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

1. Definition 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.

Simple Cases of Integration ∫ dx = x + c ∫ adx = a∫dx where a is a constant ∫ (dx + dy) = ∫ dx + ∫dy ∫ xn dx = xn+1/ (n+1) + c ; n ≠ -1 ∫ab f(x) = ∫bxo f(x) dx - ∫axo f(x) dx = F(b) – F(a)

Example: I = ∫ √(2x+1) dx I = ∫u1/2 du/2 where 2x + 1 = u I = ½ u3/2 + c 2dx = du 3/2 dx = du/2 I = 1/3 u 3/2 + c I = 1/3 (2x + 1) 3/2 + c.

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. ∫axn dx = axn+1/(n+1) +c ii. ∫eax dx = 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

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 cosn bx 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 lxl ≤ a xii. ∫dx/x = log lxl + c Note: d/dx (log x) = 1/x xiii. ∫ ax dx = ax/(log a) + c Note: d/dx (ax ) = ax log a xiv. ∫ sec2 dx = tan x + c xv. ∫ cosec2 dx = -cot x + c

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

4. INTEGRATION OF TRIGONOMETRIC FUNCTIONS 4.1 Integration of Sinusoidal Functions I = ∫sinn x 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 Sin2 x = 1 – cos2 x therefore; I = ∫sin4x sin x dx = ∫(1-cos2x)2sin x dx

I = ∫ sin4 x sin x dx = ∫(1- cos2 x)2 sin x dx I = ∫( 1 – 2 cos2 x + cos4 x) sin x dx I = ∫(sin x – 2 sin x cos2 x + cos4 x sin x) dx I = -cos x + ⅔ cos3 x – ⅕ cos5 x + c using eqn 3(ix) Case 2 for n = even, eg n = 4 I = ∫sin4 x dx Rewrite using double angle formula:

= ∫ ( ¼ – 2 * ½ * ½ cos 2x + ¼ cos2 2x) dx I = ∫ ( ½ (1-cos 2x)) 2 dx = ∫ ( ½ - ½ cos 2x)2 dx = ∫ ( ¼ – 2 * ½ * ½ cos 2x + ¼ cos2 2x) dx = ∫ ( ¼ - ½ cos 2x + ¼ cos2 2x) dx I = ¼ x + ¼ sin 2x + ¼ ∫cos2 2x dx …….* Now cos2 2x = ½ (1 + cos 4x) I2 = ∫cos2 2x = ∫ ( ½ + ½ cos 4x) dx I2 = x/2 + ⅛ sin 4x ……** sin2 x = ½ (1-cos 2x)

Substitute ** into * to get: I = ¼ x = ¼ sin 2x + ¼ [x/2 + ⅛ sin 4x] I = ¼ x + ¼ 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 = ½ (1 + cos 2 x)

5. Logarithmic Integration 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) Since the quality in numerator is the derivative of denominator I = 2 ln (x3 + sin x) + c

6. Integration by substitution 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

Example 2 I = ∫ 1 du or ∫ dx a + b cos x a + b sin 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)

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 dt [1 + 3 [( 1 – t2)(1 + t2) -1 ]] 1 + t2 = ∫ 2 dt (1 + t2) + 3 ( 1 – t2)

= ∫ (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)

7. Integration by completing of squares Example 1 = ∫ dx / (x2 + 4x + 7) = ∫ dx / (x + 2 )2 + 3 Let y = x + 2 dy = dx 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

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

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

Exercises 1. Show that ∫ sec x dx = log ∫ sec x + tan x) + c 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   3. ∫ dx / (ax + b ) where a and b are constant 4. ∫ tan x dx 5. ∫ dx/ax 6. ∫ x dx / (2x2 = 3 )