MAT238/ MAT438 : Foundation of Applied Mathematics 144 Example 3 Express 3x 11x 10 x 1 2 − + − into the sum of partial fractions. Hence, evaluate dx x x x − + − 3 11 10 1 2 . Answer : 3x 11x 10 x 1 2 − + − = 3 5 2 2 1 − − x − x dx x x x − + − 3 11 10 1 2 = x − − ln3x − 5 + C 3 2 ln 2 Solution : 3x 11x 10 x 1 2 − + − = ( 2)(3 5) 1 − − − x x x = 2 3 − 5 + − x B x A −−−−−−−−− 3x 11x 10 x 1 2 − + − = ( ) ( ) ( 2)(3 5) 3 5 2 − − − + − x x A x B x x − 1 = A(3x − 5) + B(x − 2) −−−−−−−−− Substitute x = 2 into , 2 − 1 = A(6 − 5) + 0 A = 1 Substitute x = 3 5 into , 3 5 − 1 = 0 ( 2) 3 + B 5 − 3 2 = B 3 1 − B = 3 3 2 • − = −2 Substitute A = 1, B = −2 into , 3 11 10 1 2 − + − x x x = ( 2)(3 5) 1 − − − x x x = 3 5 2 2 1 − − x − x Hence, dx x x x − + − 3 11 10 1 2 = dx x dx x − − − 3 5 1 2 2 1 = • x − − • ln 3x − 5 + C 3 1 ln 2 2 1 1 = x − − ln 3x − 5 + C 3 2 ln 2 #
Chapter 3 : Integration Techniques 145 Example 4 Express (x 2)(x 1)(x 3) 3x 1 + + − + into the sum of partial fractions. Hence, evaluate dx ( x )( x )( x ) x + + − + 2 1 3 3 1 . Answer : (x 2)(x 1)(x 3) 3x 1 + + − + = 2 1 3 1 − + + + + − x C x B x Solution : (x 2)(x 1)(x 3) 3x 1 + + − + = 2 1 − 3 + + + + x C x B x A −−−−−−−−− (x 2)(x 1)(x 3) 3x 1 + + − + = ( )( ) ( )( ) ( )( ) ( 2)( 1)( 3) 1 3 2 3 2 1 + + − + − + + − + + + x x x A x x B x x C x x 3x + 1 = A(x + 1)(x − 3) + B(x + 2)(x − 3) + C(x + 2)(x + 1) −−−−−−−−− Substitute x = −2 into , 3(− 2) + 1 = A(− 2 + 1)(− 2 − 3) + 0 + 0 − 5 = A(−1)(− 5) 5A = −5 A = −1 Substitute x = −1 into , 3(− 1) + 1 = 0 + B(−1+ 2)(−1− 3) + 0 − 2 = 0 + B(1)(− 4) + 0 − 4B = − 2 B = 2 1 Substitute x = 3 into , 3(3) + 1 = 0 + 0 + C(3 + 2)(3 + 1) 10 = C(5)(4) 20C = 10 C = 2 1
MAT238/ MAT438 : Foundation of Applied Mathematics 146 Substitute A = −1, B = ½ , C = ½ into , ( 2)( 1)( 3) 3 1 + + − + x x x x = 2 1 3 1 2 1 2 1 − + + + + − x x x = ( ) 2( 3) 1 2 1 1 2 1 − + + + + − x x x Hence, dx x x x x + + − + ( 2)( 1)( 3) 3 1 = dx x dx x dx x − + + + + − 3 1 2 1 1 1 2 1 2 1 = − • x + + • x + + • ln x − 3 + C 1 1 2 1 ln 1 1 1 2 1 ln 2 1 1 = − x + + x + + ln x − 3 + C 2 1 ln 1 2 1 ln 2 #
Chapter 3 : Integration Techniques 147 3.2.2 Type II : denominator with irreducible quadratic factors (which does not factorize) Consider the quadratic expressions ax2 + bx + c. i) b 4ac 2 − = complex numbers → irreducible (cannot factorize). So, consider in Type II (irreducible quadratic factors) ii) b 4ac 2 − = decimal numbers → irreducible (cannot factorize). So, consider in Type II (irreducible quadratic factors) iii) b 4ac 2 − = positive integer → reducible (can be factorize). When we factorize, the factors are different, then consider in Type I (different linear factors) iv) b 4ac 2 − = zero → reducible (can be factorize). When we factorize, the factors are equal, then consider in Type III (Repeated factors) − will be discuss in part 3.2.3 Example 1 Express (1 x)(4 x ) 6 x 2 − + − into the sum of partial fractions. Hence, evaluate dx ( x )( x ) x − + − 2 1 4 6 . (1 x)(4 x ) 6 x 2 − + − = 2 4 2 1 1 x x x + + + − Solution : (1 x)(4 x ) 6 x 2 − + − = 2 1 4 x Bx C x A + + + − −−−−−−−−− Standardize denominator, (1 )(4 ) 6 2 x x x − + − = ( ) ( )( ) (1 )(4 ) 4 1 2 2 x x A x Bx C x − + + + + − 6 − x = A(4 + x ) + (Bx + C)(1− x) 2 −−−−−−−−−
MAT238/ MAT438 : Foundation of Applied Mathematics 148 Substitute x = 1 into , 6 − 1 = A(4 + 1) + 0 5A = 5 A = 1 Expand 6 − x = A + Ax + Bx + C − Bx − Cx 2 2 4 0 6 2 x − x + = (A − B)x + (B −C)x +(4A + C) 2 By comparing the coefficient of x 2 : A − B = 0 B = A → B = 1 x : B − C = −1 1 − C = −1 → C = 2 Substitute A = 1, B = 1 , C = 2 into , (1 )(4 ) 6 2 x x x − + − = 2 4 2 1 1 x x x + + + − = 2 2 4 2 1 4 1 x x x x + + + + − Hence, dx x x x − + − (1 )(4 ) 6 2 = dx x dx x x dx x + + + + − 2 2 4 2 1 4 1 = dx x dx x x dx x + + + + − − − 2 2 2 2 1 2 4 2 2 1 1 1 = C x x x + − − + + + • − 2 tan 2 1 ln 4 2 2 1 ln1 2 1 = C x x x + − − + + + − 2 ln 4 tan 2 1 ln1 2 1 #
