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TABLE OF CONTENT i CHAPTER 1 : INVERSE TRIGONOMETRIC FUNCTIONS 1.0 Introduction .…..………………………………………………………………………. 1 1.0.1 Notation …………………………………………………………..…………………… 1 1.0.2 Properties of Inverse Trigonometric Function ………………………..…………… 1 1.0.3 The graph of Inverse Trigonometric Function …………………………..………… 2 1.1.1 Evaluating/Simplifying Inverse Trigonometric Function using Triangle Method (without using calculator) ……………………………..…. 2 Useful Formula ……………………………………………………………………… 3 Tutorial 1.1a : Evaluating Inverse Trigonometric Function (without using calculator) …………………………………………. 6 1.1.2 Evaluating composite inverse trigonometric functions ………………………….. 7 Tutorial 1.1b : Evaluating/Simplifying Inverse Trigonometric Function (using Triangle Method) …………………………………………..15 1.2 Derivative of Inverse Trigonometric Functions …………………………………….16 More Examples (from previous semester papers) ...……………………………18 Tutorial 1.2 : Differentiation involving Inverse Trigonometric Functions ……… 25 1.3 Integration of Inverse Trigonometric Functions ……………………………..……. 27 1.4 Integration of Inverse Trigonometric Functions involving Completing the Square Method ………………………………………………........ 35 Tutorial 1.3 : Integration of Inverse Trigonometric Functions .…………………. 46
MAT238/ MAT438 : Foundation of Applied Mathematics ii CHAPTER 2 : HYPERBOLIC AND INVERSE HYPERBOLIC FUNCTION 2.1 Hyperbolic Function ………………………………………………………………… 48 2.1.1 Definition of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function) …………………………………………………..…… 48 2.1.2 Identities of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function ………………………………………………..………. 48 2.1.3 Proving Hyperbolic Identity …………………………………………………..…….. 49 Tutorial 2.1 : Proving Hyperbolic Identity ………………………………..…..…… 53 2.1.4 Solving Hyperbolic Equation ………………………………………………..……… 54 2.1.4.1 using Hyperbolic identity (with terms containing square @ double angle) ..….. 54 2.1.4.2 using the definitions of hyperbolic function (with terms not containing square and same angle) ………………………...….. 60 Tutorial 2.1a : Solving the hyperbolic equation using Identities of Hyperbolic Functions ……………………………………………. 66 Tutorial 2.1b : Solving the hyperbolic equation using Definition of Hyperbolic Functions (Exponential Function) ………………… 66 2.1.5 Derivatives of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function) ……………………..………... 67 Tutorial 2.2 : Diffrerentiation of Hyperbolic Functions ………………………... 77 2.1.6 Integration of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Functions) ……………..…….………... 78 Tutorial 2.3 : Integration of Hyperbolic Functions …………………..…………... 86 2.2 Inverse Hyperbolic Function …………………………………………..…………... 87 2.2.1 Notation ……………………………………………………………………..………... 87 2.2.2 Properties of Inverse Trigonometric Function ………………………..………….. 87 2.2.3 Definition of Inverse Hyperbolic Function ……………………………..………….. 87 2.2.4 Proving Inverse Hyperbolic Function …………………………….……..……….. 87 2.2.5 Evaluating Inverse Hyperbolic Functions …………………………..……..……… 91 Tutorial 2.4 : Evaluate Inverse Hyperbolic Function (without using Calculator) ………………………………..……….. 95 2.2.6 Derivatives of Inverse Hyperbolic Function (comparison between Inverse Hyperbolic and Inverse Trigonometric Functions) ………………..…… 95 Tutorial 2.5 : Differentiation involving Inverse Hyperbolic Functions ………..…102 2.2.7 Integration of Inverse Hyperbolic Function (comparison between Inverse Hyperbolic and Inverse Trigonometric Functions) .,………………..…..104 2.2.8 Integration of Inverse Hyperbolic Functions involving Completing the Square Method …………………………………..…….109 Tutorial 2.6 : Integration of Inverse Hyperbolic Functions ………..…….………117 Miscellaneous Exercises .……………...…………………………………..………..118
TABLE OF CONTENT iii CHAPTER 3 : INTEGRATION TECHNIQUES 3.1 Integration using u-substitution …………………………………….…………….. 121 3.1.1 Power Rule (for n −1) with u-substitution → where u = the function in bracket .……………………….………………….. 121 3.1.2 Logarithmic Rule (for n = −1) with u-substitution → where u = denominator ………….…………………………...……………… 123 3.1.3 Exponential Rule with u-substitution → where u = exponent/power …….…… 128 3.1.4 Trigonometric Rule with u-substitution → where u = angle (if the trigonometric function is exist in the table of integrals) ..……………….. 131 3.1.5 Derivative and Integration of inverse tangent & inverse sine ………………….. 134 3.1.6 Introduction to Completing the Square …………………………………………135 3.2 Integration using Partial Fraction ………………………………………………... 141 3.2.1 Type I : denominator with different linear factors .………………………………142 3.2.2 Type II : denominator with irreducible quadratic factors (which does not factorize) ……………………… .……………………………….. 147 3.2.4 Integration using Partial Fraction (involving long division – for improper fraction) ……………………………………………………………. 158 3.2.5 Procedure to Integrate using Partial Fractions …………………………………. 174 Tutorial 3.1 : integration by partial fractions ……………….…………………… 176 Answer Tutorial 3.1 .……………………………………………………………….. 177 3.3 Integration with Trigonometric substitution ….…………………………………...178 More Examples (from previous Semester papers) …………………………….. 189 Tutorial 3.2 : Integration by Trigonometric substitution ………………………… 192 3.4 Integration by Parts (using ‘I LATE’ rule) ………………………………………… 193 3.5 A Shortcut : Tabular Integration ………………………………………………….. 197 Tutorial 3.3 : Integration by Parts ………………………………….…………….. 206 SUMMARY ………………………………………………………………………….. 208
