). ∫ Cos x dx = Sin x + C
3). ∫ tan x dx = ln | Sec x | + C = - ln | Cos x | + C
4). ∫ Cot x dx = ln | Sin x | + C
5). ∫ Cosec x dx = ln | Cosec x – Cot x | + C = ln | tan x/2 | + C
6). ∫ Sec x dx = ln | Sec x + tan x | + C = ln | tan ( x/2 + λ/2 | + C
7). ∫ Cosec2 x dx = - Cot x + C
8). ∫ Sec2x dx = tan x + C
9). ∫ Cosec x Cot x dx = - Cosec x + C
10). ∫ Sec x tan x dx = Sec x + C
11). ∫ Sin ax dx = -1/α Cos α x + C
12). ∫ Cos ax dx = 1/a Sin ax + C
13). ∫ Sin (ax + b) dx = 1/-α Cos (α x + b ) + C
14). ∫ Cos( ax + b) dx = 1/α Sin ( αx + b) + C
15). ∫ Sinⁿ x Cos x dx = 1/n+1 Sinn+1 x + C
16). ∫ Cosⁿ x Sin x dx = - 1/n+1 Cosn+1 x + C
NOTES :
Please Remember some importants Formulas to help you find the integral of Trigonometric functions !
Cos² x +Sin² x = 1
Sin² x = ½ (1 – Cos 2 x )
Cos² x = ½ (1 + Cos 2 x )
1 + tan² x = Sec² x
2 Sin αCos β = Sin (α + β ) + Sin (α - β )
2 Cos α Sin β = Sin (α + β ) - Sin (α - β )
2 Cos α Cos β = Cos (α + β ) + Cos (α- β)
- 2 Sin αCos β= Cos (α+ β) - Cos (α - β )
Examples:
1). ∫ (5 Sin x – 3Cos x + 2 Sec2x)dx
= 5 (- Cos x ) – 3Sin x + 2 tan x + C
= - 5 Cos x – 3Sin x + 2 tan x + C
2). ∫ (5 Sec2 x – 2 Sin x + 3 Cosx) dx
= 5 tan x + 2 Cos x + 3 Sin x + C
C. Integration by substitution method
Suppose by applying a substitution of u = g(x), where g is a function with derivation so that ∫f(g(x)).g’(x)dx can be changed into ∫f(u)du. If f(u) is an anti-derivative of f(x), then ∫f(g(x)).g’(x)dx = ∫f(u)du = F(u) + C = F(g(x)) + C.
To get better understanding about the above formula, study the following example:
1). Find the results of:
a) ∫x (3x2 – 5 x)10 dx
Solution: Suppose U = 3x2 – 5
du/dx = 6x
du = 6x dx
1/6 du = x dx
So, ∫x (3x2 – 5 x)10dx = ∫ U10 .1/6 du
= ∫ 1/6 .1/11U11 + C
= ∫ 1/66 U11 + C
= ∫ 1/66 ( 3x2 – 5 )10 + C
b) ∫ (Sin 7x) Cos x dx
Solution : Suppose U = sin x
du/dx = Cos x
du = Cos x dx
So, ∫ (Sin 7x) Cos x dx = ∫ U7. du
= ∫ 1/8 U8 + C
= 1/8 Sin x8 + C
C) ∫( x4 + 3x )30 ( 4 x3) dx
Solution : Suppose U = x4 + 3x
du/dx = 4 x3 + 3
du = 4 x3+ 3 dx
So, ∫ ( x4 + 3x )30( 4 x3 ) dx
= ∫ U30 .du
= ∫1/31 U31 + C
= ∫1/31 ( x4 + 3x )31 + C
d) ∫ Cos ( 3x +1). Sin ( 3x + 1 ) dx
Solution : Suppose U = Sin ( 3x + 1 )
du/dx = Cos (3x + 1).3
du = Cos (3x + 1).3 dx
1/3 du = Cos (3x + 1) dx
So, Cos ( 3x +1). Sin ( 3x + 1 ) dx
= ∫ 1/3 du . U
= ∫ U.
= ∫ ½ U2. 1/3 du
= ∫ ½ .1/3 U2 + C
= ∫ 1/6 U2 + C
= ∫ 1/6 Sin2. (3x +1)+ C
e). ∫ Sin5 x2( x . Cos )2 dx
Solution: Suppose U = Sin x2
du/dx = Cos x2 . 2x
du = 2x Cos x2.dx
½ du = x cos x2.dx
So, ∫ Sin5 x2 ( x . Cos )2 dx = ∫ U5.½ du
= ∫ 1/6 U6. ½ du
= ∫ ½. 1/6 U6+ C
= ∫ 1/12 U6 + C
= 1/12 ( Sin x2 )6 + C
Exercises :
By Using the substitution method find the following integrals !
1) ∫ ( x4 - 1 ). x2 dx = ….
2) ∫ 3y dy
√ 2y2 +53) ∫ √ (7x+4) dx = …
