Chapter VI: Integration Dr. Stéphanie Chahine 1- Riemann Sums Suppose we want to find the area under this curve We may struggle to find the exact area, but we can approximate it using rectangles And our approximation gets better if we use more rectangles These sorts of approximations are called Riemann sums Dr. Stéphanie Chahine 1- Riemann Sums Two types of Riemann sums exists: The left Riemann sum: rectangles touch the curve with their top-left corners The right Riemann sum: rectangles touch the curve with their top-right corners Neither choice is strictly better than the other Dr. Stéphanie Chahine 1- Riemann Sums We divide [𝑎, 𝑏] into 𝑛 parts 𝑎0 = 𝑎 𝑙= 𝑎𝑖 = 𝑎 + 𝑖. 𝑙 , 0 ≤ 𝑖 ≤ 𝑛 𝑎𝑛 = 𝑎 + 𝑛. 𝑏−𝑎 𝑛 𝑏−𝑎 𝑛 =𝑎+𝑏−𝑎 =𝑏 Dr. Stéphanie Chahine 1- Riemann Sums Left Riemann Sums: 𝑎𝑖 , 𝑎𝑖+1 → 𝑓 𝑎𝑖 𝑆𝑖 = 𝑎𝑖+1 − 𝑎𝑖 𝑓 𝑎𝑖 = 𝐴1 = 𝑛−1 𝑏−𝑎 𝑖=0 𝑛 𝑓 𝑎 + 𝑖. For 𝑛 → ∞, 𝐴1 → 𝐴 𝐴 = lim 𝐴= 𝑏−𝑎 𝑛→∞ 𝑛 𝑏 𝑓 𝑎 𝑛−1 𝑖=0 𝑓 𝑏−𝑎 𝑓 𝑛 𝑏−𝑎 𝑛 𝑎 + 𝑖. 𝑏−𝑎 𝑛 𝑓(𝑎 + 𝑖 𝑏−𝑎 ) 𝑛 𝑓(𝑏) 𝑓(𝑎) 𝑎 + 𝑖. 𝑏−𝑎 𝑛 𝑥 𝑑𝑥 Dr. Stéphanie Chahine 1- Riemann Sums Right Riemann Sums: 𝑎𝑖 , 𝑎𝑖+1 → 𝑓 𝑎𝑖+1 𝑆𝑖 = 𝑎𝑖+1 − 𝑎𝑖 𝑓 𝑎𝑖+1 = 𝑏−𝑎 𝑓 𝑛 𝑎 + 𝑖. 𝑏−𝑎 𝑛 𝑓(𝑏) 𝑛 𝑏−𝑎 𝑖=1 𝑛 𝑓 𝐴2 = For 𝑛 → ∞, 𝐴2 → 𝐴 𝐴 = lim 𝐴= 𝑏−𝑎 𝑛→∞ 𝑛 𝑏 𝑓 𝑎 𝑎 + 𝑖. 𝑛 𝑖=1 𝑓 𝑏−𝑎 𝑛 𝑓(𝑎 + 𝑖 𝑏−𝑎 ) 𝑛 𝑓(𝑎) 𝑎 + 𝑖. 𝑏−𝑎 𝑛 𝑥 𝑑𝑥 Dr. Stéphanie Chahine 1- Riemann Sums Recalls: 𝑛 𝑖=1(𝑎𝑖 + 𝑏𝑖 ) = 𝑛 𝑖=1 𝑎𝑖 + 𝑛 𝑖=1 𝑏𝑖 𝑛 𝑖=1(𝑎𝑖 − 𝑏𝑖 ) = 𝑛 𝑖=1 𝑎𝑖 − 𝑛 𝑖=1 𝑏𝑖 𝑛 𝑖=1(𝛼𝑎𝑖 ) 𝑛 𝑖=1 𝛼 𝑛 𝑖=1 𝑖 2 𝑛 𝑖 𝑖=1 = 𝑛(𝑛+1)(2𝑛+1) 6 3 𝑛 𝑖 𝑖=1 = 𝑛²(𝑛+1)² 4 =𝛼 𝑛 𝑖=1 𝑎𝑖 , 𝛼 = 𝑐𝑠𝑡 = 𝑛𝛼 = 𝑛(𝑛+1) 2 Dr. Stéphanie Chahine 1- Riemann Sums If 𝑎 = 0 &𝑏 = 1: 1 𝑓 0 𝑏−𝑎 𝑛→∞ 𝑛 𝑥 𝑑𝑥 = lim 𝑛 𝑖=1 𝑓 𝑎 + 𝑖. 