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

2. 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.

3. 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)

4. 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.

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. ∫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

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 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

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. 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

9. 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:

10. = ∫ ( 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)

11. 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)

12. 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

13. 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

14. 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)

15. implies cos (x/2) = 1/ √(1 + t2)
since cos x = 2 cos2 (x/2) -1
cos x = = 1 – t2
1 + t t2
Example ∫(2 / (1 + 3 cos x) dx
∫ dt
[1 + 3 [( 1 – t2)(1 + t2) -1 ]] 1 + t2
= ∫ 2 dt
(1 + t2) + 3 ( 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. Integration by completing of squares
Example 1 = ∫ dx / (x2 + 4x + 7)
= ∫ dx / (x + 2 )2 + 3
Let y = x 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

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. 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 )