MAT238/ MAT438 : Foundation of Applied Mathematics 94 Example 4/ OCT 2007/ MAT238/ Q1d (4 marks) Without using a calculator, find the value of ( 4) 1 cosh sinh ln − . Solution: Example 5/ DEC 2008/ MAT238/ Q2b (6 marks) By using the definition, find the exact value of ( 2) 1 tanh sinh ln − . [Hint : − + = − x x tanh x ln 1 1 2 1 1 ] Solution: ℎ = − − 2 ℎ( 4) = 4 − − 4 2 = 4 − 1 4 2 = 4− 1 4 2 = 15 8 Therefore, ℎ −1 [ℎ( 4)] = ℎ −1 ( 15 8 ) = ( ( 15 8 ) + √( 15 8 ) 2 −1 ) = ( 15 + √161 8 ) # = ℎ = − − 2 ℎ( 2) = 2 − − 2 2 = 2 − 1 2 2 = 2− 1 2 2 = 3 4 Therefore, ℎ −1 [ℎ( 2)] = ℎ −1 ( 3 4 ) = 1 2 | 1+ 3 4 1− 3 4 | = 1 2 (7) = (7) 1 2 = √7 # Note : Thus, 2 = 2 − 2 ≠ −2, − 2 = 1 2 = 1 2 = Since we solve without using calculator, so no need to write in decimal Since we solve without using calculator, so no need to write in decimal
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 95 TUTORIAL 2.4 : Evaluate Inverse Hyperbolic Function (without using Calculator) Without using a calculator, find the value of 1. ℎ −1 [ℎ( 0.5)] ∶ −√7 2. 2 ℎ −1 [ 1 3 ℎ( 2)] ∶ ( 5 3 ) 3. ℎ −1 [8 ℎ( 4)] ∶ (17 + √290) 4. ℎ −1 [ℎ((0.5))] ∶ ( −3 + √34 5 ) 5. ℎ −1 [ℎ( 0.2)] ∶ ( −12 + √119 5 ) 6. ℎ −1 [ℎ( 0.2)] ∶ 1 2 ( 7 17) 2.2.6 Derivatives of Inverse Hyperbolic Function (comparison between Inverse Hyperbolic and Inverse Trigonometric Function Inverse Hyperbolic Function Inverse Trigonometric Function (sinh x) dx d −1 = 2 1 1 + x (cosh x) dx d −1 = 1 1 2 x − (tanh x) dx d −1 = 2 1 1 − x (sin x) dx d −1 = 2 1 1 − x (cos x) dx d −1 = 2 1 1 − x − (tan x) dx d −1 = 2 1 1 + x
MAT238/ MAT438 : Foundation of Applied Mathematics 96 Simple Examples Differentiate with respect to x. a) y = sinh−1 (3x) b) y = tanh−1 (2x 3 ) c) y = cosh−1 2 3x d) y = sinh−1 (e 5x ) e) y = tanh−1 (ln x3 ) Solution : APPENDIX (given in Final Examination) 15. = + − − C a x sin a x dx 1 2 2 16. C a x tan a 1 a x dx 1 2 2 = + + − 17. C a x a x a x dx = + + − 1 2 2 sec 1 18. C a x x a dx = + − − 1 2 2 cosh = x + x − a + C 2 2 ln 19. C a x x a dx + = + − 1 2 2 sinh = x + x + a + C 2 2 ln 20. + + + = − + = − − − C, if x a a x coth a 1 C, if x a a x tanh a 1 C x a x a ln 2a 1 a x dx 1 1 2 2 ) = ℎ −1 (3) = 1 √1 +(3) 2 ∙ 3 = 3 √1+ 9 2 # ) = ℎ −1 (2 3 ) = 1 1 − (2 3) 2 ∙ 6 2 = 6 2 √1 − 4 6 # ) = ℎ −1 ( 3 2 ) = 1 √( 3 2 ) 2 −1 ∙ 3 2 = 1 √ 9 2 4 − 1 ∙ 3 2 = 1 √ 9 2 − 4 4 ∙ 3 2 = 1 ( √9 2 − 4 2 ) ∙ 3 2 = 2 √9 2 − 4 ∙ 3 2 = 3 √9 2 − 4 # ) = ℎ −1 ( 5 ) = 1 √1+ ( 5) 2 ∙ 5 ∙ 5 = 5 5 √1+ 10 # ) = ℎ −1 ( 3 ) = 1 1− ( 3) 2 ∙ 1 3 ∙ 3 2 = 1 1− 2 3 ∙ 3 = 3 − 2 3 #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 97 More Examples Example 1/ APR 2006/ MAT238/ Q2a(ii) (3 marks) Find the derivative of ( ) 3 2 1 y x sinh x − = Solution: Example 2/ OCT 2006/ MAT238/ Q2a (6 marks) Differentiate cos x cosh x y 2 2 2 −1 = with respect to x. Solution: = 2 (ℎ −1) 3 Differentiate with respect to x, = ′ + ′ = (ℎ −1) 3 ∙ 2 + 2 ∙ 3(ℎ −1) 2 ∙ 1 √1 + 2 = (ℎ −1) 3 ∙ 2 + 2 ∙ 3(ℎ −1) 2 ∙ 1 √1 + 2 = 2(ℎ −1) 3 + 3 2 (ℎ −1) 2 √1 + 2 # = ℎ −1 (2) 2(2) Differentiate with respect to x (using quotient rule), = ′ + ′ 2 = 2 (2) ∙ 1 √(2) 2 − 1 ∙ 2 − ℎ −1 (2) ∙ 2 (2) ∙ − (2) ∙ 2 (2(2)) 2 = 2 2 (2) √4 2 − 1 + 4 (2) (2) ℎ −1 (2) 4(2) = 2 (2) √4 2 − 1 + 4 (2) ℎ −1 (2) 3(2) #
MAT238/ MAT438 : Foundation of Applied Mathematics 98 Example 3/ NOV 2005/ MAT238/ Q2b (6½ marks) Differentiate the implicit function sinh x cosh(xy ) 2y 1 + = − with respect to x. Ans : x 1 2 x sinh(xy) 1 y x 1 sinh(xy) dx dy 2 2 + − + + = Solution: ℎ −1 + ℎ() = 2 Differentiate implicitly with respect to x, 1 √1 + 2 + ℎ() [ ′ + ′] = 2 1 √1 + 2 + ℎ() [(1) + ] = 2 1 √1 + 2 + ℎ() + ℎ() = 2 1 √1 + 2 + ℎ() = 2 − ℎ() 1 √1 + 2 + ℎ() = (2 − ℎ()) Multipying each term by √1 + 2 , 1 √1 + 2 ∙ √1 + 2 + ∙ √1 + 2 ℎ() = ∙ √1 + 2(2 − ℎ()) 1 + √1 + 2 ℎ() = √1 + 2(2 − ℎ()) = 1 + √1 + 2 ℎ() √1 + 2(2 − ℎ()) #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 99 Example 4/ APR 2007/ MAT238/ Q2c (5 marks) Find dx dy if = − 2 1 x xy sinh tan . Solution: = ℎ −1 ( 2 ) Differentiate implicitly with respect to x, ′ + ′ = 1 √1 + ( ( 2 )) 2 ∙ 2 ( 2 ) ∙ 1 2 (1) + = 1 √1 + ( ( 2 )) 2 ∙ 2 ( 2 ) ∙ 1 2 + = 1 √2 ( 2 ) ∙ 2 ( 2 ) ∙ 1 2 + = 1 ( 2 ) ∙ 2 ( 2 ) ∙ 1 2 + = 1 2 ( 2 ) = 1 2 ( 2 ) − = 1 2 ( 2 ) − # = = ′ = 1 ′ = (ℎ −1) = 1 √1 + 2 (ℎ ) = ℎ 2 1 + 2 = 2 = 1 2 ( 2 ) − 2 2 = ( 2 ) 2 − 2 2 = ( 2 ) − 2 2 #