𝑏−𝑎 𝑛 1 𝑛→∞ 𝑛 𝑖 𝑛 𝑓 𝑖=1 𝑛 𝑛 𝑖2 (1 − 2 ) 𝑛 = lim Example: 1 𝐼= 0 1 2 1 − 𝑥 𝑑𝑥 = lim 𝑛→∞ 𝑛 𝑛 𝑖=1 𝑖 𝑓 𝑛 1 = lim 𝑛→∞ 𝑛 𝑖=1 1 1 𝑛 𝑛 + 1 2𝑛 + 1 1 2𝑛3 = lim [𝑛 − 2 ] = lim [𝑛 − 2 ] 𝑛→∞ 𝑛 𝑛→∞ 𝑛 𝑛 6 6𝑛 2𝑛3 1 2 = lim 1 − 3 = 1 − = 𝑛→∞ 6𝑛 3 3 Dr. Stéphanie Chahine 2- Definite Integral 𝑏 𝑎 𝑏−𝑎 𝑓 𝑥 𝑑𝑥 = lim 𝑛→∞ 𝑛 𝑓 continuous over 𝑎, 𝑏 → 𝑓 is integrable → The integral over 𝑎, 𝑏 exists 𝑛 𝑖=1 𝑏−𝑎 𝑓 𝑎 + 𝑖. 𝑛 Dr. Stéphanie Chahine 2- Definite Integral Properties: 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 = − 𝑎 𝑓 𝑏 𝑎 𝑓 𝑎 𝑥 𝑑𝑥 = 0 𝑏 𝛼𝑓 𝑎 𝑏 𝛼 𝑎 𝑑𝑥 = 𝛼(𝑏 − 𝑎) 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 + 𝑏 𝑔 𝑎 𝑥 𝑑𝑥 = 𝑏 (𝑓 𝑎 𝑥 + 𝑔 𝑥 )𝑑𝑥 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 − 𝑏 𝑔 𝑎 𝑥 𝑑𝑥 = 𝑏 (𝑓 𝑎 𝑥 − 𝑔 𝑥 )𝑑𝑥 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 + 𝑐 𝑓 𝑏 𝑥 𝑑𝑥 = 𝑐 𝑓 𝑎 𝑥 𝑑𝑥 = 𝛼 𝑥 𝑑𝑥 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 , 𝛼 = 𝑐𝑡𝑒 𝑥 𝑑𝑥 Dr. Stéphanie Chahine 2- Definite Integral Properties: Parity: 𝑓 odd: 𝑓 even: 𝑎 𝑓 −𝑎 𝑎 𝑓 −𝑎 𝑥 𝑑𝑥 = 0 𝑎 𝑓 0 𝑥 𝑑𝑥 = 2 𝑥 𝑑𝑥 If 𝑓 is bounded : 𝑚 ≤ 𝑓 𝑥 ≤ 𝑀 on [𝑎, 𝑏]: 𝑏 𝑚(𝑏 − 𝑎) ≤ 𝑓 𝑥 𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎) 𝑎 𝑓 𝑥 ≥ 𝑔 𝑥 , ∀𝑥 ∈ [𝑎, 𝑏] → 𝑓 𝑥 ≥ 0, ∀𝑥 ∈ [𝑎, 𝑏] → 𝑏 𝑓 𝑎 𝑏 𝑓 𝑎 𝑥 𝑑𝑥 ≥ 𝑏 𝑔 𝑎 𝑥 𝑑𝑥 𝑥 𝑑𝑥 ≥ 0 Dr. Stéphanie Chahine 3- Indefinite Integral 𝑓(𝑥) continuous over 𝑎, 𝑏 ⊂ℝ thus integrable 𝐹(𝑥) : 𝑎, 𝑏 → ℝ 𝑥→𝐹 𝑥 = 𝑥 𝑓 0 𝑡 𝑑𝑡 ⇔ 𝐹′ 𝑥 = 𝑓(𝑥) 𝐹(𝑥) is continuous and is noted integrable indefinite of𝑓(𝑥)over 𝑎, 𝑥 . 