MAT238/ MAT438 : Foundation of Applied Mathematics iv CHAPTER 4 : ORDINARY DIFFERENTIAL EQUATION (ODE) 4.0 Introduction …………………………………………………………………………. 209 4.0.1 Definition of Degree ………………………………………………………………... 209 4.0.2 Definition of Order (highest derivative) …………………………………………… 210 4.1 First Order ODE ………………………………………………………………..…… 210 4.1.1 Three Methods to Solve the First Order ODE …………………………………… 210 4.1.1.1 Separable equation ……………………………………………………………….. 210 Tutorial 4.1.1 : Separable Equation …………………………………………….. 218 4.1.1.2 Homogeneous equation ………………………………………………………….. 219 Tutorial 4.1.2 : Homogeneous Equation ……………………………………….. 225 4.1.1.3 Linear equation .……………………………………………………………………226 Tutorial 4.1.3 : Linear Equation …………………………………………………. 232 Miscellaneous Exercises ………………………………………………………… 233 4.1.2 Applications of First Order ODE ………………………………………………… 225 Applications of Separable Equation ODE ……………………………………… 225 4.1.2.1 Population Growth and Bacterial Colony .……………………………………… 226 Tutorial 4.1.4 : Population growth/bacterial colony ……………………………… 232 4.1.2.2 Radioactive Decay …………………………………………………………………. 233 Tutorial 4.1.5 : Radioactive Decay ……………………………………..…………. 238 .. 4.1.2.3 Newton’s law of cooling ………………………………………………………….. 239 Tutorial 4.1.6 : Newton's Law of Cooling …………………………….………… .245 4.1.2.4 Mixing Problems …………………………………………………………….……. 246 Examples for Mixing Problem (when rate in = rate out) ………………………. 248 Examples for Mixing Problem using Linear Equation (when rate in ≠ rate out)…………………………………………………………….. 251 Tutorial 4.1.7 : Mixing Problem ………………………………………………….. 256 4.2 Second Order ODE ..………………………………………………………………. 268 4.2.1 Homogeneous equation .…………………………………………………………. 268 Tutorial 4.2.1 : 2nd order ODE (homogeneous equation) ……………………. 274 4.2.2 Non-homogeneous equation …………………………………………………….. 274 Table 4.1 : possible trial solution, yp based on the given R(x) ..……………… 275 4.2.2.1 Undetermined Coefficients ……………………………………………………….. 276 4.2.2.2 Modifying Particular Solutions (modify yp) ………………………………………. 289 Tutorial 4.2.2 : 2nd order ODE (non-homogeneous equation) . ……………….. 294 4.3 Formation of Differential Equations by eliminating arbitrary constant ……...... 295 Tutorial 4.3 : Formation of Differential Equations ………………………………. 300 Miscellaneous Exercises ………………………………………………………….. 301
Chapter 1 : Inverse Trigonometric Function 1 CHAPTER 1 INVERSE TRIGONOMETRIC FUNCTIONS 1.0 Introduction • Each of the trigonometric functions’ sine, cosine, tangent, secant, cosecant and cotangent has an inverse (with a restricted domain) • The inverse is used to obtain the measure of an angle using the ratios from basic right triangle trigonometry. • For example, the inverse of sine is denoted as sin-1 or Arcsine or asin. 1.0.1 Notation If sin x = A, then x = sin−1A If cos x = A, then x = cos−1A If tan x = A, then x = tan −1A If sec x = A, then x = sec−1A If cosec x = A, then x = cosec−1A If cot x = A, then x = cot−1A 1.0.2 Properties of Inverse Trigonometric Function sin(sin−1x) = x, sin−1 (sin x) = x cos(cos−1x) = x, cos−1 (cos x) = x tan(tan−1x) = x, tan−1 (tan x) = x sec(sec−1x) = x, sec−1 (sec x) = x cosec(cosec−1x) = x, cosec−1 (cosec x) = x cot(cot−1x) = x, cot−1 (cot x) = x List of topics Introduction (notation, properties, and their graphs) Evaluating Inverse Trigonometric Functions using Triangle Method Derivatives of Inverse Trigonometric Functions Integration of Inverse Trigonometric Functions Note: this does NOT mean sine to the power of negative one.
MAT238/ MAT438 : Foundation of Applied Mathematics 2 1.0.3 The graph of Inverse Trigonometric Function 1.1.1 Evaluating/Simplifying Inverse Trigonometric Function using Triangle Method (without using calculator) Step 1 : let A = inverse trigonometric function (or A = first inverse trigonometric function, and B = second inverse trigonometric function Step 2 : change inverse trigonometric function to ordinary trigonometric function Step 3 : draw a right angle triangle and determine all the sides based on trigonometric function obtained in Step 2. Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3 Step 5 : evaluate the given question (and refer useful formula)
Chapter 1 : Inverse Trigonometric Function 3 Useful Formula Sum & Difference Angle Identities ( ± ) = ± ( ± ) = ∓ ( ± ) = ± ∓ Double-Angle Identities = = − = − = − = − Reciprocal Trigonometric Identities = = = = = = Pythagorean Identities + = + = + = More Formulas… Quotient Identities = = Opposite Angle Identities (−) = − () (−) = () (−) = − ()
MAT238/ MAT438 : Foundation of Applied Mathematics 4 Simple Example 1 Use Calculator to evaluate Try This ! ) −1 ൬ 1 2 ൰ = ) −1 (2) = ) −1 ൬ 1 ξ2 ൰ = ) −1 ൫ξ2൯ = ) −1 ൬ 1 ξ3 ൰ = ) −1 ൫ξ3൯ = More formulas… − ( ) = − ൬ ൰ − ( ) = − ൬ ൰ − ( ) = − ൬ ൰
Chapter 1 : Inverse Trigonometric Function 5 Simple Example 2 Without using calculator, evaluate a) [−1 ( 4 3 )] Ans : 4 3 b) [−1 ( 4 3 )] Ans : 4 5 Solution : Try This !