Chapter 3 : Integration Techniques 149 Example 2 Express (x 1)(2x x 3) 4 2 + + + into the sum of partial fractions. Hence, evaluate dx ( x )( x x ) +1 2 + + 3 4 2 . Solution : (x 1)(2x x 3) 4 2 + + + = 1 2 3 2 + + + + + x x Bx C x A −−−−−−−−− Standardize denominator, (x 1)(2x x 3) 4 2 + + + = ( ) ( 1)(2 3) (2 3) ( 1) 2 2 + + + + + + + + x x x A x x Bx C x 4 = (2 3) ( )( 1) 2 A x + x + + Bx +C x + −−−−−−−−− Substitute x = −1 into , 4 = A(2 −1+ 3) + 0 4A = 4 A = 1 Expand 4 = Ax + Ax + A + Bx +Cx + Bx +C 2 2 2 3 0 0 4 2 x + x + 4 = (2A + B)x + (A + B +C)x + (3A +C) 2 By comparing the coefficient of x 2 : 2A + B = 0 B = −2A → B = −2(1) = −2 x : A + B + C = 0 1 − 2 + C = 0 → C = 1 Substitute A = 1, B = −2 , C = 1 into , ( 1)(2 3) 4 2 x + x + x + = 2 3 2 1 1 1 2 + + − + + + x x x x = 2 3 1 2 1 1 2 + + − + + x x x x
MAT238/ MAT438 : Foundation of Applied Mathematics 150 Hence, dx (x +1)(2x + x + 3) 4 2 = + + − + + dx x x x dx x 2 3 1 2 1 1 2 ------------------- (i) dx x C x = + + + ln 1 1 1 (into ) (ii) + + − + = + + − dx x x x dx x x x 2 3 2 1 2 3 1 2 2 2 = ( ) + + − − + − dx x x x 2 3 2 2 1 2 1 2 = + + − − dx x x x 2 3 4 2 2 1 2 = + + + − − dx x x x 2 3 4 1 3 2 1 2 = + + − + + + − dx x x x x x 2 3 3 2 3 4 1 2 1 2 2 = + + + + + + − dx x x dx x x x 2 3 1 2 3 2 3 4 1 2 1 2 2 *** = C x x x + + − + + + − 23 4 1 tan 23 3 ln 2 3 2 1 2 1 (into ) ***Using completing the square 2 3 2 x + x + = + + 2 3 2 2 2 x x = + − + + 2 3 4 1 4 1 2 2 2 2 2 x x = + + 16 23 4 1 2 2 x = + + 2 2 4 23 4 1 2 x (i) (ii)
Chapter 3 : Integration Techniques 151 *** + + dx 2x x 3 1 2 3 2 = + + dx x 2 2 4 23 4 1 2 1 2 3 = + + dx x 2 2 4 23 4 1 1 4 3 = + du u a 2 2 1 4 3 = C a u a + • −1 tan 1 4 3 = ( ) ( ) C x + + • − 4 23 4 1 1 4 23 tan 1 4 3 = ( ) ( ) C x + • + − 4 23 4 4 1 1 tan 23 4 4 3 = C x + − + 23 4 1 tan 23 3 1 By substituting (i) and (ii) into , thus, dx (x +1)(2x + x + 3) 4 2 = C x x x x + + + − + + + − 23 4 1 tan 23 3 ln 2 3 2 1 ln 1 2 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 152 Example 3 Express (2 x)(3 x ) 3 2x 2 − + + into the sum of partial fractions. Hence, evaluate Solution : By comparing the coefficient of dx ( x )( x ) x − + + 2 2 3 3 2 3 + 2 (2 − )(3 + 2) = 2 − + + 3 + 2 −−−−−−−−− 3 + 2 (2 − )(3 + 2) = (3 + 2 ) + ( + )(2 − ) (2 − )(3 + 2) → 3 + 2 = (3 + 2 ) + ( + )(2 − ) −−−−−−−−− Substitute = 2 into , 3 + 2(2) = (3 + 2 2 ) + 0 7 = 7 = 1 Expand RHS 3 + 2 = 3 + 2 + 2 − 2 + 2 − 0 2 + 2 + 3 = ( − ) 2 + (2 − ) + (3 + 2) ∶ − = 0 = → = 1 ∶ 2 − = 2 = 2 − 2 = 2(1) − 2 → = 0 Substitute A = 1, B = 1, C = 0 into , 3 + 2 (2 − )(3 + 2) = 1 2 − + 3 + 2 Hence, ∫ 3 + 2 (2 − )(3 + 2) = ∫ 1 2 − + ∫ 3 + 2 = − ∫ −1 2 − + 1 2 ∫ 2 3 + 2 = − ȁ2 − ȁ + 1 2 ȁ3 + 2 ȁ + #
Chapter 3 : Integration Techniques 153 3.2.3 Type III : denominator with repeated factors Example 1 Express 2 3 1 ( x ) x + + into the sum of partial fractions. Hence, evaluate dx ( x ) x + + 2 3 1 . Solution : = ( ) 2 3 + 3 + + x B x A −−−−−−−−− Standardize denominator, = ( ) 2 ( 3) 3 + + + x A x B x + 1 = A(x + 3) + B −−−−−−−−− Substitute x = −3 into , −3 + 1 = 0 + B B = −2 Substitute x = 0 into (or any value), 1 = 3A + B 1 = 3A − 2 3A = 3 A = 1 Substitute A = 1, B = −2 into , = ( ) 2 3 2 3 1 + − x + x Hence, ( ) dx x x + + 2 3 1 = ( ) dx x dx x + − + 2 3 2 3 1 = dx (x ) dx x − − + + 2 2 3 3 1 = ( ) C x x + − + + − • − 1 3 ln 3 2 1 = C x x + + + + 3 2 ln 3 # 2 (x 3) x 1 + + 2 (x 3) x 1 + + 2 (x 3) x 1 + +
MAT238/ MAT438 : Foundation of Applied Mathematics 154 Example 2 Express 2 2 (x 2)(x 1) 2x 5x 7 − − − + into the sum of partial fractions. Hence, evaluate dx ( x )( x ) x x − − − + 2 2 2 1 2 5 7 . Solution : 2 2 ( 2)( 1) 2 5 7 − − − + x x x x = ( ) 2 2 1 −1 + − + − x C x B x A −−−−−−−−− Standardize denominator, 2 2 ( 2)( 1) 2 5 7 − − − + x x x x = ( ) ( )( ) ( ) 2 2 ( 2)( 1) 1 2 1 2 − − − + − − + − x x A x B x x C x 2 5 7 2 x − x + = ( 1) ( 2)( 1) ( 2) 2 A x − + B x − x − + C x − −−−−−−−−− Substitute x = 1 into , 2 − 5 + 7 = 0 + 0 + C(1− 2) 4 = −C C = −4 Substitute x = 2 into , 8 −10 + 7 = (2 1) 0 0 2 A − + + A = 5 Substitute x = 0 into (or any value), 0 − 0 + 7 = ( 1) ( 2)( 1) ( 2) 2 A − + B − − + C − 7 = A + 2B − 2C 7 = 5 + 2B − 2(− 4) 7 − 5 − 8 = 2B 2B = −6 B = −3 Substitute A = 5, B = −3, C = −4 into , 2 2 ( 2)( 1) 2 5 7 − − − + x x x x = ( ) 2 1 4 1 3 2 5 − − − − x − x x
