Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures College of Science MATHS 101: Calculus I (University of Bahrain) 1 / 28
Indefinite Integral Given a function f, if F is a function such that then F is called antiderivative of f. F (x) = f (x) Definition 1 An antiderivative of f is simply a function whose derivative is f. Note: Any two antiderivatives of a function differ only by a constant. (University of Bahrain) 2 / 28
Indefinite Integrals If F (x) is the antiderivative of f (x), we will write where f (x) dx = F (x) +C }{{} antiderivative The symbol is called the integral sign. The function f (x) is called the integrand. The constant C is called the constant of integration. dx indicates the variable involved in the integration which is x. Note: The Fundamental Theorem of Calculus ( d d f (x) dx) = f (x) and (f (x)) dx = f (x) dx dx Integration and differentiation are reversing each other. (University of Bahrain) 3 / 28
Examples Example 2 Find 7 dx. We need to find what is the function that if we differentiate it we get 7? = 7x + C Example 3 Find x dx. We need to find what is the function that if we differentiate it we get x? = 1 2 x 2 + C (University of Bahrain) 4 / 28
Examples Example 4 Find x 9 dx. We need to find what is the function that if we differentiate it we get x 9? Example 5 Find 1 x dx. = 1 10 x 10 + C We need to find what is the function that if we differentiate it we get 1 x? = ln x + C (University of Bahrain) 5 / 28
Elementary integration formula 1 k dx = kx + C. 2 3 4 5 x n dx = 1 n+1 x n+1 + C n = 1. x 1 dx = 1 x dx = ln x + C x > 0. e x dx = e x + C. kf (x) dx = k f (x) dx. 6 (f (x) + g(x)) dx = f (x) dx + g(x) dx. (University of Bahrain) 6 / 28
Example 6 Find 5x 9 dx. Exercise 7 Find 3 x 6 dx. = 5 8 x 8 + C 3 x 6 dx = 3x 6 dx = 3 5 x 5 + C (University of Bahrain) 7 / 28
Example 8 Find 4x 6 + 3x 4 + 2x + 9 + 1 x dx. Exercise 9 = 4 7 x 7 + 3 5 x 5 + x 2 + 9x + ln x + C Find x 9.9 7x 6 + 3x 4 + x 1 + 2 dx. = 1 10.9 x 10.9 x 7 + 3 3 x 3 + ln x + 2x + C (University of Bahrain) 8 / 28
Example 10 Find x + 5 3 3 x 2 dx. Exercise 11 = Find e x + x e + e 2 dx. x 1 5 2 + 3 x 2 2 3 dx = 3 x 3 5 2 + 3 3x 1 3 + C = e x + 1 e + 1 x e+1 + e 2 x + C (University of Bahrain) 9 / 28
Example 12 Find x 2 (4x 3 + 3x + 5) dx. = Exercise 13 Find x 4 +10x x 2 dx. 4x + 3x 1 + 5x 2 dx = 2x + 3 ln x 5x 1 + C = x 2 + 10x 1 dx = 1 3 x 3 + 10 ln x + C (University of Bahrain) 10 / 28
Example 14 Find (x + 2) 2 dx. Exercise 15 Find d dx = ( 1 1+x 3 x 2 + 4x + 4 dx = 1 3 x 3 + 2x 2 + 4x + C ) dx. d dx ( ) 1 dx = 1 + x 3 1 1 + x 3 + C (University of Bahrain) 11 / 28
Exercise 16 Find (7x 3 6x 2 ln 3) dx. Exercise 17 Find e ln(x2 +1) dx. = 7 4 x 4 2x 3 (ln 3)x + C e ln(x2 +1) dx = (x 2 + 1) dx = x 3 3 + x + C Exercise 18 Find dx. dx = 1 dx = x + C (University of Bahrain) 12 / 28