MAT238/ MAT438 : Foundation of Applied Mathematics 6 Tutorial 1.1a : Evaluating Inverse Trigonometric Function (without using calculator) Without using calculator, evaluate 1. (−1 ൬ 4 3 ൰) ∶ 4 3 2. (−1 ൬ 12 13൰) ∶ 12 13 3. (−1 ൬ 15 8 ൰) ∶ 15 8 4. (−1 ൬ 4 3 ൰) ∶ 4 3 5. (−1 ൬ 13 12൰) ∶ 13 12 6. (−1 ൬ 4 5 ൰) ∶ 5 4 7. (−1 ൬ 12 13൰) ∶ 5 12 8. (−1 ൬ 15 8 ൰) ∶ 17 8 9. (−1 ൬ 4 3 ൰) ∶ 3 5 10. (−1 ൬ 13 12൰) ∶ 12 13
Chapter 1 : Inverse Trigonometric Function 7 1..1.2 Evaluating composite inverse trigonometric functions Example 1/ MAR 2013/ MAT238/ Q1a (7 marks) Without using calculator, find the value of i) [ 2 + −1 1 3 ] (Ans : − ξ8 3 ) ii) [2 −1 1 3 ] (Ans : − 2ξ8 7 ) Solution : i) Step 1 : let A = inverse trigonometric function Let = −1 ( 1 3 ) Step 2 : change inverse trigonometric function to ordinary trigonometric function Step 3 : draw a right angle triangle and determine all the sides based on trigonometric function obtained in Step 2. Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3 Step 5 : evaluate the given question (and refer useful formula) ( 2 + −1 ൬ 1 3 ൰) = ( 2 + ) = 2 − 2 = (0) − (1) = − = − ξ8 3 # 1 3 ξ8 = 1 3 = ξ8 3 Using Pythagoras theorem : ඥ3 2 − 1 2 = ξ9− 1 = ξ8 Since we solve without using calculator, so no need to write in decimal adjacent hypotenuse
MAT238/ MAT438 : Foundation of Applied Mathematics 8 ii) [2 −1 1 3 ] Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3 Step 5 : evaluate the given question (and refer useful formula) • Notice that we have the same inverse trigonometric function −1 ( 1 3 ) from question (i). • So, we use step 1,2,3 from question (i) • So, we continue with step 4. (2 −1 ൬ 1 3 ൰) = (2) = 2 1 − 2 = 2ξ8 1 − ൫ξ8൯ 2 = 2ξ8 1 − 8 = 2ξ8 −7 = − 2ξ8 7 # = ξ8 1 = ξ8 No need to write in decimal, because we solve without using calculator! Just let A=inverse-trigo (only) because constant ‘2’ here means we will use double-angle identity
Chapter 1 : Inverse Trigonometric Function 9 Example 2/ SEP 2013/ MAT238/ Q1a/ (6 marks) Simplify [−1 ( +1 ) + −1 ] in terms of x. (Ans : ξ 2−1++1 ξ2 2+2+1 ) Solution : evaluate the given question (and refer useful formula) Let = −1 ( +1 ) Let = +1 Right-angle-triangle trigo-ratios (based on right-angle-triangle in step ) = ξ2 2 + 2 + 1 = + 1 ξ2 2 + 2 + 1 + 1 Using Pythagoras theorem : ඥ( + 1) 2 + 2 = ඥ 2 + 2 + 1 + 2 = ඥ2 2 + 2 + 1 Let = −1 () Let = = 1 → = 1 Right-angle-triangle trigo-ratios (based on right-angle-triangle in step ) = 1 = ξ 2 − 1 ඥ 2 − 1 1 Using Pythagoras theorem : ඥ() 2 − 1 2 = ඥ 2 − 1 (−1 ( + 1 ) + −1 ()) = ( + ) = + = ξ2 2 + 2 + 1 ∙ ξ 2 − 1 + + 1 ξ2 2 + 2 + 1 ∙ 1 = ξ 2 − 1 + + 1 ξ2 2 + 2 + 1 # opposite adjacent opposite hypotenuse
MAT238/ MAT438 : Foundation of Applied Mathematics 10 Example 3/ MAR 2014/ MAT238/ Q1a (6 marks) Use the Right Triangle method to find the value of [−1 ξ2 3 + −1 ξ2] ans : ξ14−ξ2 3ξ3 Solution : Try This !
Chapter 1 : Inverse Trigonometric Function 11 Example 4a/ SEP 2014/ MAT238/ Q1a (4 marks) Without using calculator, evaluate [2 −1 ξ3 2 ] ans : − 1 2 Solution : Try This !
MAT238/ MAT438 : Foundation of Applied Mathematics 12 Example 4b/ SEP 2014/ MAT238/ Q1b (5 marks) Simplify [−1 2 + ] ans : − 1 ξ1−4 2 Solution : i) Step 1 : let A = inverse trigonometric function Let = −1 (2) Step 2 : change inverse trigonometric function to ordinary trigonometric function Step 3 : draw a right angle triangle and determine all the sides based on trigonometric function obtained in Step 2. Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3 Step 5 : evaluate the given question (and refer useful formula) (−1 (2) + ) = ( + ) = 1 ( + ) = 1 − = 1 (−1) − (0) = 1 − = − 1 = − 1 ξ1 − 4 2 # 1 2 ඥ1 − 4 2 = 2 = 2 1 = ξ1 − 4 2 1 = ඥ1 − 4 2 Using Pythagoras theorem : ඥ1 2 − (2) 2 = ඥ1 − 4 2 opposite hypotenuse
Chapter 1 : Inverse Trigonometric Function 13 Example 5/ MAR 2015/ MAT238/ Q1a (5 marks) Express [2 −1 ( ξ 2+ )] in terms of x. ans : 2ξ −1 Solution : Try This !
