Analytical Integration of Functions: A Comprehensive Guide

Functions and their integrals are explored in this guide, covering definitions, simple integration cases, examples, and well-known integrals. Learn about differential equations, inspection integration, and trigonometric functions. Master the techniques with detailed explanations and examples.

• Analytical Integration
• Functions
• Differential Equations
• Inspection Integration
• Trigonometric Functions

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1. ANALYTICAL INTEGRATION OF FUNCTIONS Prof. Samuel Okolie, Prof. Yinka Adekunle & Dr. Seun Ebiesuwa

2. 1. A function y = f(x) is called a solution of the differential equation dy/dx = f(x), a <x<b, if over the domain a < x < b F(x) is differentiable and d F(x)/dx = f(x) i.e, F(x) = f(x) dx + c F(x) is an integral of f(x) wrt x, c is a constant of integration. Definition

3. 2. Simple Cases of Integration dx = x + c adx = a dx (dx + dy) = dx + dy xndx = xn+1/ (n+1) + c ; n -1 abf(x) = bxof(x) dx - axof(x) dx = F(b) F(a) where a is a constant

4. Example: I = u1/2 du/2 where I = u3/2+ c I = (2x+1) dx 2x + 1 = u 2dx = du 3/2 dx = du/2 I = 1/3 u 3/2 + c I = 1/3 (2x + 1) 3/2+ c.

5. 3. Integration by Inspection Some of the simplest functions have well known integrals that should be remembered. The integrals are precisely the converse of the derivatives algebraic functions. i. axndx = axn+1/(n+1) +c ii. eaxdx = eax/ (a) + c iii. a/x dx = alog x + c iv. 1/x dx = log x + c v. a cos bx dx = a sin bx/b + c

6. vi. a tan bx dx = -a log(cos bx)/b + c vii. a cos bx sin nbx dx = a sinn+1bx / b(n+1) + c viii. a sin bx cosnbx dx = -a cos n+1bx / b(n+1) + c ix. a dx /(a2+ x2) = tan-1(x/a) + c x. -1 dx / ( a2 x2 ) = cos-1(x/a) + c lxl a xi. 1 dx / (a2 x2) = sin-1(x/a) + c xii. dx/x = log lxl + c xiii. axdx = ax/(log a) + c xiv. sec2dx = tan x + c xv. cosec2dx = -cot x + c lxl a Note: d/dx (log x) = 1/x Note: d/dx (ax) = axlog a

7. xvi. sec x tan x dx = sec x + c xvii. cos x cot x dx = -cosec x + c xviii. sec x dx = log ( sec x + tan x) + c

8. 4. 4.1 I = sinnx dx Case 1 for n = odd, e.g n = 5 I = sin5x dx Rewrite as a product of sin x and an even power of sin x and use the relation Sin2x = 1 cos2x therefore; I = sin4x sin x dx = (1-cos2x)2sin x dx INTEGRATION OF TRIGONOMETRIC FUNCTIONS Integration of Sinusoidal Functions

9. I = sin4x sin x dx = (1- cos2x)2sin x dx I = ( 1 2 cos2x + cos4x) sin x dx I = (sin x 2 sin x cos2 x + cos4x sin x) dx I = -cos x + cos3x cos5x + c using eqn 3(ix) Case 2 for n = even, eg n = 4 I = sin4x dx Rewrite using double angle formula:

10. sin2 x = (1-cos 2x) I = ( (1-cos 2x)) 2dx = ( - cos 2x)2dx = ( 2 * * cos 2x + cos22x) dx = ( - cos 2x + cos22x) dx I = x + sin 2x + cos22x dx .* Now cos22x = (1 + cos 4x) I2 = cos22x = ( + cos 4x) dx I2= x/2 + sin 4x **

11. Substitute ** into * to get: I = x = sin 2x + [x/2 + sin 4x] I = x + sin 2x + x/8 + 1/32 sin 4x 4.2 Integration of cosinusoidal functions 1 = cosn x dx Similar solution can be gotten for integration of powers of cos x for n = odd and even by using the following formula as appropriate. cos 2 x = 1 sin2 x and the double angle formula cos 2 x = (1 + cos 2 x)