Chapter 3 : Integration Techniques 155 Hence, dx x x x x − − − + 2 2 ( 2)( 1) 2 5 7 = ( ) dx x dx x dx x − − − − − 2 1 4 1 3 2 5 = dx (x ) dx x dx x − − − − − − 2 4 1 1 1 3 2 1 5 = ( ) C x x x + − − − − − − • − 1 1 5ln 2 3ln 1 4 1 = C x x x + − − − − + 1 4 5ln 2 3ln 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 156 Example 3 Given 2 3 (x 1) 5 (x 1) B (x 1) 2 x 3 A − + − + − − + = + − − + + 3 3 2 3 1 10 26 3 (x )(x ) x x x . Find the values of A and B. Hence, evaluate ( )( ) dx x x x x x + − − + + 3 3 2 3 1 10 26 3 .Solution : 2 3 (x 1) 5 (x 1) B (x 1) 2 x 3 A − + − + − − + = + − − + + 3 3 2 3 1 10 26 3 (x )(x ) x x x ---------------- ( ) ( )( ) ( )( ) ( ) 3 3 2 3 3 2 3 1 1 2 3 1 3 1 5 3 3 1 10 26 3 (x )(x ) A x x x B x x x (x )(x ) x x x + − − − + − + + − + + = + − − + + 10 26 3 ( 1) 2( 3)( 1) ( 3)( 1) 5( 3) 3 2 3 2 x − x + x + = A x − − x + x − + B x + x − + x + ----------- Substitute x = −3 into , − 27 − 90 − 78 + 3 = ( 3 1) 0 0 0 3 A − − + + + −192 = −64A A = 3 Substitute x = 0 (or any value) into , 3 ( 1) 2(3)( 1) (3)( 1) 5(3) 3 2 = A − − − + B − + 3 3( 1) 2(3)( 1) (3)( 1) 5(3) 3 2 = − − − + B − + 3 = − 3 − 6 − 3B +15 3B = 3 B = 1 A = 3, B = 1 Substitute A = 3, B = 1 into , 2 3 (x 1) 5 (x 1) 1 (x 1) 2 x 3 3 − + − + − − + = + − − + + 3 3 2 3 1 10 26 3 (x )(x ) x x x
Chapter 3 : Integration Techniques 157 Hence, dx (x )(x ) x x x + − − + + 3 3 2 3 1 10 26 3 = ( ) ( ) dx x dx x dx x dx x − + − + − − + 2 3 1 5 1 1 1 2 3 3 = dx (x ) dx (x ) dx x dx x − − + − + − − − + 2 3 1 5 1 1 1 2 3 1 3 = ( ) ( ) C x x x x + − − + • − − + − − + − − 2 1 5 1 1 3ln 3 2ln 1 1 2 = ( ) C x x x x + − − − + − − − 2 2 1 5 1 1 3ln 3 2ln 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 158 3.2.4 Integration using Partial Fraction (involving long division – for improper fraction) Improper Fractions An improper fraction is one whose numerator is of degree equal to, or greater than, that of the denominator. To express the improper fraction into partial fraction, we first divide the numerator using long division to obtain a quotient and proper fraction, and then split the latter into partial fractions. For example, let say by using long division method, (x 3)(x 1) x 2x 7 x 1 (x 3)(x 1) x 2x x 4x 4 2 2 2 4 3 2 − + − + = + + − + − − − + In this case, the expression (x 3)(x 1) x 2x 7 2 2 − + − + will be express in partial fraction. (refer page 162, Example 2) 3.2.5 Long Division Let P(x) and D(x) be polynomials with D(x) of lower degree than P(x) and D(x) of degree one or more. There exist unique polynomial Q(x) and R(x) such that D(X) R(x) Q(x) D(x) P(x) = + , or P(x) = D(x) • Q(x) + R(x) Where either R(x) = 0 or the degree of R(x) is less than the degree of D(x), and P(x) : dividend (original Polynomial) D(x) : Divisor Q(x) : Quotient R(x) : Remainder For instance, we saw Example 1, x 3 x 2x x 10 3 2 + + − + = x 2 − x + 2 + x 3 4 + We can express this result using the division algorithm: ( ) D(x) P x = ( ) ( ) D(x) R x Q x + We can divide (x3 + 2x2 − x + 10) by (x +3) using long division method. Express in partial fraction
Chapter 3 : Integration Techniques 159 Example 1 (Long Division Method) : Determine (x3 + 2x2 − x + 10) (x + 3) using long division method. Solution : we can rewrite (x3 + 2x2 − x + 10) (x + 3) as x 3 x 2x x 10 3 2 + + − + or, Step 1 : Step 2 : multiply x 2 by (x + 3), then we get (x3 + 3x2 ). Step 3 : subtract (x3 + 2x2 ) by (x3 + 3x2 ) , then we get −x 2 . Step 4 : bring down −x Step 5 : multiply (−x) by (x + 3), then we get (−x 2 −3x). x + 3 x3 + 2x2 − x + 10 x + 3 x 3 + 2x2 − x + 10 x 3 + 3x2 x 2 x + 3 x 3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 x 2 x + 3 x 3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x −x 2 − 3x x 2 − x x + 3 x3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x x 2