MAT238/ MAT438 : Foundation of Applied Mathematics 14 Example 6/ SEP 2015/ MAT238/ Q1a (6 marks) Simplify (−1 + −1 2 ) Solution : Therefore, [−1 + −1 2] = ( + ) = − = ඥ1 − 2 ∙ 1 ξ1 + 4 2 − ∙ 2 ξ1 + 4 2 = ξ1 − 2 − 2 2 ξ1 + 4 2 # let A = first inverse trigonometric function Let A = x 1 sin− change to ordinary trigonometry draw a right angle triangle and find all the three sides find the relevant trigo-ratios sin A = x cos A = 2 1− x let A = first inverse trigonometric function Let B = −1 2 change to ordinary trigonometry tan B = 2 = 2 1 draw a right angle triangle and find all the three sides find the relevant trigo-ratios → = 2 ξ1 + 4 2 → = 1 ξ1 + 4 2 opposite hypotenuse = = 1 opposite adjacent B 2 1 ඥ1 + 4 2 A 1 ඥ1 − 2
Chapter 1 : Inverse Trigonometric Function 15 Tutorial 1.1b : Evaluating/Simplifying Inverse Trigonometric Function (using Triangle Method) Use triangle Method, to find the value of 1. ൬2−1 ൬ 4 3 ൰ − ൰ ∶ − 25 24 2. (−1 ൬ 5 13൰ − −1 ൬ 3 5 ൰) ∶ − 33 65 3. [−1 3 4 + −1 7 ξ130 ] ∶ 5ξ130 4. ( − 2 −1 ൬ 4 5 ൰) ∶ − 24 7 5. ൬−1 ൬ 4 5 ൰ + −1 ൬ 12 13൰൰ ∶ 65 16 Use triangle Method, to simplify the following 6. ( + −1 (ඥ1 − 9 2)) ∶ − 1 3 7. ൫ − 2 −1 (2)൯ ∶ −1 8 2 − 1 1 1 − 8 2 8. (−1 () − −1 ()) ∶ 1 − 2 2 2ξ1 − 2 9. ൬−1 (ඥ1 − 2) − −1 ൬ 1 2 ൰൰ ∶ − 2ξ1 − 2 ξ1 + 4 2 10. ൬−1 ൬ − 1 ൰ + −1 ()൰ ∶ − ( − 1)ξ 2 − 1 ξ2 2 − 2 + 1
MAT238/ MAT438 : Foundation of Applied Mathematics 16 1.2 Derivative of Inverse Trigonometric Functions 1. (−1 )= 1 ξ1 − 2 2. (−1 )= −1 ξ1 − 2 3. (−1 )= 1 1 + 2 4. (−1 )= −1 1 + 2 5. (−1 )= 1 ξ 2 − 1 6. (−1 )= −1 ξ 2 − 1 Simple Examples Differentiate with respect to x. a) y = sin−1 (2x) b) y = tan−1 (3x 2 ) c) y = cos−1( 2 ) d) y = sin−1 (e 4x ) e) y = sec−1 (ln x 2 ) solution : a) y = sin−1 (2x) = b) y = tan−1 (3x 2 ) = 1 ඥ1 − (2) 2 2 (−1) = 1 ξ1 − 2 (2) = 2 = 2 ξ1 − 4 2 1 1 + (3 2) 2 6 (−1) = 1 1 + 2 (3 2 ) = 6 = 6 1 + 9 4
Chapter 1 : Inverse Trigonometric Function 17 c) y = cos−1( 2 ) d) y = sin−1 (e 4x ) = e) y = sec−1 (ln x 2 ) = −1 √1 − ( 2 ) 2 1 2 (−1) = −1 ξ1 − 2 ൬ 1 2 ൰ = 1 2 = −1 √ 4 − 2 4 ∙ 1 2 = = −1 ( ξ4 − 2 2 ) ∙ 1 2 = −2 ξ4 − 2 ∙ 1 2 = −1 ξ4 − 2 # 1 ඥ1 − ( 4) 2 4 (−1) = 1 ξ1 − 2 ( ) = = 4 4 ξ1 − 8 4 # (4) = 4 1 2ඥ( 2) 2 − 1 1 2 (−1) = 1 ξ 2 − 1 () = 1 = 2 2ඥ( 2) 2 − 1 2 # ( 2 ) = 2
MAT238/ MAT438 : Foundation of Applied Mathematics 18 More Examples (from previous semester papers) Example 1/ OCT 2007/ MAT238/ Q1b (5 marks) Differentiate = ξ−1 2 + −1 2 with respect to x. ans : 1 ඥ(−1 2)(1−4 2) + −1 2 + 2 1+ 4 Solution : = ඥ−12 + −1 2 = + Differentiate wrt x: = ′ + ′ → = 1 ඥ(−12)(1 − 4 2) + −1 2 + 2 1 + 4 # = ඥ−12 = (−12) 1 2 Using generalized power rule (combine with inverse-trigo rule) ′ = 1 2 (−12) − 1 2 ∙ 1 ඥ1 − (2) 2 ∙ 2 ′ = 1 2 ∙ 1 ξ−12 ∙ 1 ξ1 − 4 2 ∙ 2 ′ = 1 ඥ(−12)(1 − 4 2) = −1 2 Using product rule (combine with exponential rule and inverse-trigo rule) ′ = ′ + ′ ′ = −1 2 + ∙ 1 1 + ( 2) 2 ∙ 2 ′ = −1 2 + 2 1 + 4 = = −1 2 ′ = ′ = 1 1 + ( 2) 2 ∙ 2