Indefinite Integrals involving Trigonometric and inverse trigonometric functions Example 19 Find csc 2 x dx. Exercise 20 Find 4 sin x + 3 cos x dx. = cot x + C = 4 cos x + 3 sin x + C (University of Bahrain) 13 / 28
Example 21 Find x 3 x 4 cos x+5 x 4 dx. Exercise 22 Find 7 1 x 2 dx. = ( ) 1 x cos x + 5x 4 dx = ln x sin x 5 2 x 3 + C = 7 sin 1 x + C (University of Bahrain) 14 / 28
Example 23 Find 1+cos 2 x cos 2 x dx. 1 + cos 2 x cos 2 x ( ) 1 = cos 2 x + 1 dx = sec 2 x + 1 dx = tan x + x + C (University of Bahrain) 15 / 28
Exercise 24 Find sin 2x sin x dx. sin 2x 2 sin x cos x sin x dx = sin x = 2 cos x dx = 2 sin x + C dx (University of Bahrain) 16 / 28
Example 25 Find t 2 1 t 4 1 dx. t 2 1 t 4 1 = (t 2 1) (t 2 + 1)(t 2 1) dx 1 = t 2 + 1 dx = tan 1 t + C (University of Bahrain) 17 / 28
Exercise 26 Find cos x(tan x + sec x) dx. cos x(tan x + sec x) dx = cos x tan x + cos x sec x dx = cos x sin x cos x + cos x 1 cos x dx = (sin x + 1) dx = cos x + x + C (University of Bahrain) 18 / 28
Note: If f (x) dx = F (x) + C, then f (kx) dx = 1 k F (kx) + C. Example 27 1 e 2x dx = 1 2 e2x + C. 2 e x dx = 1 1 e x + C. 3 e 2x dx = 1 2 e 2x + C. (University of Bahrain) 19 / 28
Example 28 Find (e x + 5) 2 dx. Exercise 29 Find (e x e x ) 2 dx. (e x + 5) 2 dx = (e 2x + 10e x + 25) dx = 1 2 e2x + 10e x + 25x + C (University of Bahrain) 20 / 28
Example 30 Find 1+e x e x dx. 1 + e x ( ) 1 e x dx = e x + ex e x dx = (e x + 1) dx = e x + x + C (University of Bahrain) 21 / 28
Example 31 1 sin(2x) dx = 1 2 cos(2x) + C. 2 cos( x) dx = 1 1 sin( x) + C. 3 sec(πx) tan(πx) dx = 1 π sec(πx) + C. 4 sec 2 (7x) dx = 1 7 tan(7x) + C. (University of Bahrain) 22 / 28
Recall: The trigonometric Identitites: (cos and sin functions) 1 cos 2 x + sin 2 x = 1. 2 1 sin 2 x = cos 2 x. 3 1 cos 2 x = sin 2 x. (tan and sec functions) 1 sec 2 x tan 2 x = 1. 2 sec 2 = 1 + tan 2 x. 3 tan 2 x = sec 2 x 1. (Double Angle Formula) 1 sin(2x) = 2 sin x cos x. 2 cos(2x) = cos 2 x sin 2 x. 3 cos 2 x = 1 2 (1 + cos(2x)). 4 sin 2 x = 1 2 (1 cos(2x)). (University of Bahrain) 23 / 28
Example 32 Find sec 2 x dx. Example 33 Find tan 2 x dx. sec 2 x dx = tan x + C tan 2 x dx = (sec 2 x 1) dx = tan x x + C (University of Bahrain) 24 / 28
Example 34 Find sin 2 x dx. Exercise 35 Find cos 2 x dx 1 sin 2 x dx = (1 cos(2x)) dx 2 = 1 (1 cos(2x)) dx 2 = 1 (x 12 ) 2 sin(2x) + C (University of Bahrain) 25 / 28
Exercise 36 Find 6 cot 2 x dx. 6 cot 2 x dx = 6 (csc 2 x 1) dx = 7 csc 2 x dx = 7x cot x + C (University of Bahrain) 26 / 28
Example 37 Find (sec x + tan x) 2 dx. (sec x + tan x) 2 dx = sec 2 x + 2 sec x tan x + tan 2 x dx = sec 2 x + 2 sec x tan x + sec 2 x 1 dx = 2 sec 2 x + 2 sec x tan x 1 dx = 2 tan x + 2 sec x x + C (University of Bahrain) 27 / 28
Example 38 Find 1 4 x 2 dx. Exercise 39 Find 1 9+x 2 dx. 1 ( dx = x ) 4 x 2 sin 1 2 1 9 + x 2 dx = 1 ( x ) 3 tan 1 3 (University of Bahrain) 28 / 28