MAT238/ MAT438 : Foundation of Applied Mathematics 100 Example 5/ SEP 2011/ MAT238/ Q2b (7 marks) Find dx dy if ( ) cosh x ln x = y tanh x + e − 3 3 1 . Solution: ( 3 ) = ℎ −13 + ℎ Differentiate implicitly with respect to x, 1 3 ∙ 3 2 = ′ + ′ + ℎ ∙ ℎ 3 = (ℎ −13) ( ) + ∙ 1 1 − (3) 2 ∙ 3 + ℎ ∙ ℎ 3 = (ℎ −13) ( ) + 3 1 − 9 2 + ℎ ∙ ℎ 3 − 3 1 − 9 2 − ℎ ℎ = (ℎ −13) ( ) = 3 − 3 1 − 9 2 − ℎ ℎ ℎ −13 #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 101 Example 6/ DEC 2008/ MAT238/ Q2c (6 marks) Find dx dy if ln (x y ) (x tanh x) 2 2 1 3 1 − + = . Solution: ( 2 + 2 ) = 1 3 ( ℎ −1) Multiplying by 3, 3 ( 2 + 2 ) = ℎ −1 Differentiate implicitly with respect to x, 3 ∙ 1 2 + 2 (2 + 2 ) = ′ + ′ 3 ∙ 1 2 + 2 (2 + 2 ) = (ℎ −1)(1) + ∙ 1 1 − 2 6 + 6 2 + 2 = ℎ −1 + 1 − 2 6 + 6 = ( 2 + 2 ) (ℎ −1 + 1 − 2 ) 6 = ( 2 + 2 ) (ℎ −1 + 1 − 2 ) − 6 = ( 2 + 2 ) 6 (ℎ −1 + 1 − 2 ) − 6 6 = ( 2 + 2 ) 6 (ℎ −1 + 1 − 2 ) − #
MAT238/ MAT438 : Foundation of Applied Mathematics 102 Tutorial 2.5 : Differentiation involving Inverse Hyperbolic Functions ℎ . 1. = √2 ℎ −1 ( 3 ) 2. = (ℎ −1 3) 2 + 2 ℎ −1 √ 3. = ℎ −1 ( 4 4 ) 4. = (√ℎ) + ℎ −1 () 5. = 3 2 ℎ −1 √ − (2) ℎ . 6. (√) = √ℎ −1() + 2 ℎ −1 (√) 7. (ℎ ) ℎ = ℎ −1 ( 2 ) 8. ℎ 2 () = ℎ −1 (√) − () 9. ℎ −1() = ℎ −1 (()) 10. ℎ( 2 ) = ℎ −1 ( )
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 103 Answer Tutorial 2.5 1. = ℎ −1 ( 3 ) √2 − 3√2 2 − 9 2. = 6 ℎ −1 3 √9 2 − 1 + 1 √(1 − ) 3. = − 4 (1 + 4 4) √16 2 + 2 4 4. = 1 2 ℎ + 5. = 6 2 ℎ −1 √ + 3 2 2√√1 + + 2 2 (2) 6. = 2 2 ℎ −1 (√) + 2 2√( − 1) ( 1 2 − 1 2(1 − 2)√ℎ −1() ) 7. = 2 2 (1 − 4) ℎ −1( 2) − ℎ ℎ (ℎ (ℎ ) − 1 ) 8. = − 2 (ℎ()) ℎ 2 () ( 1 2√(1 + ) − ) 9. = ( ℎ −1 () ) ( 1 (1 − 2())ℎ −1(()) − √1 + 2 ) 10. = − 2( 2 − 2 ) 2 ℎ 2 ( 2 )
MAT238/ MAT438 : Foundation of Applied Mathematics 104 2.2.7 Integration of Inverse Hyperbolic Function (comparison between Inverse Hyperbolic and Inverse Trigonometric Functions) Inverse Hyperbolic Function Inverse Trigonometric Function du a u + 2 2 1 = c a u sinh + −1 du u a − 2 2 1 = c a u cosh + −1 du a u − 2 2 1 = c a u tanh a + 1 −1 = c a u a u ln a + − + 2 1 du u a − 2 2 1 = c a u coth a + 1 −1 = c a u a u ln a + − + 2 1 du a u − 2 2 1 = c a u sin + −1 du a u + 2 2 1 = c a u tan a + 1 −1 Example 1 Evaluate + dx x 2 4 9 1 . Solution : Appendix (given in Final Examination) 18. C ln x x a C a x cosh x a dx 1 2 2 2 2 = + = + − + − − 19. C x x a C a x x a dx + = + + + = + − 1 2 2 2 2 sinh ln 20. + + + = − + = − − − C, if x a a x coth a 1 C, if x a a x tanh a 1 C x a x a ln 2a 1 a x dx 1 1 2 2 ∫ 1 √4 + 9 2 = ∫ 1 √(2) 2 + (3) 2 = ∫ 1 √ 2 + 2 ∙ 3 = 1 3 ∫ 1 √ 2 + 2 = 1 3 ℎ −1 ( ) + = 1 3 ℎ −1 ( 3 2 ) + # = 2 = 3 = 3 = 3
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 105 Example 2 Evaluate − dx e e x x 3 3 2 . Solution : Example 3 Evaluate ( ) − 2 4 2 x ln x dx . Solution : ∫ 2 [4 − ( ) 2] = 2 ∫ 1 [(2) 2 − ( ) 2] ∙ 1 = 2 ∫ 1 2 − 2 ∙ 1 ∙ = 2 ∫ 1 2 − 2 = 2 ∙ 1 2 ln | + − |+ = 2 ∙ 1 2(2) ln | + 2 − 2 |+ = 1 2 ln | + 2 − 2 |+ # ∫ 3 √ 2 − 3 = 3 ∫ 1 √( ) 2 − (√3) 2 ∙ = 3 ∫ 1 √ 2 − 2 ∙ ∙ = 3 ∫ 1 √ 2 − 2 = 3 ℎ −1 ( ) + = 3 ℎ −1 ( √3 ) + # = √3 = = = ∫ 2 [4 − ( ) 2] = 2 ∫ 1 [(2) 2 − ( ) 2] ∙ 1 = 2 ∫ 1 2 − 2 ∙ 1 ∙ = 2 ∫ 1 2 − 2 = 2 ∙ 1 ℎ −1 ( )+ = 2 ∙ 1 2 ℎ −1 ( 2 )+ = ℎ −1 ( 2 )+ # = 2 = = 1 =