Chapter 1 : Inverse Trigonometric Function 19 Example 2/ APR 2007/ MAT238/ Q1b (6 marks) Find the derivative of the function = −1 ( 2 2 ) ans : 2−2 2 2 ξ4 2−2 2 Solution : = −1 ൬ 2 2 ൰ Differentiate wrt x: = −1 √1 − ( 2 2 ) 2 ∙ ′ − ′ 2 = −1 √1 − 22 4 2 ∙ 4 22 − 2 2 (2) 2 = −1 √ 4 2 − 22 4 2 ∙ 4 22 − 2 2 4 2 = −1 ( ξ4 2 − 22 2 ) ∙ 4 22 − 2 2 4 2 = −2 ξ4 2 − 22 ∙ 4 22 − 2 2 2 ∙ 2 = −1 ξ4 2 − 22 ∙ 4 22 − 2 2 2 = −(4 22 − 2 2) 2ξ4 2 − 22 = −2(2 22 − 2) 2ξ4 2 − 22 = −(2 22 − 2) ξ4 2 − 22 = 2 − 2 22 ξ4 2 − 22 # = 2 = 2 ′ = 22 ∙ 2 ′ = 2 ′ = 2 22
MAT238/ MAT438 : Foundation of Applied Mathematics 20 Example 3/ OCT 2006/ MAT238/ Q1b (6 marks) Find if = 3 −1 (2) ans : 3 3(1+4 2 ) −1 2 1+42−2 3 Solution : = 3 −1 (2) Differentiate implicitly wrt x: = ′ + ′ = 3 3 −1 (2) + 2 3 1 + 4 2 − 2 3 1 + 4 2 = 3 3 −1 (2) (1 − 2 3 1 + 4 2 ) = 3 3 −1 (2) ( 1 + 4 2 − 2 3 1 + 4 2 ) = 3 3 −1 (2) = 3 3 (1 + 4 2 )−1 (2) 1 + 4 2 − 2 3 # = 3 = −1 (2) ′ = 3 3 ′ = 1 1 + (2) 2 ∙ 2 ′ = 2 1 + 4 2
Chapter 1 : Inverse Trigonometric Function 21 Example 4/ SEP 2011/ MAT238/ Q1b (5 marks) Differentiate = (−1 5 ) 2 + 4 −1 ξ2 with respect to x. ans : = 10 −1 5 1+25 2 + 4 ξ2−4 2 Solution : = (−15) 2 + 4 −1ξ2 = + Differentiate wrt x: = ′ + ′ → = 10 −1 (5) 1 + 25 2 + 4 ξ2 − 4 2 # = 4 −1ξ2 Using product rule (combine with exponential rule and inverse-trigo rule) ′ = 4 ∙ 1 √1 − ൫ξ2൯ 2 ∙ 1 ξ2 ′ = 4 ξ2ξ1 − 2 ′ = 4 ඥ2(1 − 2) ′ = 4 ξ2 − 4 2 = (−15) 2 Using generalized power rule (combine with inverse-trigo rule) ′ = 2(−15) 1 ∙ 1 1 + (5) 2 ∙ 5 ′ = 10 −1 (5) 1 + 25 2 = ξ2 = (2) 1 2 ′ = 1 2 (2) − 1 2 ∙ 2 ′ = (2) − 1 2 ′ = 1 ξ2
MAT238/ MAT438 : Foundation of Applied Mathematics 22 Example 5/ MAR 2013/ MAT238/ Q1b (4 marks) Find if = −1 ൫ξ 3 ൯ • () + −1 2 ans : = 3 2 3√1− 2 3 + −1 ൫ ξ 3 ൯ − 2 ξ1−4 2 Solution : = −1 ൫ξ 3 ൯ ∙ + −12 = + Differentiate wrt x: = ′ + ′ = 3 2 3√1 − 2 3 + −1 ൫ξ 3 ൯ + −2 ξ1 − 4 2 → = 3 2 3√1 − 2 3 + −1 ൫ξ 3 ൯ − 2 ξ1 − 4 2 # = −1 ൫ξ 3 ൯ ∙ Using product rule (combine with inverse-trigo rule and ln rule) ′ = ′ + ′ ′ = 3 2 3√1 − 2 3 + −1 ൫ξ 3 ൯ = −12 Using inverse-trigo rule ′ = −1 ඥ1 − (2) 2 ∙ 2 ′ = −2 ξ1 − 4 2 = −1 ൫ξ 3 ൯ = −1 ( 1 3) = ′ = 1 √1 − ( 1 3) 2 ∙ 1 3 − 2 3 ′ = 1 ′ = 1 3 2 3√1 − 2 3 = ξ 3 = 1 3 ′ = 1 3 − 2 3
Chapter 1 : Inverse Trigonometric Function 23 Example 6/ SEP 2014/ MAT238/ Q1c (5 marks) Find the derivative of if = 2 −1 ( 4 ) ans : = 2 −1 ( 4 ) − 8 ξ 2−16 Solution : Try This ! = 2 −1 ൬ 4 ൰ = = ′ + ′ = 2 −1 ൬ 4 ൰ + (2) ൬− 4 ξ 2 − 16 ൰ = 2 −1 ൬ 4 ൰ − 8 ξ 2 − 16 # = 2 = −1 ൬ 4 ൰ ′ = 2 ′ = 1 √1 − ( 4 ) 2 ∙ − 4 2 ′ = 1 √1 − 16 2 ∙ − 4 2 ′ = 1 √ 2 − 16 2 ∙ − 4 2 ′ = 1 ( ξ 2 − 16 ) ∙ − 4 2 ′ = ξ 2 − 16 ∙ − 4 2 ′ = − 4 ξ 2 − 16 = 4 = 4 −1 ′ = −4 −2 = − 4 2
MAT238/ MAT438 : Foundation of Applied Mathematics 24 Example 7/ MAR 2015/ MAT238/ Q1b (5 marks) Differentiate = 2 3 −1 ξ − 3 ans : = 3 ξ(1+) + 6 3 −1 ξ − 3 2 3 Solution : Substitute A’ & B’ into ① = 6 3 −1 ξ + 2 3 2ξ(1 + ) − 3 2 3 # Try This ! = 2 3 −1 ξ = 2 3 = −1 ξ ′ = 2 3 ∙ 3 ′ = 1 1 + ൫ξ൯ 2 ∙ 1 2ξ ′ = 6 3 ′ = 1 2ξ(1 + ) ′ = ′ + ′ ′ = 6 3 −1 ξ + 2 3 2ξ(1 + ) = 2 3 −1 ξ − 3 = − = ′ − ′ − − − − ① = 3 ′ = 2 3 ∙ 3 ′ = 3 2 3
