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Kalkulus II

INTEGRAL

1. Integral as Anti-Derivative

Integration is notated ∫ , introduced by Leibniz (1646-1716) from German. The relationship between Integral and Differentiation can be written :

.

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, , 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 + C, with n ≠ - 1

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

= 5 x¹⁰ – 3 x⁶ + 2x^5/2 + Ln x + C

10 6 5/2

= 5 x¹⁰ – 3 x⁶ + 4 x^5/2+ Lnx + C

10 6 5

= 1 x¹⁰ – 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

= Cos² x + Sin² x

Cos² x

= 1/ Cos² x = Sec² x

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

Sin² x

= - Sin² x – Cos² x

Sin² x

= - ( Sin² x + Cos² x)

Sin² x

= -1/ Cos² x = - 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). ∫ Sec² x 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 α Cos 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 α Cos β = Cos (α + β ) - Cos (α - β )

Examples:

1). ∫ (5 Sin x – 3Cos x + 2 Sec² x)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)¹⁰ dx = ∫ 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. 1/3 du

= ∫ 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 +5

3) ∫ √ (7x+4) dx = …