MAT238/ MAT438 : Foundation of Applied Mathematics 106 Example 4 Evaluate − 9 3 8 3 x x dx . Solution : Example 5/ page 107/ OCT 2012/ MAT238/ Q2c (ii) (4 marks) Evaluate dx sec h x sec h x tanh x 25 − 5 5 5 2 . Solution : ∫ 3 3 8 − 9 = 3 ∫ 1 ( 4) 2 −(3) 2 ∙ 3 = 3 ∫ 1 2 − 2 ∙ 3 ∙ 4 3 = 3 4 ∫ 1 2 − 2 = 3 4 ∙ 1 ℎ −1 ( )+ = 3 4 ∙ 1 3 ℎ −1 ( 4 3 )+ = 1 4 ℎ −1 ( 4 3 ) + # ∫ ℎ 5 ℎ 5 25 − ℎ 25 = ∫ 1 (5) 2 −(ℎ 5) 2 ∙ ℎ 5 ℎ 5 = ∫ 1 2 − 2 ∙ ℎ 5 ℎ 5 ∙ −5 ℎ 5 ℎ 5 = ∫ 1 2 − 2 ∙ −5 = − 1 5 ∫ 1 2 − 2 = − 1 5 ∙ 1 ℎ −1 ( )+ = − 1 5 ∙ 1 5 ℎ −1 ( ℎ 5 5 ) + = − 1 25 ℎ −1 ( ℎ 5 5 )+ # ∫ 3 3 8 − 9 = 3 ∫ 1 ( 4) 2 −(3) 2 ∙ 3 = 3 ∫ 1 2 − 2 ∙ 3 ∙ 4 3 = 3 4 ∫ 1 2 − 2 = 3 4 ∙ 1 2 ln | + − |+ = 3 4 ∙ 1 2(3) ln | 4 +3 4 −3 | + = 1 8 ln | 4 +3 4 −3 |+ # = 3 = 4 = 4 3 = 4 3 ∫ ℎ 5 ℎ 5 25 − ℎ 25 = ∫ 1 (5) 2 −(ℎ 5) 2 ∙ ℎ 5 ℎ 5 = ∫ 1 2 − 2 ∙ ℎ 5 ℎ 5 ∙ −5 ℎ 5 ℎ 5 = ∫ 1 2 − 2 ∙ −5 = − 1 5 ∫ 1 2 − 2 = − 1 5 ∙ 1 2 ln | + − |+ = − 1 5 ∙ 1 2(5) | ℎ 5 + 3 ℎ 5 − 3 | + = − 1 50 | ℎ 5 + 3 ℎ 5 − 3 | + # = 5 = ℎ 5 = −5 ℎ 5 ℎ 5 = −5 ℎ 5 ℎ 5
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 107 Example 6 Evaluate ( ) dx x + − 2 4 2 1 1 . Solution : Example 7 Evaluate ( ) 2 − 3 − 8 2 x dx . Solution : ∫ 1 √4 +2( −1) 2 = ∫ 1 √2[2 + ( − 1) 2] = 1 √2 ∫ 1 √2+ ( − 1) 2 = 1 √2 ∫ 1 √(√2) 2 + ( −1) 2 = 1 √2 ∫ 1 √ 2 + 2 = 1 √2 ℎ −1 ( )+ = 1 √2 ℎ −1 ( − 1 √2 ) + # = √2 = − 1 = 1 = ∫ √2( − 3) 2 − 8 = ∫ √2[( − 3) 2 −4] = 1 √2 ∫ 1 √( −3) 2 − 4 = 1 √2 ∫ 1 √( −3) 2 − (2) 2 = 1 √2 ∫ 1 √ 2 − 2 = 1 √2 ℎ −1 ( )+ = 1 √2 ℎ −1 ( − 3 2 ) + # = 2 = − 3 = 1 =
MAT238/ MAT438 : Foundation of Applied Mathematics 108 Example 8/ Page 6/ MAR 2005/ MAT238/ Q2b(ii) (4½ marks) Evaluate ( ) 25 − 2 − 9 2 x dx . Solution : ∫ √25( − 2) 2 − 9 = ∫ √25 [( − 2) 2 − 9 25] = 1 √25 ∫ 1 √( − 2) 2 − ( 3 5 ) 2 = 1 5 ∫ 1 √( − 2) 2 − ( 3 5 ) 2 = 1 5 ∫ 1 √ 2 − 2 = 1 5 ℎ −1 ( ) + = 1 5 ℎ −1 ( − 2 ( 3 5 ) ) + = 1 5 ℎ −1 ( 5( − 3) 3 ) + # = 3 5 = − 2 = 1 =
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 109 2.2.8 Integration of Inverse Hyperbolic Functions involving Completing the Square Method Procedure completing the square method, arrange based on the highest power of x, such that ax + bx + c 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 + + a c x a b a x 2 put + ( )2 − ( )2 after the term containing x, where the expression inside ( ) must be • 2 1 coefficient of x = a b • 2 1 = a b 2 (as follow) + − + + a c a b a b x a b a x 2 2 2 2 2 write the first three terms into the form of ( )2 and simplify the next terms Or, using formula, ax + bx + c 2 = + − + a c a b a b a x 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 110 Example 1/ Page 18/ OCT 2006/ MAT238/ Q2b (6 marks) Integrate 2 − 4 + 6 2 x x dx . Solution : Using completing the square method, 2 2 − 4 + 6 = 2[ 2 − 2 + 3] = 2[ 2 − 2 + (−1) 2 − (−1) 2 + 3] = 2[( − 1) 2 + 2] Therefore, ∫ √2 2 − 4 + 6 = ∫ 1 √2(2 + ( − 1) 2) = 1 √2 ∫ 1 √2 + ( − 1) 2 = 1 √2 ∫ 1 √(√2) 2 + ( − 1) 2 = 1 √2 ∫ 1 √ 2 + 2 = 1 √2 ℎ −1 ( ) + = 1 √2 ℎ −1 ( − 1 √2 ) + # (factorize the coefficient of x2 ) write the first three terms into the form of ( )2 and simplify the next terms = √2 = − 1 = 1 = put + ( )2 − ( )2 after the term containing x, where the expression inside ( ) must be ( 1 2 × )
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 111 Example 2 Integrate ∫ √2 2+12−14 . Solution : Using completing the square, Hence, 2 2 + 12 − 14 = 2( 2 + 6 − 7) = 2( 2 + 6 + (+3) 2 − (+3) 2 − 7) = 2(( + 3) 2 − 16) ∫ √2 2 + 12 − 14 = ∫ 1 √2[( + 3) 2 − 16] = 1 √2 ∫ 1 √( + 3) 2 − 16 = 1 √2 ∫ 1 √( + 3) 2 − (4) 2 = 1 √2 ∫ 1 √ 2 − 2 = 1 √2 ℎ −1 ( ) + = 1 √2 ℎ −1 ( + 3 4 ) + # = 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
MAT238/ MAT438 : Foundation of Applied Mathematics 112 Example 3 Evaluate 4 + 24 +11 2 x x dx . Solution : Using completing the square, Hence, 4 2 + 24 + 11 = 4 ( 2 + 6 + 11 4 ) = 4 ( 2 + 6 + (+3) 2 − (+3) 2 + 11 4 ) = 4 (( + 3) 2 − 25 4 ) = 4 (( + 3) 2 − ( 5 2 ) 2 ) ∫ √4 2 + 24 + 11 = ∫ 1 √4 (( + 3) 2 − ( 5 2 ) 2 ) = 1 √4 ∫ 1 √( + 3) 2 − ( 5 2 ) 2 = 1 2 ∫ 1 √( + 3) 2 − ( 5 2 ) 2 = 1 2 ∫ 1 √ 2 − 2 = 1 2 ℎ −1 ( + 3 5 2 ) + = 1 2 ℎ −1 ( 2( + 3) 5 ) + # = 5 2 = + 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