Chapter 1 : Inverse Trigonometric Function 25 Tutorial 1.2 : Differentiation involving Inverse Trigonometric Functions ℎ . 1. = ξ −1 ൬ 2 ൰ 2. = (−1 2) 3 + 2 −1 ξ2 3. = −1 ൬ 3 3 ൰ 4. = ൫ξ൯ + −1 () 5. = 2 2 −1 ξ − (2) ℎ . 6. () = ඥ−1(2) + 2 −1 ൫ξ൯ 7. ( ) = 2 −1 (2) 8. 2 ൫ 2 ൯ = −1 ൫ξ3൯ − ( 2) 9. −1() = −1 ൫(2)൯ 10. ( 2 ) = −1 ൬ ൰
MAT238/ MAT438 : Foundation of Applied Mathematics 26 Answer Tutorial 1.2 1. = −1 ( 2 ) 2ξ − 2ξ 2 + 4 2. = 6(−1 2) 2 ξ1 − 4 2 + ξ2 ξ(1 + 2) 3. = 3 (3 3 − 1) ξ9 2 − 2 3 4. = − 1 2 − 1 ඥ1 − (2) 5. = 4 2 −1 ξ + 2 2 2ξξ1− + 2 (2) (2) 6. = 2 2 −1 ൫ξ൯ − ൬ 2 2ξξ1 − ൰ ( 1 − 1 (1 + 4 2)ඥ−1(2) ) 7. = ( ) − ൬ 2 (1 + 4 2) −1(2) ൰ ( 2 + ) 8. = 3 2ඥ3(1 − 3) − 2 − 4 2 ൫ 2 ൯ 9. = ξ1 − 2 − ()൫1 + 2 (2)൯ −1 ൫(2)൯ ξ1 − 2൫1 + 2(2)൯ −1() −1൫(2)൯ 10. = + 2( 2 + 2 ) 2 ( 2 )( 2 )
Chapter 1 : Inverse Trigonometric Function 27 1.3 Integration involving Inverse Trigonometric Functions Example 1/ MAR 2014/ MAT238/ Q1cii (4 marks) Evaluate ∫ 2 ξ1 − 2 2 ans : ξ2 −1 ൫ξ2൯ + Solution : 1. ∫ 1 ξ 2 − 2 = −1 ( ) + 2. ∫ 1 2 + 2 = 1 −1 ( ) + 3. ∫ 1 ξ 2 − 2 = 1 −1 ( ) + Appendix (given in Final Examination) 14. ∫ ξ 2 − 2 = −1 ( ) + 15. ∫ 2 + 2 = 1 −1 ( ) + 16. ∫ ξ 2 − 2 = 1 −1 ( ) + ∫ 2 ξ1 − 2 2 = 2 ∫ 1 √(1) 2 − ൫ξ2൯ 2 = 2 ∫ 1 ξ 2 − 2 ∙ ξ2 = 2 ξ2 ∫ 1 ξ 2 − 2 = ξ2 −1 ( ) + = ξ2 −1 ( ξ2 1 ) + = ξ2 −1 ൫ξ2൯ + # = 1 = ξ2 = ξ2 = ξ2 Where = = () Where = = () 2 ξ2 = ξ2 Because, 2 ξ2 = ൫ξ2൯ 2 ξ2 = ξ2ξ2 ξ2 = ξ2
MAT238/ MAT438 : Foundation of Applied Mathematics 28 Example 2/ SEP 2014/ MAT238/ Q2c (5 marks) Solve the following integral ∫ 2 4 + 3 2 ans : 1 ξ3 −1 ( ξ3 2 ) + Solution : ∫ 2 4 + 3 2 = 2 ∫ 1 (2) 2 + ൫ξ3൯ 2 = 2 ∫ 1 2 + 2 ∙ ξ3 = 2 ξ3 ∫ 1 2 + 2 = 2 ξ3 ∙ 1 −1 ( ) + = 2 ξ3 ∙ 1 2 −1 ( ξ3 2 ) + = 1 ξ3 −1 ( ξ3 2 ) + # = 2 = ξ3 = ξ3 = ξ3 Where = = ()
Chapter 1 : Inverse Trigonometric Function 29 Example 3/ MAR 2013/ MAT238/ Q1c (4 marks) Evaluate the integral ∫ 5 ( − 5) 2 + 5 Ans : ξ5 −1 ( −5 ξ5 ) + Solution : ∫ 5 ( − 5) 2 + 5 = 5 ∫ 1 ( − 5) 2 + ൫ξ5൯ 2 = 5 ∫ 1 2 + 2 = 5 ∙ 1 −1 ( ) + = 5 ∙ 1 ξ5 −1 ൬ − 5 ξ5 ൰ + = 5 ξ5 −1 ൬ − 5 ξ5 ൰ + = ξ5 −1 ൬ − 5 ξ5 ൰ + # = ξ5 = − 5 = 1 = 5 ξ5 = ξ5 Because, 5 ξ5 = ൫ξ5൯ 2 ξ5 = ξ5ξ5 ξ5 = ξ5 Where = = ()
MAT238/ MAT438 : Foundation of Applied Mathematics 30 Example 4/ MAR 2015/ MAT238/ Q1c (5 marks) Use the substitution u = e 2x to solve ∫ 4 2 4 + 9 Ans : 2 3 −1 ( 2 3 ) + Solution : ∫ 4 2 4 + 9 = 4 ∫ 2 ( 2) 2 + (3) 2 = 4 ∫ 2 2 + 2 ∙ 2 2 = 2 ∫ 1 2 + 2 = 2 ∙ 1 −1 ( ) + = 2 ∙ 1 3 −1 ( 2 3 ) + = 2 3 −1 ( 2 3 ) + # = 3 = 2 = 2 2 = 2 2 ( 2 ) 2 = 4 Because, ( ) = Where = = ()
Chapter 1 : Inverse Trigonometric Function 31 Example 5 : Evaluate ∫ 2 ඥ4 − ( ) 2 Solution : Example 6 : Evaluate ∫ 3 3 8 + 9 Solution : ∫ 2 ඥ4 − ( ) 2 = 2 ∫ 1 ඥ(2) 2 − ( ) 2 = 2 ∫ 1 ξ 2 − 2 ∙ = 2 ∫ 1 ξ 2 − 2 = 2 −1 ( ) + = 2 −1 ൬ 2 ൰ + # = 2 = = 1 = Where = = () ∫ 3 3 8 + 9 = 3 ∫ 3 ( 4) 2 + (3) 2 = 3 ∫ 3 2 + 2 ∙ 4 3 = 3 4 ∫ 1 2 + 2 = 3 4 ∙ 1 −1 ( ) + = 3 4 ∙ 1 3 −1 ( 4 3 ) + = 1 4 −1 ( 4 3 ) + # = 3 = 4 = 4 3 = 4 3 Where = = ()