𝑓(𝑥) continuous at𝑥0 ⇔∀𝜀 > 0, ∃𝑙 > 0/∀𝑧 ∈ 𝑎, 𝑏 , 𝑧 − 𝑥0 < 𝑙 → 𝑓(𝑧) − 𝑓(𝑥0 ) < 𝜀 Dr. Stéphanie Chahine 3- Indefinite Integral 𝑥→𝐹 𝑥 = 𝑥 𝑓 0 𝑡 𝑑𝑡 ⇔ 𝐹′ 𝑥 = 𝑓(𝑥) Prove: 𝑥0 𝑥0 𝑥0 (𝑓(𝑧) − 𝑓(𝑥 ))𝑑𝑧 ≤ 𝑓(𝑧) − 𝑓(𝑥 ) 𝑑𝑧 ≤ 𝜀𝑑𝑧 0 0 𝑥 𝑥 𝑥 𝑥0 𝑥0 𝑥0 (𝑓(𝑧) − 𝑓(𝑥 ))𝑑𝑧 = 𝑓(𝑧)𝑑𝑧 − 𝑓(𝑥0 )𝑑𝑧 0 𝑥 𝑥 𝑥 𝑎 𝑥0 𝑥0 = 𝑓 𝑧 𝑑𝑧 + 𝑥 𝑓 𝑧 𝑑𝑧 − 𝑎 = 𝜀 𝑥 − 𝑥0 𝑓(𝑥0 )𝑑𝑧 𝑥 = 𝐹 𝑎 − 𝐹 𝑥 + 𝐹 𝑥0 − 𝐹 𝑎 − 𝑓(𝑥0 )(𝑥0 −𝑥) = −𝐹 𝑥 + 𝐹 𝑥0 − 𝑓(𝑥0 )(𝑥0 −𝑥) 𝑥0 (𝑓(𝑧) 𝑥 𝐹 𝑥0 −𝐹 𝑥 (𝑥0 −𝑥) − 𝑓(𝑥0 ))𝑑𝑧 = −𝐹 𝑥 + 𝐹 𝑥0 − 𝑓(𝑥0 )(𝑥0 −𝑥) ≤ 𝜀 𝑥 − 𝑥0 − 𝑓(𝑥0 ) ≤ 𝜀 𝐹 𝑥 −𝐹 𝑥0 𝑥→𝑥0 (𝑥−𝑥0 ) lim = 𝐹 ′ 𝑥 = 𝑓(𝑥) Dr. Stéphanie Chahine 3- Indefinite Integral Primitive: 𝑓 : 𝑎, 𝑏 → ℝ continuous thus integrable 𝑏 𝑓(𝑥) 𝑎 = 𝐹 𝑏 − 𝐹(𝑎) 𝐹 : 𝑎, 𝑏 → ℝ is a primitive of 𝑓 if and only if:𝐹 ′ 𝑥 = 𝑓(𝑥) If 𝐹1 , 𝐹2 are two primitives of 𝑓 → 𝐹′1 𝑥 − 𝐹 ′ 2 𝑥 = 0 → 𝐹1 𝑥 − 𝐹2 𝑥 = 𝑐𝑠𝑡 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction Rational fraction: numerator and denominator are polynomials: If the degree of the numerator is smaller than the degree of the denominator: Factorize the denominator and decompose it into a sum of simple elements. The factors of the denominator are all of the first degree and do not repeat. 𝐴 (𝑥−𝑎) The factors of the denominator are first-order but repeated: 𝐴𝑛 𝐴𝑛−1 𝐴1 + + ⋯+ (𝑥 − 𝑎)𝑛 (𝑥 − 𝑎)𝑛−1 (𝑥 − 𝑎)1 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction Rational fraction: numerator and denominator are polynomials: The denominator contains second-order factors that are not repeated. For each second-order factor 𝑥 2 + 𝑏𝑥 + 𝑐 → this form by completing the square: 𝑥 + suppose 𝑡 = 𝑥 + 𝑏 2 2 𝐴𝑥+𝐵 𝑥 2 +𝑏𝑥+𝑐 + 1 4 .We integrate 4𝑐 − 𝑏 2 And we 𝑏 2 The denominator contains second-order factors that are repeated. 