12. 5. Integrals for which the integrand may be written as a fraction in which number is the derivation of the denominator may be easily evaluated using the formula Examples: 1/x dx = ln x + c I = (6 x2 + 2 cos x) dx x3 + sin x = 2(3x2 + cos x) x3 + sin x Since the quality in numerator is the derivative of denominator I = 2 ln (x3 + sin x) + c Logarithmic Integration

13. 6. Sometimes a substitution of variables may be made that turns a complicated integral into a simple one which can then be integrated by a standard method. There are many useful substitutions and knowing what to use comes from experience. Example: 1 I = 1/ (1-x2 ) dx Let x = sin u dx = cos u d u I = 1/ 1-sin 2 u (cos u) du = cos u / cos u du = du = u + c = sin-1 x + c Integration by substitution

14. Example 2 a + b cos x we can make use of the substitution of the form t = tan (x/2) x = 2 tan -1 t dt/dx = sec2 (x/2) = [ 1 + tan2 x/2] = ( 1 + t2) dx = dt 1 + t2 Further since 1 + t2 = sec2 (x/2) I = 1 du or dx a + b sin x

15. implies cos (x/2) = 1/ (1 + t2) since cos x = 2 cos2 (x/2) -1 cos x = 2 -1 = 1 t2 1 + t2 1 + t2 Example (2 / (1 + 3 cos x) dx 2 [1 + 3 [( 1 t2)(1 + t2) -1 ]] = 2 dt (1 + t2) + 3 ( 1 t2) dt 1 + t2

16. = (1/(2- t2) dt = dt ( 2 t)( 2 + t) = [(1/ 2 - t) + 1/( 2 + t)] dt = [ln ( 2 t) + ln ( 2 + t)] + c = ln [( 2 t)( 2 + t)] + c = ln [( 2 - tan (x/2))(( 2 + tan (x/2))] + c Integrals of similar form as dx / (a + b cos x ) or dx / (a + b sin x) but involving sin 2x, cos 2x, tan 2x, sin2 x cos 2 x or tan2x instead of sin x and cos 2 x should be evaluated by the substitution t = tan x Hence: sin x = 2t/( 1 + t2 ) ; cos x = 1- t2/(1 + t2)

17. 7. Example 1 = dx / (x2 + 4x + 7) = dx / (x + 2 )2 + 3 Let y = x + 2 I = ( 1/ (y2 + 3) dy = dy / (3 + y2) = 1/ 3 3 dy / (( 3)2 + y2 ) = 3/ 3 tan-1(y/ 3) + c = 3/3 tan -I ( x + 2 / 3) + c. 8. Integration by parts Since d/dx (uv) = u dv/dx + vdu/ dx, v, u are functions of x . uv = u dv/dx dx + du/dx vdx Integration by completing of squares dy = dx

18. therefore u dv/dx dx = uv - du/dx v dx Integral of a product of two functions, u and dv/dx is { the 1st * integral of the second integral of (derivative of the 1st * integral of the second] In this case the integral of the second must be determinable by inspection Example: x sin x dx u dv/ dx u = x du/dx = 1 dv/dx = sin x, v = cos x I = -x cos x + sin x

19. Notes: the separation of the functions is not always apparent. Example 1: x3 e-2 dx = x2 (xe-2) dx (2) A trick that is sometimes useful is to take 1 as one factor of the product. eg, I = log x dx = (log x) . l dx. u dv/dx I = log x (x) - (1/2) x dx I = x log x x + c

20. Exercises 1. Hint: sec x = sec x (sec x + tan x) / (tan x + sec x) = (sec x tan x = sec2 x ) / (sec x + tan x) = f/ (x) / f (x) 2. find (6x2 + 2 cos x / (x3 + sin x) dx Show that sec x dx = log sec x + tan x) + c 3. dx / (ax + b ) where a and b are constant 4. tan x dx 5. dx/ax x dx / (2x2 = 3 ) 6.