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 113 Example 4 Evaluate 2 − 4 − 7 2 x x dx . Solution : Using completing the square, Hence, 2 2 − 4 − 7 = 2 ( 2 − 2 − 7 2 ) = 2 ( 2 − 2 + (−1) 2 − (−1) 2 − 7 2 ) = 2 (( − 1) 2 − 9 2 ) ∫ 2 2 − 4 − 7 = ∫ 1 2 (( − 1) 2 − 9 2 ) = 1 2 ∫ 1 ( −1) 2 − 9 2 = 1 2 ∫ 1 ( −1) 2 − ( 3 √2 ) 2 = 1 2 ∫ 1 2 − 2 = 1 2 ∙ 1 ℎ −1 ( )+ = 1 2 ∙ 1 ( 3 √2 ) ℎ −1 ( − 1 3 √2 )+ = 1 2 ∙ 1 ( 3 √2 ) ℎ −1 ( − 1 3 √2 )+ = 1 (√2) 2 ∙ √2 3 ℎ −1 ( √2( − 1) 3 ) + = 1 3√2 ℎ −1 ( √2( − 1) 3 )+ # = 3 √2 = − 1 = 1 = ∫ 2 2 − 4 − 7 = ∫ 1 2 (( − 1) 2 − 9 2 ) = 1 2 ∫ 1 ( −1) 2 − 9 2 = 1 2 ∫ 1 ( −1) 2 − ( 3 √2 ) 2 = 1 2 ∫ 1 2 − 2 = 1 2 ∙ 1 2 | + − |+ = 1 2 ∙ 1 2 ∙ 3 √2 | − 1 + 3 √2 − 1 − 3 √2 |+ = 1 2 ∙ 1 (√2) 2 ∙ 3 √2 | √2( − 1)+ 3 √2 √2( − 1)− 3 √2 |+ = 1 2 ∙ 1 √2 ∙ 3 | √2( − 1)+ 3 √2( − 1)− 3 |+ = 1 6√2 | √2( − 1)+ 3 √2( − 1)− 3 | + #
MAT238/ MAT438 : Foundation of Applied Mathematics 114 Example 5 Evaluate ∫ −2 2+12−17 Solution : Using completing the square, Hence, −2 2 + 12 − 17 = −2 ( 2 − 6 + 17 2 ) = −2 ( 2 − 6 + (−3) 2 − (−3) 2 + 17 2 ) = −2 (( − 3) 2 − 1 2 ) = 2 ( 1 2 − ( − 3) 2) ∫ −2 2 + 12 − 17 = ∫ 1 2 ( 1 2 − ( − 3) 2) = 1 2 ∫ 1 ( 1 √2 ) 2 − ( − 3) 2 = 1 2 ∫ 1 2 − 2 = 1 2 ∙ 1 ℎ −1 ( ) + = 1 2 ∙ 1 ( 1 √2 ) ℎ −1 ( − 3 ( 1 √2 ) ) + = 1 2 ∙ √2 ℎ −1 (√2( − 3)) + = √2 2 ℎ −1 (√2( − 3)) + # = 1 √2 = − 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
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 115 Example 6 Evaluate ∫ 3 √ 2+4+13 . Solution : Using completing the square, Hence, 2 + 4 + 13 = 2 + 4 + (+2) 2 − (+2) 2 + 13 = ( + 2) 2 + 9 ∫ 3 √ 2 + 4 + 13 = 3 ∫ 1 √9 + ( + 2) 2 = 3 ∫ 1 √(3) 2 + ( + 2) 2 = 3 ∫ 1 √ 2 + 2 = 3 ℎ −1 ( ) + = 3 ℎ −1 ( + 2 3 ) + # = 3 = + 2 = 1 =
MAT238/ MAT438 : Foundation of Applied Mathematics 116 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 −( − 8) = 3 ∫ 1 16 − ( − 4) 2 = 3 ∫ 1 (4) 2 − ( − 4) 2 = 3 ∫ 1 2 − 2 = 3 ∙ 1 ℎ −1 ( ) + = 3 ∙ 1 4 ℎ −1 ( − 4 4 ) + # = 3 4 ℎ −1 ( − 4 4 ) + # = 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 2 : Hyperbolic and Inverse Hyperbolic Functions 117 Tutorial 2.6 : Integration of Inverse Hyperbolic Functions ℎ . 1. ∫ 4 √4 + 3 2 ∶ 4 √3 ℎ −1 ( √3 2 ) + 2. ∫ 1 √9 + ( + 2) 2 ∶ ℎ −1 ( + 2 3 ) + 3. ∫ 1 25 − (2 + 1) 2 ∶ 1 10 ℎ −1 ( 2 + 1 5 ) + 4. ∫ 5 − (2 − 3) 2 ∶ 1 2√5 ℎ −1 ( 2 − 3 √5 ) + 5. ∫ √2 2 − 8 ∶ 1 √2 ℎ −1 ( 2 ) + 6. ∫ 12 − 9 2 ∶ 1 6 ℎ −1 ( 3 − 2 2 ) + 7. ∫ 12 + 4 − 2 ∶ 1 4 ℎ −1 ( − 2 4 ) + 8. ∫ −2 √ 2 + 6 + 34 ∶ −2 ℎ −1 ( + 3 5 ) + 9. ∫ 1 √2 2 − 12 + 13 ∶ 1 √2 ℎ −1 ( √2( − 3) √5 ) + 10. ∫ 1 1 + 8 − 2 2 ∶ 1 3√2 ℎ −1 ( √2( − 2) 3 ) + 11. ∫ 2 √ 4 − 3 ∶ 1 2 ℎ −1 ( 2 √3 ) + 12. ∫ 1 (4 − 2(3)) ∶ 1 2 ℎ −1 ( (3) 2 ) + 13. ∫ (5) (5) √25 + 2(5) ∶ 1 5 ℎ −1 ( (5) 5 ) + 14. ∫ (4) 9 − 2(4) ∶ − 1 12 ℎ −1 ( (4) 3 ) + 15. ∫ 2 3 3 − 8 ∶ 1 2√3 ℎ −1 ( 4 √3 ) + 16. ∫ 4 2 √ 6 − 16 ∶ 4 3 ℎ −1 ( 3 4 ) + END OF CHAPTER 2
MAT238/ MAT438 : Foundation of Applied Mathematics 118 A. Derivatives of Trigonometric Function/ Inverse Trigonometric Function and Hyperbolic Function/ Inverse Hyperbolic Function Find dx dy for : 1. ( ) 2 1 3 tanh x y e sinh y x − + = 2. xy sinh (tan y ) −1 = 3. y e y x sinh3 cosh 2 3 −1 = − 4. i) y = ln (tanh(4x)) ii) y e ( x) x sinh 3 2 −1 = 5. ( ) ( ) 1 3 sinh xy tan x − = B. Integration of Trigonometric Function/ Inverse Trigonometric Function and Hyperbolic Function/ Inverse Hyperbolic Function Solve the following integrals. 1. ( ) x x dx 2 cosh10 2. i) tanh 2x sec h 2x dx 2 ii) − dx x 2 1 2 2 3. dx x x + 2 0 8 3 4 4 4. ( x) x dx 2 0 2 sec cos sin h , hint : u = cos x 5. x (x )dx 2 2 sech 6. i) dx x x + 4 1 3 ii) coth x dx 4 4 6. sinh x cosh(xy) 2y 1 + = − 7. ( ) x x y x 2 2 cosh2 + = sinh2 8. ( ) ( ) y y x e 2 1 3 cosh = tan − − 9. ( ) ( ) 1 2 ln xy = cos siny + 5y − 10. 10. ( ) y x x y e 3 1 2 sinh 2 + tan = − 7. + 2 1 3 dx cosh x sinh x 8. − dx x 2 49 4 1 9. − 3 9 1 3 3 2 dx cos x sin x 10. + dx x 2 4 3 2 11. − x cosh x dx 1 2 2 12. ( ) 9 + 3 + 4 2 x dx 13. i) ( ) − 1 2 x ln x dx ii) ii) − − − x x e e dx 2 1