MAT238/ MAT438 : Foundation of Applied Mathematics 32 Example 7 : Evaluate ∫ 5 5 ξ25 − 2 5 Solution : Example 8 : Evaluate ∫ ඥ25 − ( + 2) 2 Solution : ∫ 5 5 ξ25 − 2 5 = ∫ 5 5 ඥ(5) 2 − ( 5 ) 2 = ∫ 5 5 ξ 2 − 2 ∙ 5 5 5 = 1 5 ∫ 1 ξ 2 − 2 = 1 5 −1 ( ) + = 1 5 −1 ൬ 5 5 ൰ + # = 5 = 5 = ( 5 5 ) ∙ 5 = 5 5 5 Where = = () ∫ ඥ25 − ( + 2) 2 = ∫ 1 ඥ(5) 2 − ( + 2) 2 = ∫ 1 ξ 2 − 2 = −1 ( ) + = −1 ൬ + 2 5 ൰ + # = 5 = + 2 = 1 = Where = = ()
Chapter 1 : Inverse Trigonometric Function 33 Example 9 : Evaluate ∫ 3 ξ1 − 2 3 3 9 Solution : ∫ 3 ξ1 − 2 3 = ∫ 3 ඥ(1) 2 − ( 3) 2 = ∫ 3 ξ 2 − 2 ∙ −3 3 = − 1 3 ∫ 1 ξ 2 − 2 = − 1 3 −1 ( ) + = − 1 3 −1 ൬ 3 1 ൰ + = − 1 3 −1 ( 3) + # Hence, ∫ 3 ξ1 − 2 3 3 9 = − 1 3 [ −1 ( (3 ∙ 3 )) − −1 ( (3 ∙ 9 ))] = − 1 3 [ −1 (()) − −1 ( ( 3 ))] = − 1 3 [ −1 (−1) − −1 ൬ 1 2 ൰] = 0.6981 # ( 2 9 ) = 1 = 3 = (− 3) ∙ 3 = −3 3 Where = = () Set your calculator in mode ‘radian’
MAT238/ MAT438 : Foundation of Applied Mathematics 34 Example 10 : Evaluate ∫ 3 3 16 + 2 3 3 12 Solution : ∫ 3 3 16 + 2 3 = ∫ 3 3 (4) 2 + ( 3) 2 = ∫ 3 3 2 + 2 ∙ −3 3 3 = − 1 3 ∫ 1 2 + 2 = − 1 3 ∙ 1 −1 ( ) + = − 1 3 ∙ 1 4 −1 ൬ 3 4 ൰ + = − 1 12 −1 ൬ 3 4 ൰ + # Hence, ∫ 3 3 16 + 2 3 3 12 = − 1 12 [ −1 ( (3 ∙ 3 )) − −1 ( (3 ∙ 12))] = − 1 12 [ −1 (()) − −1 ( ( 4 ))] = − 1 12 [ −1 (0) − −1 (1)] = − 1 12 [ 0 − 4 ] = 48 # = 4 = 3 = − 3 3 ∙ 3 = −3 3 3 Where = = () Set your calculator in mode ‘radian’
Chapter 1 : Inverse Trigonometric Function 35 1.4 Integration of Inverse Trigonometric Functions involving Completing the Square Method Procedure completing the square method, arrange based on the highest power of x, such that 2 + + coefficient of x 2 must positive 1. If coefficient of x 2 not equal to positive 1, then factorize coefficient of x as follow [ 2 + + ] put + ( )2 − ( )2 afterthe term containing x, where the expression inside ( ) must be 1 2 •coefficient of x = 1 2 • = 2 (as follow) [ 2 + + ൬ 2 ൰ 2 − ൬ 2 ൰ 2 + ] write the first three terms into the form of ( )2 and simplify the next terms Or, using formula, 2 + + = [( + 2 ) 2 − ( 2 ) 2 + ] When we use completing the square method? When: i) numerator is a constant, and ii) denominator is in the form of quadratic expression with the term containing ‘x’
MAT238/ MAT438 : Foundation of Applied Mathematics 36 Example 1 Evaluate ∫ ξ21 − 4 − 2 Solution : Using completing the square, 21 − 4 − 2 = − 2 − 4 + 21 = −( 2 + 4 − 21) = −( 2 + 4 + (2) 2 − (2) 2 − 21) = −(( + 2) 2 − 25) = 25 − ( + 2) 2 Hence, ∫ ξ21 − 4 − 2 = ∫ ඥ25 − ( + 2) 2 Complete this integral. You can Refer page 27, no. Re-arrange : 2 + + coefficient of x 2 must positive 1 (factorize coefficient of x 2 if necessary) put + ( )2 − ( )2 after the term containing x, where the expression inside ( ) must be 1 2 •coef x write the first three terms into the form of ( )2 and simplify the remaining terms
Chapter 1 : Inverse Trigonometric Function 37 Example 2 Evaluate ∫ 2 2 − 12 + 26 ans : 1 4 −1 ( −3 2 ) + Solution : Try This !