MAT238/ MAT438 : Foundation of Applied Mathematics 160 Step 6 : subtract (−x 2 − x) by (−x 2 − 3x) , then we get 2x. Step 7 : bring down 10 Step 8 : multiply 2 by (x + 3), then we get (2x + 6). Step 9 : subtract (2x + 10) by (2x + 6) , the we get the remainder 4. (final step) x + 3 x3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x − (−x 2 − 3x) 2x x 2 − x x + 3 x 3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x − (−x 2 − 3x) 2x + 10 2x + 6 x 2 − x + 2 x + 3 x3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x − (−x 2 − 3x) 2x + 10 − (2x + 6) 4 x 2 − x + 2 x + 3 x3 + 2x2 − x + 10 − (x3 + 3x2 ) −x 2 − x − (−x 2 − 3x) 2x + 10 x 2 − x
Chapter 3 : Integration Techniques 161 From the final step, x 3 x 2x x 10 3 2 + + − + = x 2 − x + 2 + x + 3 4 or x 3 + 2x2 − x + 10 = (x + 3) (x2 − x + 2) + 4 Check : x2 − x + 2 + x 3 4 + = x 3 (x x 2)(x 3) 4 2 + − + + + = x 3 x x 2x 3x 3x 6 4 3 2 2 + − + + − + + = x 3 x 2x x 10 3 2 + + − + Understanding long division method is important to solve Integration using Partial Fraction (for improper fraction). { Divisor, D(x) }x + 3 x3 + 2x2 − x + 10 { dividend, P(x) } − (x3 + 3x2 ) −x 2 − x − (−x 2 − 3x) 2x + 10 − (2x + 6) 4 { Remainder, R(x) } x 2 − x + 2 { Quotient, Q(x) }
MAT238/ MAT438 : Foundation of Applied Mathematics 162 Example 2 (integration using Long Division) Express ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x into the sum of partial fractions. Hence, evaluate ( )( ) dx x x x x x x − + − − − + 3 1 2 4 4 2 4 3 2 . Solution : Step 1 : doing long division ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x = 3 3 2 4 4 3 2 4 3 2 − + − − − − + x x x x x x x ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x = Quotient + Divisor Remainder ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x = ( ) ( 3)( 1) 2 7 x 1 2 2 − + − + + + x x x x --------------- Step 2 : Express Divisor Remainder into the sum of partial fraction ( 3)( 1) 2 7 2 2 − + − + x x x x = 3 1 2 + + + − x Bx C x A --------------- = ( ) ( )( ) ( 3)( 1) 1 3 2 2 − + + + + − x x A x Bx C x = Remainder Quotient Divisor
Chapter 3 : Integration Techniques 163 Equating numerator for both sides, 2 7 2 x − x + = ( 1) ( )( 3) 2 A x + + Bx + C x − Substitute the values of A, B, and C into , ( 3)( 1) 2 7 2 2 − + − + x x x x = ( ) 1 0 2 3 1 2 + − + − x x x ( 3)( 1) 2 7 2 2 − + − + x x x x = 1 2 3 1 2 + − x − x (substitute into ) Hence, ( 3)( 1) 2 4 4 2 4 3 2 − + − − − + x x x x x x = ( ) 1 2 3 1 1 2 + − − + + x x x Step 3 : Integrate the given question dx x x x x x x − + − − − + ( 3)( 1) 2 4 4 2 4 3 2 = ( ) dx x x x + − − + + 1 2 3 1 1 2 = ( ) dx x dx x x dx + − − + + 1 2 3 1 1 2 = ( ) dx x dx x x dx + − − + + 1 1 2 3 1 1 2 = x x x C x 2 + + − − + −1 ln 3 2 tan 2 # When = 3 3 2 − 2(3) + 7 = (3 2 + 1) + 0 10 = 10 = 1 When = 0, = 1 0 2 − 2(0) + 7 = (1)(0 2 + 1) + (−3) 7 = 1 − 3 3 = −6 = −2 When = 1 ( ), = 1, = −2 1 2 − 2(1) + 7 = (1)(1 2 + 1) + ( − 2)(1 − 3) 6 = 2 + ( − 2)(−2) 4 = −2 + 4 2 = 0 = 0
MAT238/ MAT438 : Foundation of Applied Mathematics 164 Example 3 Express ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x into the sum of partial fractions. Hence, evaluate ( )( ) dx x x x x x − + + − + 1 3 2 2 7 3 2 . Solution : Step 1 : doing long division ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x = 2 3 2 2 7 2 3 2 + − + − + x x x x x ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x = Quotient + Divisor Remainder ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x = ( 1)( 3) 7 − + + + x x x x --------------- Step 2 : Express Divisor Remainder into the sum of partial fraction ( 1)( 3) 7 − + + x x x = 1 + 3 + − x B x A --------------- = ( ) ( ) ( 1)( 3) 3 1 − + + + − x x A x B x Equating numerator for both sides, x + 7 = A(x + 3) + B(x −1) = Remainder Quotient Divisor
Chapter 3 : Integration Techniques 165 Substitute the values of A and B into , ( 1)( 3) 7 − + + x x x = 3 1 1 2 + − x − x (substitute into ) Hence, ( 1)( 3) 2 2 7 3 2 − + + − + x x x x x = 3 1 1 2 + − − + x x x Step 3 : Integrate the given question ( )( ) dx x x x x x 1 3 2 2 7 3 2 − + + − + = dx x x x + − − + 3 1 1 2 = dx x dx x x dx + − − + 3 1 1 1 2 = x x C x 2 + 2ln −1 − ln + 3 + 2 # When x = 1, 1+ 7 = A(1+ 3) + 0 8 = 4A A = 2 When x = −3 − 3 + 7 = 0 + B(− 3 −1) 4 = − 4B B = −1