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 119 C. Integration of Inverse Trigonometric Function and Inverse Hyperbolic Function by Completing the Square. 1. − + dx x 2x 11 4 2 2. 2 − 4 + 6 2 x x dx 3. + + dx 4x 4x 5 1 2 4. − + − 3 2 2 x 4x 3 dx 5. dx x x − 6 + 5 5 2 . A. Derivatives of Trigonometric Function/ Inverse Trigonometric Function and Hyperbolic Function/ Inverse Hyperbolic Function 1. = ℎ −1 √ − − 2 2√1 − 2. = − − 3. = 3 3√4 2 − 1 3√4 2 − 1 ℎ 3 + 2 4() = 4 ℎ 4 ℎ 2 4 4() = 2 2 ℎ −1 3 + 3 2 √1 + 9 2 5. = 1 [ 3 2 (1 + 6) ℎ − ] 6. = 1 + √ 2 + 1 ℎ( ) √ 2 + 1[2 − ℎ( )] 7. = 1 ( 2 + 2 )[ℎ 2 ( ℎ 2 ) + ℎ 2 ℎ 2 ] − 6. dx x x − 2 + 5 1 2 7. dx x x − − 2 + 4 1 2 . 8. dx x x x + 2 + 3 4 2 . 9. dx x x 2 +3 + 7 3 2 . 10. 10. dx x x 2 − 8 −10 2 2
MAT238/ MAT438 : Foundation of Applied Mathematics 120 8. = 1 2√(1 + )(2 ℎ 2 + 3 3) 9. = (10 2 − − 1) 10. = (1 + 2 )(6 ℎ 2 2 ℎ 2 + ℎ −1 ) 2 2 + 2 2 2 − B. Integration of Trigonometric Function/ Inverse Trigonometric Function and Hyperbolic Function/ Inverse Hyperbolic Function 1. 1 20 ℎ(10 2 )+ 2() 1 3 (ℎ 2 ) 2 3 ⁄ + () √2 −1 (√2)+ 3. 0.7232 4. 0.7616 5. 1 2 ℎ 2 + 6() 3 2 −1 ( 2 )+ () |ℎ 4 | + 7. 0.398 C. Integration of Inverse Trigonometric Function and Inverse Hyperbolic Function by Completing the Square. 1. 4 ℎ −1 ( − 1 √10 ) + 2. 1 √2 ℎ −1 ( − 1 √2 ) + 3. 1 4 −1 ( 2 + 1 2 ) + 4. 1.5708 5. 5 ℎ −1 ( − 3 2 ) + 8. 1 28 | 2 + 7 2 − 7 |+ 9. 2 9 10. 2 √3 ℎ −1 ( √3 2 )+ 11. 1 2 2 ℎ −1 2 2 − 1 4 √4 4 − 1 + 12. 1 3 ℎ −1 [ 3 2 ( + 3)]+ 13(i) cosh−1 (ln x) + C (ii) sin−1 (e−x ) + C 6. ℎ −1 ( − 1 2 ) + 7. −1 ( + 1 √5 ) + 8. 4√ 2 + 2 + 3 − 4 ℎ −1 ( + 1 √2 ) + 9. 6 √47 −1 ( 4 + 3 √47 ) + 10. √2 ℎ −1 ( − 2 3 ) +
Chapter 3 : Integration Techniques 121 CHAPTER 3 INTEGRATION TECHNIQUES List of Topics • Integration using u-substitution • Integration using partial fraction • Integration using trigonometric substitution • Integration by parts 3.1 Integration using u-substitution 3.1.1 Power Rule (for n −1) with u-substitution → where u = the function in bracket From c n x x dx n n + + = + 1 1 , therefore (u) du n = c n u n + + + 1 1 Or in general (u) u' dx n • = c n u n + + + 1 1 because (u) u' dx n • = ( ) dx dx du u n • • = (u) du n = c n u n + + + 1 1
MAT238/ MAT438 : Foundation of Applied Mathematics 122 Examples Evaluate : a) ( ) dx x + 2 1 1 b) ( ) dx x − 3 1 1 c) dx x 3 +1 2 d) dx x x 3 + 1 2 2 Solution : How about dx x 3 + 1 2 2 ?? ) ∫ 2 √3 + 1 = 2 ∫(3 + 1) − 1 2 = 2 ∫(3 + 1) − 1 2 ∙ 3 = 2 3 ∫ − 1 2 = 2 3 • 1 2 ( 1 2 ) + = 2 3 • 2 1 √ + = 4 3 √3 + 1 + # ) ∫ 1 ( + 1) 2 = ∫( + 1) −2 = ∫ −2 = −1 −1 + = − 1 + = −1 + 1 + # ) ∫ 2 √3 2 + 1 = 2 ∫ (3 2 + 1) −1 2 ⁄ = 2 ∫ • − 1 2 • 6 = 1 3 ∫ − 1 2 = 1 3 • 1 2 ( 1 2 ) + = 1 3 • 2 1 √ + = 2 3 √3 2 + 1 + # ) ∫ 1 (1 − ) 3 = ∫(1 − ) −3 = ∫ −3 (−) = − ∫ −3 = − −2 −2 + = 1 2 2 + = 1 2(1 − ) 2 + # = 1 − = −1 = − = 3 +1 = 3 = 3 = 3 2 +1 = 6 = 6 = + 1 = 1 =
Chapter 3 : Integration Techniques 123 3.1.2 Logarithmic Rule (for n = −1) with u-substitution → where u = denominator From dx ln x c x = + 1 , therefore : du u 1 = Or in general, = ln u + c because dx u u' = u' dx u • • 1 = dx dx du u • • 1 = du u 1 = ln u + c dx u u' ln u + c Reminder : Caution : Don’t ever do this!! du u 1 = − u du 1 = c u + 0 0 = undefined (using power rule) But, always use du u 1 = ln u + c Hot Tips!! • dx ln x c x = + 1 • dx ln x c x dx x x = = + 1 2 • dx ln x c x dx x x = = + 1 3 2 • dx ln x c x dx x x = = + 1 4 3 • dx ln x c x dx x x = = + 1 5 4 and so on…