MAT238/ MAT438 : Foundation of Applied Mathematics 38 Example 3 Evaluate ∫ 5 2 2 + 4 + 9 ans : 5 ξ14 −1 ( ξ2 ξ7 ( + 1)) + Solution : Using completing the square, 2 2 + 4 + 9 = 2 [ 2 + 2 + 9 2 ] = 2 [ 2 + 2 + (1) 2 − (1) 2 + 9 2 ] = 2 [( + 1) 2 + 7 2 ] = 2 [( + 1) 2 + ( ξ7 ξ2 ) 2 ] Hence, ∫ 5 2 2 + 4 + 9 = 5 ∫ 1 2 [( + 1) 2 + ( ξ7 ξ2 ) 2 ] = 5 2 ∫ 1 ( + 1) 2 + ( ξ7 ξ2 ) 2 = 5 2 ∫ 1 2 + 2 = 5 2 ∙ 1 −1 ( ) + = 5 2 ∙ 1 ( ξ7 ξ2 ) −1 ( + 1 ( ξ7 ξ2 ) ) + = 5 ξ14 −1 ( ξ2( + 1) ξ7 ) + = 5 ξ14 −1 ( ξ2 ξ7 ( + 1)) + # = ξ7 ξ2 = + 1 = 1 = Where = = () Note : ξξ = ൫ξ൯ = 5 2 ∙ 1 ൬ ξ7 ξ2 ൰ = 5 2 ∙ ξ2 ξ7 = 5 ξ2ξ2 ∙ ξ2 ξ7 = 5 ξ2 ∙ 1 ξ7 = 5 ξ14 Note : + 1 ( ξ7 ξ2 ) = ( + 1) ÷ ( ξ7 ξ2 ) = ( + 1) 1 ∙ ( ξ2 ξ7 ) = ξ2 ξ7 ( + 1)
Chapter 1 : Inverse Trigonometric Function 39 Example 4 Evaluate ∫ ξ6 − 2 Solution : Using completing the square, − 2 + 6 = −( 2 − 6) = −( 2 − 6 + (−3) 2 − (−3) 2 ) = −(( − 3) 2 − 9) = 9 − ( − 3) 2 Hence, ∫ ξ6 − 2 = ∫ 1 ඥ9 − ( − 3) 2 = ∫ 1 ඥ(3) 2 − ( − 3) 2 = ∫ 1 ξ 2 − 2 = −1 ( ) + = −1 ൬ − 3 3 ൰ + # = 3 = − 3 = 1 = Where = = ()
MAT238/ MAT438 : Foundation of Applied Mathematics 40 Example 5 Evaluate ∫ 2 − 6 + 13 4 2 ans : 1 2 −1 ( −3 2 ) + Solution : Using completing the square, 2 − 6 + 13 = 2 − 6 + (−3) 2 − (−3) 2 + 13 = ( − 3) 2 + 4 therefore, ∫ 2 − 6 + 13 = ∫ 1 ( − 3) 2 + 4 = ∫ 1 ( − 3) 2 + 4 = ∫ 1 ( − 3) 2 + (2) 2 = ∫ 1 2 + 2 = 1 −1 ( ) + = 1 2 −1 ൬ − 3 2 ൰ + Hence, ∫ 2 − 6 + 13 4 2 = [ 1 2 −1 ൬ − 3 2 ൰] 2 4 = 1 2 [−1 ൬ − 3 2 ൰] 2 4 = 1 2 [−1 ൬ 4 − 3 2 ൰ − −1 ൬ 2 − 3 2 ൰] = 1 2 [−1 ൬ 1 2 ൰ − −1 ൬ −1 2 ൰] = 0.4636 # = 2 = − 3 = 1 = Where = = () Set your calculator in mode ‘radian’
Chapter 1 : Inverse Trigonometric Function 41 Example 6/ OCT 2012/ MAT238/ Q2c (ii) (5 marks) Integrate ( ) dx x x − + 8 3 . Solution : Using completing the square, Hence, −( + 8) = − 2 − 8 = −( 2 + 8) = −( 2 + 8 + (+4) 2 − (+4) 2 ) = −(( + 4) 2 − 16) = 16 − ( + 4) 2 ∫ 3 ඥ−( + 8) = 3 ∫ 1 ඥ16 − ( + 4) 2 = 3 ∫ 1 ඥ(4) 2 − ( + 4) 2 = 3 ∫ 1 ξ 2 − 2 = 3 −1 ( ) + = 3 −1 ൬ + 4 4 ൰ + # = 4 = + 4 = 1 = Re-arrange : 2 + + if coefficient of 2 not equal to positive one, then factorize the coefficient of 2 i) Put +( ) − ( ) after the term containing x ii) inside bracket must ( × coeffiecient of x) The first three terms → write ( ) The rest → simplify to avoid square root negative→ multiply negative sign with term inside bracket
MAT238/ MAT438 : Foundation of Applied Mathematics 42 Example 7 Integrate ( ) dx x x − − 8 3 Solution : Using completing the square, Hence, −( − 8) = − 2 + 8 = −( 2 − 8) = −( 2 − 8 + (−4) 2 − (−4) 2 ) = −(( − 4) 2 − 16) = 16 − ( − 4) 2 ∫ 3 ඥ−( − 8) = 3 ∫ 1 ඥ16 − ( − 4) 2 = 3 ∫ 1 ඥ(4) 2 − ( − 4) 2 = 3 ∫ 1 ξ 2 − 2 = 3 −1 ( ) + = 3 −1 ൬ − 4 4 ൰ + # = 4 = − 4 = 1 = Re-arrange : 2 + + if coefficient of 2 not equal to positive one, then factorize the coefficient of 2 i) Put +( ) − ( ) after the term containing x ii) inside bracket must ( × coeffiecient of x) The first three terms → write ( ) The rest → simplify to avoid square root negative→ multiply negative sign with term inside bracket
Chapter 1 : Inverse Trigonometric Function 43 Example 8 Integrate − − 2 14 12x 2x dx . Solution : Using completing the square, Hence, 14 − 12 − 2 2 = −2 2 − 12 + 14 = −2( 2 + 6 − 7) = −2( 2 + 6 + (+3) 2 − (+3) 2 − 7) = −2(( + 3) 2 − 16) = 2[16 − ( + 3) 2 ] ∫ ξ14 − 12 − 2 2 = ∫ 1 ඥ2[16 − ( + 3) 2] = 1 ξ2 ∫ 1 ඥ16 − ( + 3) 2 = 1 ξ2 ∫ 1 ඥ(4) 2 − ( + 3) 2 = 1 ξ2 ∫ 1 ξ 2 − 2 = 1 ξ2 −1 ( ) + = 1 ξ2 −1 ൬ + 3 4 ൰ + # = 4 = + 3 = 1 = Re-arrange : 2 + + if coefficient of 2 not equal to positive one, then factorize the coefficient of 2 i) Put +( ) − ( ) after the term containing x ii) inside bracket must ( × coeffiecient of x) The first three terms → write ( ) The rest → simplify to avoid square root negative→ multiply negative sign with term inside bracket