MAT238/ MAT438 : Foundation of Applied Mathematics 166 Example 4 Express ( 2)( 1) 2 2 − + + x x x into the sum of partial fractions. Hence, evaluate ( )( ) dx x x x 2 1 2 2 − + + . Solution : Step 1 : doing long division ( 2)( 1) 2 2 − + + x x x = 2 2 2 2 − − + x x x ( 2)( 1) 2 2 − + + x x x ( 2)( 1) 2 2 − + + x x x = Quotient + Divisor Remainder ( 2)( 1) 2 2 − + + x x x = ( 2)( 1) 4 1 − + + + x x x --------------- Step 2 : Express into the sum of partial fraction ( 2)( 1) 4 − + + x x x = 2 +1 + − x B x A --------------- = ( ) ( ) ( 2)( 1) 1 2 − + + + − x x A x B x Equating numerator for both sides, x + 4= A(x +1) + B(x − 2) When x = 2, 2 + 4= A(2 +1) + 0 6 = 3A A = 2 When x = −1 −1+ 4 = 0 + B(−1− 2) 3 = − 3B B = −1 = Remainder Quotient Divisor
Chapter 3 : Integration Techniques 167 Substitute the values of A and B into , ( 2)( 1) 4 − + + x x x = 1 1 2 2 + − x − x (substitute into ) Hence, ( 2)( 1) 2 2 − + + x x x = 1 1 2 2 1 + − − + x x Step 3 : Integrate the given question ( )( ) dx x x x − + + 2 1 2 2 = dx x x + − − + 1 1 2 2 1 = dx x dx x dx + − − + 1 1 2 1 1 2 = x + 2ln x − 2 − ln x +1 + C #
MAT238/ MAT438 : Foundation of Applied Mathematics 168 More Examples (from previous Semester papers) Example 5/ page 3/ MAT238/ OCT 2004/ Q4a (10 marks) Evaluate the integral dx x x x x + + + 3 3 2 2 . Solution : Note : x x x x + + + 3 3 2 2 is an improper integral because the degree of numerator to the degree of denominator so, we need to doing long division and express x x x x + + + 3 3 2 2 = Quotient + Divisor Remainder before we express into the sum of partial fraction Step 1 : doing long division x x x x + + + 3 3 2 2 x x x x + + + 3 3 2 2 = Quotient + Divisor Remainder x x x x + + + 3 3 2 2 = x x x + + + 3 2 1 --------------- Step 2 : Express Divisor Remainder into the sum of partial fraction x x x + + 3 2 = ( 1) 2 2 + + x x x = 1 2 + + + x Bx C x A --------------- = ( ) ( ) ( 1) 1 2 2 + + + + x x A x x Bx C = ( ) 1 2 0 0 0 2 2 3 2 3 2 3 2 + − + + + + + + + x x x x x x x x x x Remainder Quotient Divisor
Chapter 3 : Integration Techniques 169 Equating numerator for both sides, x + 2 = A(x + 1) + x(Bx + C) 2 -------------- Expand : x + 2 = Ax + A + Bx + Cx 2 2 0 2 2 x + x + = (A + B)x + Cx + A 2 By comparing the coefficient of Constant : A = 2 x : C = 1 x 2 : A + B = 0 2 + B = 0 → B = −2 Substitute the values of A, B and C into , x x x + + 3 2 = 1 2 2 1 2 + − + + x x x x x x + + 3 2 = 1 1 1 2 2 2 2 + + + − x x x x (substitute into ) Hence, x x x x + + + 3 3 2 2 = 1 1 1 2 2 1 2 2 + + + + − x x x x Step 3 : Integrate the given question dx x x x x + + + 3 3 2 2 = dx x x x x + + + + − 1 1 1 2 2 1 2 2 = dx x dx x x dx x dx + + + + − 1 1 1 1 2 1 2 2 2 = x + x − x + + x + C 2 −1 2ln ln 1 tan #
MAT238/ MAT438 : Foundation of Applied Mathematics 170 Example 6/ page 11/ MAT238/ NOV 2005/ Q4c (9½ marks) Evaluate dx x x x x x − − − − 3 2 4 3 1 using integration by partial fractions. Solution : Note : 3 2 4 3 1 x x x x x − − − − is an improper integral because the degree of numerator to the degree of denominator so, we need to doing long division and express 3 2 4 3 1 x x x x x − − − − = Quotient + Divisor Remainder before we express into the sum of partial fraction Step 1 : doing long division 3 2 4 3 1 x x x x x − − − − 3 2 4 3 1 x x x x x − − − − = Quotient + Divisor Remainder 3 2 4 3 1 x x x x x − − − − = 3 2 1 x x x x − − − + --------------- Step 2 : Express Divisor Remainder into the sum of partial fraction 3 2 1 x x x − − − = x ( 1) 1 2 − − − x x = 1 2 − + + x C x Ax B --------------- = ( )( ) x ( 1) 1 2 2 − + − + x Ax B x Cx = ( ) x x x x x x x x x x 1 0 1 4 3 3 2 4 3 2 − − − − − − + − − Remainder Quotient Divisor
Chapter 3 : Integration Techniques 171 Equating numerator for both sides, − x −1 = ( )( ) 2 Ax + B x − 1 + Cx -------------- Expand : − x −1 = 2 2 Ax + Bx − Ax − B + Cx 0 1 2 x − x − = (A + C)x + (B − A)x − B 2 By comparing the coefficient of Constant : −B = −1 → B = 1 x : B − A = −1 1 − A = −1 → A = 2 x 2 : A + C = 0 2 + C = 0 → C = −2 Substitute the values of A, B and C into , 3 2 1 x x x − − − = 1 2 1 2 2 − − + x x x 3 2 1 x x x − − − = 1 2 1 2 2 2 − + − x x x x 3 2 1 x x x − − − = 1 2 1 2 2 − + − x x x (substitute into ) Hence, 3 2 4 3 1 x x x x x − − − − = 1 2 1 2 2 − + + − x x x x Step 3 : Integrate the given question − − − − dx x x x x x 3 2 4 3 1 = dx x x x x − + + − 1 2 1 2 2 = dx x dx x dx x x dx − + + − − 1 1 2 1 2 2 = x C x x x − − + − + + − 2ln 1 1 2ln 2 2 1 = x C x x x + − − 2ln − 1 + 1 2ln 2 2 #