MAT238/ MAT438 : Foundation of Applied Mathematics 124 Example 1 Evaluate : a) dx x +1 1 b) dx 1− x 1 Solution : a) dx x +1 1 = du u 1 = ln u + C = ln x + 1 + C # b) dx 1− x 1 = ( du) u • − 1 = du u − 1 = − ln u + C = − ln1− x + C #
Chapter 3 : Integration Techniques 125 Example 2 Evaluate : a) dx x x −1 2 2 b) dx x x 4 + 3 3 2 Solution : a) dx x x −1 2 2 = x du u x 2 2 • = du u 1 = ln u + C = ln x − 1 + C 2 # b) dx x x 4 + 3 3 2 = x du u x 8 3 • = du u 1 8 3 = lnu + C 8 3 = ln 4x + 3 + C 8 3 2 #
MAT238/ MAT438 : Foundation of Applied Mathematics 126 Example 3 a) dx x x x − + − 2 1 2 2 2 b) dx x x x − + − 2 1 1 2 c) dx x x x − 2 +1 2 2 Solution : a) dx x x x − + − 2 1 2 2 2 = ( ) ( ) dx x x − − 2 1 2 1 = ( ) dx x −1 1 2 = du u 1 2 = 2lnu + C = 2ln x −1 + C # b) dx x x x − + − 2 1 1 2 = ( ) dx x x − − 2 1 1 = ( ) dx x −1 1 = du u 1 = lnu + C = ln x −1 + C # c) dx x x x − 2 +1 2 2 = ( ) dx x x − 2 1 2 = du u x 2 2 = du u u + 2 1 2 = du u u u + 2 2 1 2 = u du u + 1 −2 2 = C u u + − + − 1 2 ln 1 = C u u − + 2 2ln = C x x + − − − 1 2 2ln 1 #
Chapter 3 : Integration Techniques 127 Example 4 a) tan x dx b) cot x dx 2 c) dx cos x sin x 2 + 3 3 Solution : a) tan x dx = dx x x cos sin = − • x du u x sin sin = − du u 1 = − ln u + C = − ln cos x + C # b) cot x dx 2 = dx x x sin2 cos2 = • x du u x 2cos2 cos2 = du u 1 2 1 = lnu + C 2 1 = ln sin2x + C 2 1 # c) dx cos x sin x 2 + 3 3 = − • x du u x 3 sin3 sin3 = − du u 1 3 1 = − lnu + C 3 1 = − ln 2 + cos3x + C 3 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 128 Use 3.1.3 Exponential Rule with u-substitution → where u = exponent/power From e dx e c x x = + therefore : e du e c u u = + Or in general, e u' dx e c u u • = + because e u' dx u • = dx dx du e u • • = e du e c u u = + Example 1 Evaluate a) e dx 2x b) − e dx x c) e dx 5x c a e e dx ax ax = + (this formula only suitable for the exponent with x the power of positive one) Solution : a) e dx 2x = C e x + 2 2 # b) − e dx x = C e x + − − 1 = e C x − + − # c) e dx 5x = C e x + 5 5 # Reminder :
Chapter 3 : Integration Techniques 129 Example 2 Evaluate a) x e dx x 2 b) x e dx x 3 2 2 c) dx x e x Solution : a) x e dx x 2 = • x du x e u 2 = e du u 2 1 = e C u + 2 1 = e C x + 2 2 1 # b) x e dx x 3 2 2 = • 2 2 6x du x e u = e du u 6 1 = e C u + 6 1 = e C x + 3 2 6 1 # c) dx x e x = • x du x e u 2 = e du u 2 = e C u 2 + = e C x 2 + #
MAT238/ MAT438 : Foundation of Applied Mathematics 130 Example 3 Evaluate a) e cos x dx sinx b) e sec xdx tan x 3 3 2 c) e ec x x dx ec x cos 2 cot2 cos 2 Solution : a) e cos x dx sinx = • • x du e x u cos cos = e du u = e C u + = e C x + sin # b) e sec xdx tan x 3 3 2 = • • x du e x u 3sec 3 sec 3 2 2 = e du u 3 1 = e C u + 3 1 = e C x + tan3 3 1 # c) e ec x x dx ec x cos 2 cot2 cos 2 = − • • ec x x du e ec x x u 2cos 2 cot2 cos 2 cot2 = − e du u 2 1 = e C u − + 2 1 = e C ec x − + cos 2 2 1 #
Chapter 3 : Integration Techniques 131 3.1.4 Trigonometric Rule with u-substitution → where u = angle (if the trigonometric function is exist in the table of integrals) Trigonometric Rule Trigonometric Rule with u-substitution sin x dx = − cos x + c cos x dx = sin x + c sec x dx = tan x + c 2 cos ec x dx = − cot x + c 2 sec x tan x dx = sec x + c cosec x cot x dx = − cosec x + c sinu du = − cosu + c cos u du = sinu + c sec u du = tanu + c 2 cos ec u du = − cot u + c 2 sec u tanu du = sec u + c cos ec u cot u du = − cos ec u + c Example 1 Evaluate a) sin2x dx b) cos 3x dx c) cos ec 4x dx 2 d) sec 3x tan 3x dx Solution : a) sin2x dx = C x + − 2 cos2 # b) cos 3x dx = C x + 3 sin3 # c) cos ec 4x dx 2 = C x + − 4 cot 4 # d) sec 3x tan 3x dx = C x + 3 sec 3 # Useful formula, c a cos ax sinax dx + − = c a sinax cos ax dx = + c a tan ax sec ax dx = + 2 c a cot ax cos ec ax dx + − = 2 c a sec ax sec ax tan ax dx = + c a cos ec ax cos ec ax cot ax dx + − = Important : this formula only suitable for the exponent with x the power of positive one How about : i) tanx dx ii) cotx dx ?? (try this!!)