𝐴𝑛 𝑥 + 𝐵𝑛 𝐴𝑛−1 𝑥 + 𝐵𝑛−1 𝐴1 𝑥 + 𝐵1 + + ⋯+ 2 (𝑥 2 + 𝑏𝑥 + 𝑐)𝑛 (𝑥 2 + 𝑏𝑥 + 𝑐)𝑛−1 (𝑥 + 𝑏𝑥 + 𝑐)1 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction Rational fraction: numerator and denominator are polynomials: If the degree of the numerator is greater than the degree of the denominator. The fraction can be reduced to the sum of a polynomial and a fraction with the degree of the numerator less than that of the denominator, by dividing the numerator by the denominator. Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑥 2 +1 𝑑𝑥 (𝑥−1)(𝑥−2)(𝑥−3) Multiplying by 𝑥 − 1 = 𝐴 𝐵 𝐶 + + 𝑑𝑥 (𝑥−1) (𝑥−2) (𝑥−3) 𝑥 2 +1 : (𝑥−2)(𝑥−3) =𝐴+ 𝐵(𝑥−1) 𝐶(𝑥−1) + (𝑥−2) (𝑥−3) 2 (−1)(−2) Multiplying by 𝑥 − 2 : 5 (1)(−1) Multiplying by 𝑥 − 3 : 10 (2)(1) 1 5 5 − + 𝑑𝑥 (𝑥−1) (𝑥−2) (𝑥−3) =1=𝐴 = −5 = 𝐵 =5=𝐶 = ln 𝑥 − 1 − 5 ln 𝑥 − 2 + 5 ln 𝑥 − 3 + 𝑐𝑠𝑡 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑑𝑥 𝑥²(𝑥−1) = 𝐴 𝑥² 𝐵 𝑥 + + Multiplying by 𝑥 − 1 : Multiplying by 𝑥² : Multiplying by 𝑥 : −1 1 − 𝑥² 𝑥 + 1 (𝑥−1) 𝐶 (𝑥−1) 1=𝐶 −1 = 𝐴 0 = 0 + 𝐵 + 𝐶; 𝐵 = −𝐶 = −1 1 𝑥 = − ln 𝑥 + ln 𝑥 + 1 + 𝑐𝑠𝑡 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑑𝑥 (𝑥−1)²(𝑥 2 +1) = 𝐴 𝐵 𝐶𝑥+𝐷 + + 2 𝑥−1 (𝑥−1)² (𝑥 +1) 1 2 Multiplying by 𝑥 − 1 ²: Multiplying by 𝑥 : 𝐹𝑜𝑟 𝑥 = 0 : 1 = −A + 𝐵 + 𝐷 𝐹𝑜𝑟 𝑥 = 2 : 1 5 A = − ;B = ;C = ;𝐷 = 0 1 2 1 2 − =𝐵 0=𝐴+𝐶 =A+𝐵+ 2𝐶+𝐷 5 1 2 −1 1 𝑥 − + 2 2(𝑥−1) 2(𝑥−1)² 2(𝑥 +1) 1 2 = − ln 𝑥 − 1 − 1 1 + ln 2(𝑥−1) 4 𝑥² + 1 + 𝑐𝑠𝑡 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑥−1 𝑑𝑥 (𝑥 2 +2𝑥+2)² = 𝐴𝑥+𝐵 𝐶𝑥+𝐷 + (𝑥 2 +2𝑥+2)² (𝑥 2 +2𝑥+2) Suppose 𝑢 = 𝑥 + 1; 𝑦= 𝑥−1 𝑑𝑥 ((𝑥+1)²+1)² 𝐿= (𝑢)𝑑𝑢 (𝑢2 +1)² 𝑄= 𝑑𝑢 𝑢2 +1 2 𝑢2 𝑑𝑢 𝑢2 +1 2 = = = 𝑑𝑣 2𝑣² 𝐴𝑥+𝐵 𝐶𝑥+𝐷 + ((𝑥+1)²+1)² ((𝑥+1)²+1) = 𝑑𝑢 = 𝑑𝑥 (𝑢−2)𝑑𝑢 (𝑢2 +1)² =− = (𝑢)𝑑𝑢 − (𝑢2 +1)² 2 𝑑𝑢 