MAT238/ MAT438 : Foundation of Applied Mathematics 172 Example 7/ page 19/ MAT238/ OCT 2006/ Q4c (10 marks) Evaluate the integral of the improper rational function ( ) dx x x − 2 2 2 1 16 . Solution : Solution : Note : ( ) 2 2 2 1 16 x − x is an improper integral because the degree of numerator to the degree of denominator so, we need to doing long division and express ( ) 2 2 2 1 16 x − x = Quotient + Divisor Remainder before we express into the sum of partial fraction Step 1 : doing long division ( ) 2 2 2 1 16 x − x = 4 4 1 16 2 2 x − x + x ( ) 2 2 2 1 16 x − x = Quotient + Divisor Remainder ( ) 2 2 2 1 16 x − x = 4 4 1 16 4 4 2 − + − + x x x ( ) 2 2 2 1 16 x − x = ( ) 2 2 1 16 4 4 − − + x x --------------- Step 2 : Express Divisor Remainder into the sum of partial fraction ( ) 2 2 1 16 4 − − x x = ( ) 2 2 1 2 −1 + − x B x A --------------- = ( ) ( ) 2 2 1 2 1 − − + x A x B (note : set the denominator of RHS exactly equals to denominator of LHS) = Remainder Quotient Divisor
Chapter 3 : Integration Techniques 173 Equating numerator for both sides, 16x − 4 = A(2x −1) + B -------------- Expand : 16x − 4 = 2Ax − A+B 16x − 4 = 2Ax + (B − A) By comparing the coefficient of x : 2A = 16 → A = 8 Constant : B − A = −4 B − 8 = −4 → B = 4 Substitute the values of A and B into , ( ) 2 2 1 16 4 − − x x = ( ) 2 2 1 2 −1 + − x B x A ( ) 2 2 1 16 4 − − x x = ( ) 2 2 1 4 2 1 8 − + x − x (substitute into ) Hence, ( ) 2 2 2 1 16 x − x = ( ) 2 2 1 16 4 4 − − + x x ( ) 2 2 2 1 16 x − x = ( ) 2 2 1 4 2 1 8 4 − + − + x x Step 3 : Integrate the given question ( ) dx x x − 2 2 2 1 16 = ( ) dx x x − + − + 2 2 1 4 2 1 8 4 = ( ) dx x dx x dx − + − + 2 2 1 1 4 2 1 1 4 8 = dx ( x ) dx ( x ) dx − − + − + − 1 2 4 8 2 1 4 2 1 = ( ) C x x + − + • − + • 3 2 1 2 1 ln 2 1 4 2 1 4x 8 3 = + x − + ( x − ) + C 3 2 1 3 2 4x 4ln 2 1 #
MAT238/ MAT438 : Foundation of Applied Mathematics 174 3.2.5 Procedure to Integrate using Partial Fraction ( ) ( ) dx D x P x Check whether the degree of P(x) D(x)? Yes (The degree of P(x) D(x)) ) No (The degree of P(x) D(x)) ) Use long division method and write into the form of ( ) ( ) ( ) ( ) D(x) R x Q x D x P x = + , where the degree of R(x) D(x) Express ( ) D(x) R x into the sum of partial fraction (using Type I @ Type II @ Type III Integrate ( ) ( ) dx D x P x = ( ) ( ) ( ) dx + D x R x Q x Express ( ) D(x) P x into the sum of partial fraction (using Type I @ Type II @ Type III Integrate ( ) ( ) dx D x P x end end
Chapter 3 : Integration Techniques 175 Hot Tips!! (for integration using partial fraction) In general, • (Type I) dx = ln f(x) + c • (Type II) dx = ln f(x) and/or without tan−1 f(x) + c • (Type III) dx = ln f(x) and power rule (with u-subs) + c Long division not always comes together with partial fractions. Sometimes we need to use long division method without using partial fractions as follows, dx x x + 2 2 1 (try this!!)
MAT238/ MAT438 : Foundation of Applied Mathematics 176 TUTORIAL 3.1 : INTEGRATION BY PARTIAL FRACTIONS 1. Decompose 22− 2+−6 into its partial fractions. Hence, evaluate ∫ 22− 2+−6 . 2. Decompose 1+ 2 2(−1) into its partial fractions. Hence, find ∫ 1+ 2 2(−1) . 3. Decompose 2+8 (+1) 2 into its partial fractions. Hence, evaluate ∫ 2+8 (+1) 2 . 4. Decompose 1+3 ( 2+9) into its partial fractions. Hence, evaluate ∫ 1+3 ( 2+9) . 5. Decompose 3+5 (+3) 2 into its partial fractions. Hence, evaluate ∫ 3+5 (+3) 2 . 6. Decompose 2 3+1 (−1)( 2+3) into its partial fractions. Hence, evaluate ∫ 2 3+1 (−1)( 2+3) . 7. Use partial fraction to show that 2+1 (+1) 2 = 1 − 1 +1 + 1 (+1) 2 into its partial fractions. Hence, evaluate ∫ 2+1 (+1) 2 . 8. Use partial fraction to show that 3+11 2−−6 = 4 −3 − 1 +2 into its partial fractions. Hence, evaluate ∫ 3+11 2−−6 . 9. Use partial fraction to show that 3+1 (+3)(−1) = 2 +3 + 1 −1 into its partial fractions. Hence, evaluate ∫ 3+1 (+3)(−1) . 10. Decompose 1+2 ( 2+4) into its partial fractions. Hence, evaluate ∫ 1+2 ( 2+4) . 11. Evaluate ∫ 2+4 2(−4) . 12. Write 5 (−1) 3 in the form of partial fractions. Hence, evaluate ∫ 5 (−1) 3 .