MAT238/ MAT438 : Foundation of Applied Mathematics 132 Example 2 Evaluate a) ( ) x sin x dx 2 b) ( ) x cos x dx 2 3 2 c) dx x cos ec x cot x Solution : a) ( ) x sin x dx 2 = • x du x u 2 sin = sinu du 2 1 = (− cosu) + C 2 1 = − cosu + C 2 1 = − (x )+ C 2 cos 2 1 # b) ( ) x cos x dx 2 3 2 = • 2 2 6 cos x du x u = cosu du 6 1 = sinu + C 6 1 = ( x )+ C 3 sin 2 6 1 # c) dx x cos ec x cot x = • x du x ecu u 2 cos cot = 2 cosecu cotu du = 2 (− cos ecu) + C = − cos ecu + C = − cosec x + C #
Chapter 3 : Integration Techniques 133 Example 3 Evaluate a) ( ) dx x cos ln x b) ( ) e sin e dx 3x 3x Solution : a) ( ) dx x cos ln x = • x du x cosu = cosu du = sinu + C = sin(ln x) + C # b) ( ) e sin e dx 3x 3x = • x x e du e u 3 3 3 sin = sinu du 3 1 = (− cosu) + C 3 1 = − cosu + C 3 1 = (e ) C x − + 3 cos 3 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 134 3.1.5 Derivative and Integration of inverse tangent & inverse sine Differentiation Integration (sin x) dx d −1 = 2 1 1 − x − a u dx d 1 sin = 2 2 1 a − u du x − 2 1 1 = x + c −1 sin du a u − 2 2 1 = c a u sin + −1 (tan x) dx d −1 = 2 1 1 + x − a u dx d 1 tan = 2 2 a u a + du + x 2 1 1 = x + c −1 tan du a u + 2 2 1 = c a u tan a + 1 −1 Simple Examples Differentiate with respect to x. a) y = sin−1 (2x) b) y = tan−1 (5x) c) y = sin−1 (3x) d) y = tan−1 (4x) Solution : a) y = sin−1 (2x) dx dy = ( ) 2 1 2 1 2 • − x = 2 1 4 2 − x # b) y = tan−1 (5x) dx dy = ( ) 5 1 5 1 2 • + x = 2 1 25 5 + x # c) y = sin−1 (3x) = = # d) y = tan−1 (4x) = = #
Chapter 3 : Integration Techniques 135 3.1.6 Introduction to Completing the Square We need to use completing the square method to evaluate the integrals i) + + dx ax bx c d 2 ii) + + dx ax bx c d 2 Where a, b, c, d are constants and a,b,d ≠ 0 Integration of Inverse Trigonometricc Functions involving Completing the Square Method Procedure completing the square method, arrange based on the highest power of x, such that ax + bx + c 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 + + a c x a b a x 2 put + ( )2 − ( )2 after the term containing x, where the expression inside ( ) must be • 2 1 coefficient of x = a b • 2 1 = a b 2 (as follow) + − + + a c a b a b x a b a x 2 2 2 2 2 write the first three terms into the form of ( )2 and simplify the next terms Or, using formula, ax + bx + c 2 = + − + a c a b a b a x 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 136 Example 1 Evaluate − − 2 21 4x x dx . Solution : Using completing the square, 2 21− 4x − x = 4 21 2 − x − x + = ( 4 21) 2 − x + x − = 4 (2) (2) 21 2 2 2 − x + x + − − = ( 2) 25 2 − x + − = ( ) 2 25 − x + 2 Hence, − − 2 21 4x x dx = ( ) dx x − + 2 25 2 1 = ( ) dx x − + 2 2 5 2 1 = du a u − 2 2 1 = C a u + −1 sin = C x + − + 5 2 sin 1 #
Chapter 3 : Integration Techniques 137 Example 2 Evaluate 2 −12 + 26 2 x x dx . Solution : Using completing the square, 2 12 26 2 x − x + = 2( 6 13) 2 x − x + = 2 6 ( 3) ( 3) 13 2 2 2 x − x + − − − + = 2( 3) 4 2 x − + Hence, 2 −12 + 26 2 x x dx = ( ) dx x 2 − 3 + 4 1 2 = ( ) dx x − 3 + 4 1 2 1 2 = ( ) dx x + − 2 2 2 3 1 2 1 = du a + u 2 2 1 2 1 = C a u a + • −1 tan 1 2 1 = C x + − • − 2 3 tan 2 1 2 1 1 = C x + − − 2 3 tan 4 1 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 138 Example 3 Evaluate dx 2x + 4x + 9 5 2 ans : ( ) C x + − − 14 2 1 tan 14 5 1 Solution : Using completing the square, 2 4 9 2 x + x + = − + 2 9 2 2 2 x x = ( ) ( ) − + − − − + 2 9 2 2 1 1 2 2 2 x x = ( ) − − + 2 9 2 1 1 2 x = ( ) − + 2 7 2 1 2 x Hence, dx 2x + 4x + 9 5 2 = ( ) dx x − + 2 7 2 1 1 5 2 = ( ) dx x − + 2 7 1 1 2 5 2 = ( ) ( ) dx x + − 2 2 2 7 1 1 2 5 = du a u + 2 2 1 2 5 = C x + − • − 2 7 1 2 7 1 tan 1 2 5 = ( ) C x + − • − 7 2 1 tan 7 2 2 5 1 = ( ) C x + − • − 2 7 2 2 1 tan 7 2 2 2 5 1 = ( ) C x + − − 14 2 1 tan 14 5 1 #