𝑢2 +1 2 = 𝐿 − 2𝑄 1 2 𝑢2 +1 (1+𝑢2 −𝑢2 )𝑑𝑢 𝑢2 +1 2 = (1+𝑢2 )𝑑𝑢 𝑢2 +1 2 − 𝑢2 𝑑𝑢 𝑢2 +1 2 = 𝑑𝑢 𝑢2 +1 − = arctan 𝑢 − 𝑅 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑥−1 𝑑𝑥 (𝑥 2 +2𝑥+2)² = 𝐴𝑥+𝐵 𝐶𝑥+𝐷 + (𝑥 2 +2𝑥+2)² (𝑥 2 +2𝑥+2) 𝐴𝑥+𝐵 𝐶𝑥+𝐷 + ((𝑥+1)²+1)² ((𝑥+1)²+1) 𝑢2 𝑑𝑢 𝑢2 +1 2 𝑅= Suppose 𝑣 = 𝑢; 𝑣 ′ = 1 and 𝑤 ′ = 𝑅=− 𝑦=− = 𝑢 2 𝑢2 +1 1 2 𝑢2 +1 + 1 2 𝑢2 +1 − 2 arctan 𝑢 − 𝑑𝑢 = − 𝑢 𝑢2 +1 𝑢 ;𝑤 𝑢2 +1 2 =− 𝑢 2 𝑢2 +1 arctan 𝑢 2 + 1 2 𝑢2 +1 + arctan 𝑢 Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑦= 𝑥 5 𝑑𝑥 (𝑥 2 +1)(𝑥−1) = 𝑥 5 𝑑𝑥 (𝑥 3 −𝑥 2 +𝑥−1) 𝑥5 𝑥3 − 𝑥2 + 𝑥 − 1 −(𝑥 5 − 𝑥 4 + 𝑥 3 − 𝑥 2 ) 𝑥2 + 𝑥 𝑥4 − 𝑥3 + 𝑥2 −(𝑥 4 − 𝑥 3 + 𝑥 2 − 𝑥) 𝑥 𝑥5 = 𝑥3 − 𝑥2 + 𝑥 − 1 𝑥2 + 𝑥 + 𝑥 𝑦= 𝑥 3 −𝑥 2 +𝑥−1 𝑥 2 +𝑥 +𝑥 (𝑥 3 −𝑥 2 +𝑥−1) 𝑑𝑥 = 𝑥2 + 𝑥 + 𝑥𝑑𝑥 (𝑥 3 −𝑥 2 +𝑥−1) Dr. Stéphanie Chahine 4- Techniques of integration a- Integration of rational fraction 𝑥 5 𝑑𝑥 (𝑥 2 +1)(𝑥−1) 𝑦= 𝑥 (𝑥 2 +1)(𝑥−1) = = 𝑥 5 𝑑𝑥 (𝑥 3 −𝑥 2 +𝑥−1) − 1): Multiplying by 𝑥: 0 𝑦= 𝑥2 +𝑥 𝑥2 + 𝑥 + 𝑥𝑑𝑥 (𝑥 3 −𝑥 2 +𝑥−1) 𝐴 𝐵𝑥+𝐶 + 2 (𝑥−1) (𝑥 +1) Multiplying by (𝑥 For (𝑥 = 1 2 =𝐴 = 𝐴 + 𝐵; 𝐵 = − = 0): 0 = −𝐴 + C; C = 𝑥𝑑𝑥 + (𝑥 3−𝑥2+𝑥−1) = 𝑥3 3 + 𝑥2 2 1 1 2 1 2 1 1 + 2 ln 𝑥 − 1 − 4 ln 𝑥 2 + 1 + 2 arctan 𝑥 + 𝑐𝑠𝑡 Dr. Stéphanie Chahine 4- Techniques of integration b- The substitution Rule 𝑓 𝑔 𝑥 . 𝑔′ 𝑥 𝑑𝑥 = 𝑓 𝑢 𝑑𝑥 𝑢=𝑔 𝑥 𝑑𝑢 = 𝑔′ 𝑥 𝑑𝑥 𝛽 𝑔(𝛽) 𝑓 𝑔 𝑥 . 𝑔′ 𝑥 𝑑𝑥 = 𝛼 𝑓 𝑢 𝑑𝑥 𝑔(𝛼) Dr. Stéphanie Chahine 4- Techniques of integration b- The substitution Rule Example: I= 𝑢 = 𝑥 2 − 9𝑥 + 1 I= 2𝑥 − 9 𝑥2 ; − 9𝑥 + 1 𝑑𝑥 𝑑𝑢 = (2𝑥 − 9)𝑑𝑥 𝑑𝑢 = 2 𝑢 ∗ +𝑐 = 2 𝑥 2 − 9𝑥 + 1 + c 𝑢 Dr. Stéphanie Chahine 4- Techniques of integration c- Integration by parts 𝑓(𝑥) . 