Chapter 3 : Integration Techniques 177 ANSWER TUTORIAL 3.1 1. 22− 2+−6 = 4 −2 − 5 +3 ; ∫ 22− 2+−6 = 4 ȁ − 2ȁ − 5 ȁ + 3ȁ + 2. 1+ 2 2(−1) = − 1 − 1 2 + 2 −1 ; ∫ 1+ 2 2(−1) = − + 1 + 2 ȁ − 1ȁ + 3. 2+8 (+1) 2 = 8 − 8 +1 − 6 (+1) 2 ; ∫ 2+8 (+1) 2 = 8 − 8 ȁ + 1ȁ + 6 +1 + 4. 1+3 ( 2+9) = 1 9 + 27− 9( 2+9) ; ∫ 1+3 ( 2+9) = 1 9 + −1 ( 3 ) − 1 18 ȁ 2 + 9ȁ + 5. 3+5 (+3) 2 = 1 3 − 1 3(+3) + 4 (+3) 2 ; ∫ 3+5 (+3) 2 = 1 3 − 1 3 ȁ + 3ȁ − 4 +3 + 6. 2 3+1 (−1)( 2+3) = 2 + 3 4(−1) + 5−19 4( 2+3) ; ∫ 2 3+1 (−1)( 2+3) = 2 + 3 4 ȁ − 1ȁ + 5 8 ȁ 2 + 3ȁ − 19 4√3 −1 ( √3 ) + 7. ∫ 2+1 (+1) 2 = − ȁ + 1ȁ − 1 +1 + 8. ∫ 3+11 2−−6 = 4 ȁ − 3ȁ − ȁ + 2ȁ + 9. ∫ 3+1 (+3)(−1) = 2 ȁ + 3ȁ + ȁ − 1ȁ + 10. 1+2 ( 2+4) = 1 4 + 8− 4( 2+4) ; ∫ 1+2 ( 2+4) = 1 4 + −1 ( 2 ) − 1 8 ȁ 2 + 4ȁ + 11. 2+4 2(−4) = − 1 4 − 1 2 + 5 4(−4) ; ∫ 2+4 2(−4) = − 1 4 + 1 + 5 4 ȁ − 4ȁ + 12. 5 (−1) 3 = 5 (−1) 2 + 5 (−1) 3 ; ∫ 5 (−1) 3 = − 5 −1 − 5 2 (−1) 2 +
MAT238/ MAT438 : Foundation of Applied Mathematics 178 3.3 Integration with Trigonometric substitution To eliminate radicals : i) 2 2 a − x → take x = asin Reason : from x = asin , then 2 2 2 x = a sin Therefore, 2 2 a − x = 2 2 2 a − a sin = ( ) 2 2 a 1− sin = 2 2 a cos = a cos ii) 2 2 a + x → take x = atan Reason : from x = atan , then 2 2 2 x = a tan Therefore, 2 2 a + x = 2 2 2 a + a tan = ( ) 2 2 a 1+ tan = 2 2 a sec = a sec iii) 2 2 x − a → take x = asec Reason : from x = asec , then 2 2 2 x = a sec Therefore, 2 2 x − a = 2 2 2 a sec − a = ( 1) 2 2 a sec − = 2 2 a tan = a tan
Chapter 3 : Integration Techniques 179 Example 1a : Evaluate dx x x − 2 2 16 Solution for Example 1a : Integration by Trigonometric-substitution Substitute into the given question ∫ 2 √16 − 2 = ∫ 16 2 4 ∙ 4 = 16 ∫ 2 = 16 ∫ 1 2 (1 − 2) = 8 ∫(1 − 2) = 8 ( − 1 2 2) + = 8 − 4 2 + = 8 (−1 ( 4 ))− 4 ( √16 − 2 8 )+ = 8 −1 ( 4 )− 1 2 √16 − 4 2 + # 4 √16− 2 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ √16 − 2 = √16 − 16 2 = √16(1 − 2) = √16 2 = 4 ❶ = 4 = 4 → = −1 ( 4 ) = √16− 2 4 2 = 2 sin = 2 ( 4 ) ( √16− 2 4 ) = √16 − 2 8
MAT238/ MAT438 : Foundation of Applied Mathematics 180 Example 1b : Evaluate dx x x − 2 16 Solution for Example 1b : Method I (Trigonometric-substitution) Substitute into the given question Solution for Example 1b : Method II (Integration using u-substitution) ∫ √16 − 2 = ∫ 4 4 ∙ 4 = 4 ∫ = 4(−) + = −4 + # = −4 ( √16− 2 4 ) + = −√16− 2 + # 4 √16− 2 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ √16 − 2 = √16 − 16 2 = √16(1 − 2) = √16 2 = 4 ❶ = 4 = 4 → = √16 − 2 4 ∫ √16− 2 = ∫ √ ∙ −2 = − 1 2 ∫ 1 √ = − 1 2 ∫ − 1 2 = −( 1 2 ) 1 2 ( 1 2 ) + = − 1 2 + = −√ + = −√16 − 2 + # = 16 − 2 = −2 = −2
Chapter 3 : Integration Techniques 181 Example 1c : Evaluate dx x − 2 16 1 Solution for Example 1c : Method I (Trigonometric-substitution) Substitute into the given question Solution for Example 1c : Method II (Integration of Inverse Trigonometric Function) ∫ 1 √16 − 2 = ∫ 1 4 ∙ 4 = ∫ 1 = + = −1 ( 4 ) + # 4 √16 − 2 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ √16 − 2 = √16 − 16 2 = √16(1 − 2) = √16 2 = 4 ❶ = 4 = 4 → = −1 ( 4 ) ∫ 1 √16 − 2 = ∫ 1 √4 2 − 2 = ∫ 1 √ 2 − 2 = −1 ( ) + = −1 ( 4 ) + # Let's look at Examples 2a, 2b & 2c. These examples involve integration for terms containing − ; but there are no radicals. In the solution, we show you, that we can still use Trigonometric Substitution, even without radicals. Sometimes there is more than one method. You can choose which method is most convenient for you. However, if the question mentioned 'use Trigonometric- substitution' then you need to follow the requirements of the question. ∫ √ − = − ( ) +