Chapter 3 : Integration Techniques 139 Example 4 Evaluate − 2 6x x dx . Solution : Using completing the square, 2 6x − x = x 6x 2 − + = (x 6x) 2 − − = ( ) ( ) 2 2 2 − x − 6x + − 3 − − 3 = ( 3) 9 2 − x − − = ( ) 2 9 − x − 3 Hence, − 2 6x x dx = ( ) dx x − − 2 9 3 1 = ( ) dx x − − 2 2 3 3 1 = du a u − 2 2 1 = C a u + −1 sin = C x + − − 3 3 sin 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 140 Example 5 Evaluate − 6 + 13 2 x x dx ans : C x + − − 2 3 tan 2 1 1 Solution : Using completing the square, 6 13 2 x − x + = 6 ( 3) ( 3) 13 2 2 2 x − x + − − − + = ( 3) 4 2 x − + Hence, − 6 + 13 2 x x dx = ( ) dx x − 3 + 4 1 2 = ( ) dx x + − 2 2 2 3 1 = du a u + 2 2 1 = C a u a + −1 tan 1 = C x + − − 2 3 tan 2 1 1 #
Chapter 3 : Integration Techniques 141 3.2 Integration using Partial Fraction Expressing a Fractional Function In Partial Fractions Before a fractional function can be expressed directly in partial fractions, the numerator must be of at least one degree less than the denominator. What is partial fraction? Early training in algebra teaches us how to ‘simplify’ an expression such as x 1 1 x 1 1 + − − by reducing it to x 1 2 2 − . We have now reached the stage when we reversed the process. Given a fraction such as x 1 2 2 − whose denominator factorizes, we may split it up into its component fractions, writing as x 1 1 x 1 1 + − − ; it now said to be in partial fraction. In general, there are 3 types of partial fraction i) denominator with different linear factors ii) denominator with irreducible quadratic factors (which does not factorize) iii) denominator with repeated factors useful formula : Summary Denominator containing… Expression Form of Partial Fraction i) Different Linear Factors ( ) (ax b)(cx d) f x + + cx d B ax b A + + + ii) Irreducible Quadratic Factors ( ) (ax bx c)(gx h) f x + + + 2 gx h C ax bx c Ax B + + + + + 2 iii) Repeated Linear Factors ( ) ( ) 3 ax b f x + ( ) ( ) 2 3 ax b C ax b B ax b A + + + + + iv) Repeated Quadratic Factors ( ) ( ) 3 2 ax bx c f x + + ( ) ( ) 3 2 2 2 2 ax bx c Ex F ax bx c Cx D ax bx c Ax B + + + + + + + + + + + Note : In each of the above cases, f(x) must be of less degree than the relevant denominator. ∫( + ) = ە ۖ ۔ ۖ ( + )∫ → − =/ ۓ = ( + ) + ( + ) + ⬚ = − → ∫( + ) − = ∫ + = ȁ + ȁ +
MAT238/ MAT438 : Foundation of Applied Mathematics 142 3.2.1 Type I : denominator with different linear factors Example 1 Express ( 3)( 3) 6 x − x + into the sum of partial fractions. Hence, evaluate ( )( ) dx x − 3 x + 3 6 . Answer : ( 3)( 3) 6 x − x + = 3 1 3 1 + − x − x ( )( ) dx x − 3 x + 3 6 = ln x − 3 − ln x + 3 + C Solution : ( 3)( 3) 6 x − x + = 3 + 3 + − x B x A −−−−−−−−− Standardize denominator, ( 3)( 3) 6 x − x + = ( ) ( ) ( 3)( 3) 3 3 − + + + − x x A x B x 6 = A(x + 3) + B(x − 3) −−−−−−−−− Substitute x = 3 into , 6 = A(3 + 3) + 0 6 = 6A A = 1 Substitute x = −3 into , 6 = 0 + B(− 3 − 3) 6 = − 6B B = −1 Substitute A = 1, B = −1 into , ( 3)( 3) 6 x − x + = 3 1 3 1 + − x − x Hence, ( )( ) dx x − 3 x + 3 6 = dx x dx x + − − 3 1 3 1 = ln x − 3 − ln x + 3 + C #
Chapter 3 : Integration Techniques 143 Example 2 Express (2 x)(2 x) x + − into the sum of partial fractions. Hence, evaluate dx ( x )( x ) x 2 + 2 − . Answer : (2 x)(2 x) x + − = ( x) ( − x) + + − 2 2 1 2 2 1 dx ( x )( x ) x 2 + 2 − = − + x − ln2 − x + C 2 1 ln2 2 1 Solution : (2 x)(2 x) x + − = x B x A − + 2 + 2 −−−−−−−−− Standardize denominator, (2 x)(2 x) x + − = ( ) ( ) ( x)( x) A x B x + − − + + 2 2 2 2 x = A(2 − x) + B(2 + x) −−−−−−−−− Substitute x = −2 into , −2 = A(2 − (− 2)) + 0 −2 = 4A A = 2 1 − Substitute x = 2 into , 2 = 0 + B(2 + 2) 2 = 4B B = 2 1 Substitute A = 2 1 − , B = 2 1 into , (2 x)(2 x) x + − = x − x + + − 2 2 2 1 2 1 = ( x) ( − x) + + − 2 2 1 2 2 1 Hence, dx x x x (2 + )(2 − ) = dx x dx x − + + − 2 1 2 1 2 1 2 1 = x − x + C − − • + + • ln 2 1 1 2 1 ln 2 1 1 2 1 = − + x − ln 2 − x + C 2 1 ln 2 2 1 #