𝑔′ 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑔 𝑥 − 𝑢 𝑑𝑣 = uv − 𝑓′ 𝑥 𝑔(𝑥)𝑑𝑥 𝑣𝑑𝑢 𝑏 𝑏 𝑓(𝑥) . 𝑔′ 𝑥 𝑑𝑥 = 𝑓 𝑏 𝑔 𝑏 − 𝑓 𝑎 𝑔 𝑎 − 𝑎 𝑓′ 𝑥 𝑔(𝑥)𝑑𝑥 𝑎 Dr. Stéphanie Chahine 4- Techniques of integration c- Integration by parts Example1: I= 𝑢 = ln 𝑥 ; 𝑑𝑢 = dv = 𝑑𝑥 ; v=𝑥 I= ln 𝑥 𝑑𝑥 1 𝑥 𝑢𝑑𝑣 = 𝑥 ln 𝑥 − 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥 + 𝑐 Dr. Stéphanie Chahine 4- Techniques of integration c- Integration by parts Example2: I= 𝑢=𝑥 𝑥 𝑥 + 1𝑑𝑥 𝑑𝑢 = 𝑑𝑥 ; dv = 𝑥 + 1𝑑𝑥 ; v= (𝑥+1)3/2 3/2 3/2 I= (𝑥 + 1) 𝑢𝑑𝑣 = 𝑥 3/2 − 𝑥+1 3 2 3 2 𝑑𝑥 2 2 (𝑥 + 1)5/2 3/2 = 𝑥(𝑥 + 1) − +𝑐 3 3 5/2 Dr. Stéphanie Chahine 4- Techniques of integration c- Integration by parts Example3: (2𝑥 4 + 1) cos 𝑥 𝑑𝑥 I= 𝑢 = 2𝑥 4 + 1 ; 𝑑𝑢 = 8𝑥 3 𝑑𝑥 dv = cos 𝑥 𝑑𝑥 ; v = sin 𝑥 I = 2𝑥 4 + 1 sin 𝑥 − 𝑢 = 8𝑥 3 dv = sin 𝑥 𝑑𝑥 8𝑥 3 sin 𝑥 𝑑𝑥 𝑑𝑢 = 24𝑥 2 𝑑𝑥 ; ; v = −𝑐𝑜𝑠 𝑥 I = 2𝑥 4 + 1 sin 𝑥 + 8𝑥 3 cos 𝑥 + 24𝑥 2 cos 𝑥 𝑑𝑥 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions Product of sin 𝑥 cos 𝑥: We use integration by parts We use the formulas 1 𝑠𝑖𝑛 𝑎 𝑠𝑖𝑛 𝑏 = [𝑐𝑜𝑠 𝑎 − 𝑏 − 𝑐𝑜𝑠 𝑎 + 𝑏 ] 2 1 𝑠𝑖𝑛 𝑎 𝑐𝑜𝑠 𝑏 = [𝑠𝑖𝑛 𝑎 − 𝑏 + 𝑠𝑖𝑛 𝑎 + 𝑏 2 1 𝑐𝑜𝑠 𝑎 𝑐𝑜𝑠 𝑏 = [𝑐𝑜𝑠 𝑎 − 𝑏 + 𝑐𝑜𝑠 𝑎 + 𝑏 2 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions 𝑠𝑖𝑛𝑚 𝑥 𝑐𝑜𝑠 𝑛 𝑥𝑑𝑥 If 𝑚 is odd: 𝑚 = 2𝑘 + 1, we use sin²𝑥 = 1 − cos²𝑥, then we perform the variable change 𝑢 = cos 𝑥. If 𝑚 is even and 𝑛 is odd, 𝑛 = 2𝑘 + 1, we use cos²𝑥 = 1 − sin²𝑥, and we set 𝑢 = 𝑠𝑖𝑛𝑥. If both m and n are even, we substitute: 1 + cos 2𝑥 2 1 − cos 2𝑥 ² 𝑠𝑖𝑛 𝑥 = 2 𝑐𝑜𝑠 ² 𝑥 = Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions Examples: 𝐼= 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 3 𝑥𝑑𝑥 