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 ) +