Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

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1 Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures College of Science MATHS : Calculus I (University of Bahrain) Integrals / 7

2 The Substitution Method Idea: To replace a relatively complicated integral by a simpler one (one from the list). This is done by adding an extra variable which we will call it u. Theorem f (g(x)) g (x) dx = f (u) du Two properties we are looking for in u: u should be an inner function. 2 Almost the derivative of u appears in the integral. (University of Bahrain) Integrals 2 / 7

3 Example 2 Find xe x2 dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = x 2 du = 2xdx dx = du 2x xe x2 dx = xe u du 2x = e u du 2 = 2 eu + C = 2 ex2 + C (University of Bahrain) Integrals 3 / 7

4 Exercise 3 Find 3x + 5 dx. (University of Bahrain) Integrals 4 / 7

5 Example 4 Find x 2 x 3 dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = x 3 du = 3x 2 dx dx = x 2 x dx = 2 x 3 u du 3x 2 = du 3x 2 3 = 3 u du (u) 2 du = 2 3 u 2 + C = 2 3 ( x 3 ) 2 + C (University of Bahrain) Integrals 5 / 7

6 Exercise 5 Find cos x + sin x dx. (University of Bahrain) Integrals 6 / 7

7 Example 6 Find (ln x) 2 x dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = ln x du = dx dx = xdu x (ln x) 2 (u) 2 dx = xdu = (u) 2 du x x = 3 u3 + C = 3 (ln x)3 + C (University of Bahrain) Integrals 7 / 7

8 Exercise 7 Find sec 2 x tan x dx. (University of Bahrain) Integrals 8 / 7

9 Example 8 Find ax+b dx (a = ) substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = ax + b du = adx dx = du a ax + b dx = u du a = a = ln u + C a u du = ln ax + b + C a (University of Bahrain) Integrals 9 / 7

10 Exercise 9 Find sin(2x) cos 2 x dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = cos 2 x du = ( 2 cos x( sin x))dx dx = du 2 cos x sin x sin(2x) sin(2x) cos 2 x dx = du 2 cos x sin x u 2 cos x sin x = u = du = ln u + C u du 2 cos x sin x = ln cos 2 x + C (University of Bahrain) Integrals / 7

11 (Integral of tan x) Example Find tan x dx substitution. Note that tan x = sin x cos x. Hence we are looking for an inner function with almost the derivative is somewhere in the integral. Let u = cos x du = sin xdx dx = du sin x sin x sin x cos x dx = du u sin x = = ln u + C = ln cos x + C u du (University of Bahrain) Integrals / 7

12 (Double Substitution) Example Find x 3 x 2 + dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let u = x 2 + du = 2xdx dx = du 2x x 3 x 2 + dx = x 3 u du 2x = x 2 u du 2 Note that x 2 = u. (University of Bahrain) Integrals 2 / 7

13 = x 2 u du 2 = (u )u 2 du 2 = u 3 2 u 2 + C 2 = ( ) u u C = 5 (x 2 + ) (x 2 + ) C (University of Bahrain) Integrals 3 / 7

14 Exercise 2 Find 4x 7 (x 4 + 4) dx. (University of Bahrain) Integrals 4 / 7

15 (Definite Integral and the substitution method) Example 3 Find π x cos(x 2 ) dx substitution. We are looking for an inner function with almost the derivative is somewhere in the integral. Let π x cos(x 2 ) dx = u = x 2 du = 2x dx dx = du 2x if x =, then u = if x = π, then u = ( π) 2 = π = π x cos(u) du 2x = 2 [ 2 sin(u) ] π = π ( 2 sin(π) cos(u) du ) ( 2 sin() ) = (University of Bahrain) Integrals 5 / 7

16 (Definite Integral and the substitution method) Exercise 4 Find 2 sin x x 2 dx substitution. Let 2 u = sin x du = dx dx = x 2 du x 2 if x =, then u = sin = if x = 2, then u = sin 2 = π 6 sin x dx = x 2 [ = π 6 u x 2 x 2 du = ] π 6 = π 6 u du ( π 2 u ()2 (University of Bahrain) Integrals 6 / 7 ( ) 2 ) ( ) = π2 72

17 (Definite Integral and the substitution method) Example 5 Find ln 3 e x +e 2x dx substitution. Let ln 3 e x + e 2x dx = u = e x du = e x dx dx = du e x if x =, then u = e = if x = ln 3, then u = e ln 3 = 3 3 e x du + (u) 2 e x = [ tan (u) ] 3 = = 3 ( tan ( 3) + (u) 2 du ) ( tan () ) = π 2 (University of Bahrain) Integrals 7 / 7