𝑢 = sin 𝑥 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥 cos 𝑥 𝑑𝑥 𝐼= 𝐼= d𝑢 = cos 𝑥 𝑑𝑥 ; = 𝑢2 1− 𝑢2 𝑑𝑢 = (𝑢2 −𝑢4 )𝑑𝑢 𝑢3 𝑢5 = − +𝑐 3 5 𝑠𝑖𝑛5 𝑥 𝑑𝑥 𝑢 = cos 𝑥 𝐼= d𝑢 = −𝑠𝑖𝑛 𝑥 𝑑𝑥 ; (𝑠𝑖𝑛2 𝑥)² sin 𝑥 𝑑𝑥 = 1 − 𝑢2 2 −𝑑𝑢 = − 1 + 𝑢4 − 2𝑢2 𝑑𝑢 𝑢5 2𝑢3 = −𝑢 − + +𝑐 5 3 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions 𝑡𝑎𝑛𝑛 𝑥 𝑑𝑥 𝑐𝑜𝑠 𝑚 𝑥 If 𝑛 is odd: we perform the variable change 𝑢 = If 𝑚 Example:I 𝑢 = 1 cos 𝑥 I= ; 1 cos 𝑥 is even we set 𝑢 = tan 𝑥 𝑡𝑎𝑛3 𝑥 = 𝑑𝑥 𝑐𝑜𝑠 5 𝑥 sin 𝑥 d𝑢 = 𝑑𝑥 cos² 𝑥 2 𝑡𝑎𝑛 𝑥 tan 𝑥 𝑑𝑥 = 4 𝑐𝑜𝑠 𝑥 cos 𝑥 = tan 𝑥 𝑑𝑥 cos 𝑥 ; 1 cos² 𝑥 = 1 + 𝑡𝑎𝑛2 𝑥 7 5 𝑢 𝑢 (𝑢2 − 1) 𝑢4 𝑑𝑢 = − +c 7 5 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions 𝑃(cos 𝑥,sin 𝑥) 𝑑𝑥 𝑄(cos 𝑥,sin 𝑥) we perform the variable change t = 1 2 1 + 𝑡 2 𝑑𝑥 sin 𝑥 = 2𝑡 1+𝑡² dt = tan 𝑐𝑜𝑠 𝑥 = ; 𝑥 2 1−𝑡² 1+𝑡² Example: I= 𝑑𝑥 = sin 𝑥 2𝑑𝑡 2𝑡 (1 + 𝑡 2 ) 1 + 𝑡² = 𝑑𝑡 = ln |𝑡| + 𝑐 𝑡 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions 𝑃(cos² 𝑥,sin² 𝑥) 𝑑𝑥 𝑄(cos² 𝑥,sin² 𝑥) we perform the variable change t dt = 1 + 𝑡 2 𝑑𝑥 sin² 𝑥 = tan 𝑥 = 𝑡² 1+𝑡² 𝑐𝑜𝑠² 𝑥 = ; 1 1+𝑡² Example: I= = 𝑑𝑥 = 4 − sin² 𝑥 𝑑𝑡 1+ 𝑡2 𝑡2 4− 1 + 𝑡2 = 𝑑𝑡 𝑑𝑡 = 4𝑡 2 + 4 − 𝑡 2 4 1+ 3 3 1 2 1 𝑎𝑟𝑐𝑡𝑎𝑛 𝑡+𝑐 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝑡+𝑐 4 3 2 2 2 3 3 2 𝑡 2 Dr. Stéphanie Chahine 4- Techniques of integration d- Integrals involving trigonometric functions 𝑐𝑜𝑠²𝑥 + 𝑠𝑖𝑛²𝑥 = 1 𝐼= 𝐼= = 1 − 𝑦² 𝑑𝑦 𝑦 = sin 𝑥 1 − 𝑦² 𝑑𝑦 = ; ch2 𝑥 − 𝑠ℎ2 𝑥 = 1 ; d𝑦 = 𝑐𝑜𝑠 𝑥 𝑑𝑥 𝑐𝑜𝑠 2 𝑥 cos 𝑥 𝑑𝑥 = 𝑐𝑜𝑠² 𝑥𝑑𝑥 1 + cos 2𝑥 𝑥 1 𝑑𝑥 = + sin 2𝑥 + 𝑐 2 2 4 arcsin 𝑦 1 = + sin 2 arcsin 𝑦 + 𝑐 2 4 arcsin 𝑦 1 = + (2 sin arcsin 𝑦 cos arcsin 𝑦 ) + 𝑐 2 4 Dr. Stéphanie Chahine