MAT238/ MAT438 : Foundation of Applied Mathematics 182 Example 2a Evaluate dx x x − 2 2 16 answer : − x + 2ln 4 + x + 2ln 4 − x + C Solution for Example 2a : Method I (Long Division) When = −4 When = 4 2 16 − 2 = + 2 16 − 2 = −1 + 16 16 − 2 − 2 + 0 + 16 2 + 0 + 0 −1 −( 2 − 0 − 16) 16 quotient remainder divisor 2 16 − 2 = ∫ 2 16 − 2 = ∫ (−1 + 16 16 − 2 ) = ∫ −1 + ∫ 16 16 − 2 = − + 4 ℎ −1 ( 4 ) + # = − + 2 ȁ4 + ȁ − 2 ȁ4 − ȁ + # ∫ 16 16− 2 = 16 ∫ 1 4 2 − 2 = 16 ∙ 1 ℎ −1 ( ) + = 16 ∙ 1 4 ℎ −1 ( 4 )+ = 4 ℎ −1 ( 4 )+ 16 16 − 2 = 16 (4 + )(4 − ) = 4 + + 4 − = (4 − ) + (4 + ) (4 + )(4 − ) 16 = (4 − ) + (4 + ) 16 = (8) + 0 8 = 16 = 2 16 16 − 2 = 2 4 + + 2 4 − ∫ 16 16 − 2 = 2 ∫ 1 4 + + 2 ∫ 1 4 − = 2 ∫ 1 4 + − 2 ∫ −1 4 − = 2 ȁ4 + ȁ − 2 ȁ4 − ȁ + 16 = 0 + (8) 8 = 16 = 2 ∫ 16 16 − 2 = 16 ∫ 1 4 2 − 2 = 16 ∙ 1 2 | + − |+ = 16 ∙ 1 2(4) | 4+ 4− |+ = 2 | 4+ 4− |+ = 2 ȁ4 + ȁ −2 ȁ4 − ȁ + Refer Method I, II, III Method I Method II Method III ∫ − = − ( ) + ∫ − = | + − | + − ( ) = | + − |
Chapter 3 : Integration Techniques 183 Solution for Example 2a : Method II (Trigonometric-substitution) Substitute into the given question ∫ 2 16 − 2 = ∫ 16 2 16 2 ∙ 4 = 4 ∫ 2 = 4 ∫ 1 − 2 = 4 ∫ ( 1 − 2 ) = 4 ∫( − ) = 4 (ȁ + ȁ − ) + = 4 ȁ + ȁ − 4 + = 4 | 4 √16 − 2 + √16 − 2 | − 4 ( 4 ) + = 4 | 4 + √16 − 2 | − + = 4 (ȁ4 + ȁ − |√16 − 2|) − + = 4 (ȁ4 + ȁ − 1 2 ȁ16 − 2 ȁ) − + = 4 ȁ4 + ȁ − 2 ȁ16 − 2 ȁ − + = 4 ȁ4 + ȁ − 2 ȁ(4 + )(4 − )ȁ − + = 4 ȁ4 + ȁ − 2 (ȁ4 + ȁ + ȁ4 − ȁ) − + = 4 ȁ4 + ȁ − 2 ȁ4 + ȁ − 2ȁ4 − ȁ − + = 2 ȁ4 + ȁ − 2ȁ4 − ȁ − + # 4 √16 − 2 ❶ = 4 , = 4 , = √16 − 2 4 → = 4 √16− 2 = √16 − 2 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ 16 − 2 = 16 − 16 2 = 16(1 − 2) = 16 2 ∫ = ȁ + ȁ +
MAT238/ MAT438 : Foundation of Applied Mathematics 184 Example 2b Evaluate dx x x − 2 16 answer :− 1 2 ȁ16 − 2 ȁ + or − 1 2 ȁ4 + ȁ − 1 2 ȁ4 − ȁ + Solution for Example 2b : Method I (u-substitution) Solution for Example 2b : Method II (Trigonometric-substitution) Substitute into the given question ∫ 16 − 2 = ∫ ∙ −2 = − 1 2 ∫ 1 = − 1 2 ȁȁ + = − 1 2 ȁ16 − 2 ȁ + # = 16 − 2 = −2 = −2 ∫ 16 − 2 = ∫ 4 16 2 ∙ 4 = ∫ = ∫ ∙ − = − ∫ 1 = −ȁȁ + = −ȁȁ + = − | √16− 2 4 | + = −(√16− 2 − (4)) + = −√16− 2 +(4) + = − 1 2 ȁ16 − 2 ȁ+ # 4 √16− 2 ❶ = 4 , = 4 , = √16 − 2 4 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ 16 − 2 = 16 − 16 2 = 16(1 − 2) = 16 2 = = − = −
Chapter 3 : Integration Techniques 185 Example 2c Evaluate dx x − 2 16 1 answer : + x − ln 4 − x + C 8 1 ln 4 8 1 Solution for Example 2c : Method I (inverse hyperbolic function) ∫ 1 16 − 2 = ∫ 1 4 2 − 2 = 1 ℎ −1 ( ) + = 1 4 ℎ −1 ( 4 ) + ∫ 1 16 − 2 = ∫ 1 4 2 − 2 = 1 2 | + − | + = 1 2(4) | 4 + 4 − | + = 1 8 | 4 + 4 − | + = 1 8 ȁ4 + ȁ − 1 8 ȁ4 − ȁ + Method I Method II ∫ − = − ( ) + ∫ − = | + − | + − ( ) = | + − |
MAT238/ MAT438 : Foundation of Applied Mathematics 186 Solution for Example 2c : Method II (Trigonometric-substitution) Substitute into the given question ∫ 1 16 − 2 = ∫ 1 16 2 ∙ 4 = 1 4 ∫ 1 = 1 4 ∫ = 1 4 ȁ + ȁ + = 1 4 | 4 √16 − 2 + √16 − 2 | + = 1 4 | 4 + √16 − 2 | + = 1 4 (ȁ4 + ȁ − |√16 − 2|) + = 1 4 (ȁ4 + ȁ − 1 2 ȁ16 − 2 ȁ) + = 1 4 ȁ4 + ȁ − 1 8 ȁ16 − 2 ȁ + = 1 4 ȁ4 + ȁ − 1 8 ȁ(4 + )(4 − )ȁ + = 1 4 ȁ4 + ȁ − 1 8 (ȁ4 + ȁ + ȁ4 − ȁ) + = 1 4 ȁ4 + ȁ − 1 8 ȁ4 + ȁ − 1 8 ȁ4 − ȁ + = 1 8 ȁ4 + ȁ − 1 8 ȁ4 − ȁ + # 4 √16 − 2 ❶ = 4 , ❷ 2 = 16 2 ❸ = 4 ➍ 16 − 2 = 16 − 16 2 = 16(1 − 2) = 16 2 ∫ = ȁ + ȁ + ❶ = 4 , = 4 , = √16 − 2 4 → = 4 √16 − 2 = √16 − 2
Chapter 3 : Integration Techniques 187 Example 3 Evaluate + 2 2 x 25 x dx Solution : Substitute into the given question ∫ 2√25 + 2 = ∫ 5 2 25 2 ∙ 5 = 1 25 ∫ 2 = 1 25 ∫ 2 = 1 25 ∫ 1 ∙ 2 2 = 1 25 ∫ 2 = 1 25 ∫ 2 ∙ = 1 25 ∫ 1 2 = 1 25 ∫ −2 = 1 25 ( −1 −1 ) + = − 1 25 + = − 1 25 + = − 1 25 + = − 1 25 ( √25 + 2 ) + = − √25 + 2 25 + # 5 ❶ = 5 , = 5 = √25 + 2 → = √25 + 2 ❶ = 5 , ❷ 2 = 25 2 ❸ = 5 2 ➍√ 25 + 2 = √25 + 25 2 = √25(1 + 2) = √252 = 5 = = =
MAT238/ MAT438 : Foundation of Applied Mathematics 188 Example 4 Evaluate ( ) − 2 3 4 9 2 x dx Solution : Substitute into the given question ∫ (4 2 − 9) 3 2 = ∫ ((4 2 − 9) 1 2) 3 = ∫ (√4 2 − 9) 3 = ∫ 3 2 (3 ) 3 = ∫ 3 2 27 3 = 1 18 ∫ 2 = 1 18 ∫ 2 = 1 18 ∫ 1 ∙ 2 2 = 1 18 ∫ 2 = 1 18 ∫ 2 ∙ = 1 18 ∫ 1 2 = 1 18 ∫ −2 = 1 18 ( −1 −1 )+ = − 1 18 + = − 1 18 + = − 1 18 + = − 1 18 ( 2 √4 2 − 9 )+ = − 9√4 2 − 9 + # 3 2 √4 2 − 9 ❶ = 3 2 , = 2 3 → = 3 2 = √4 2 − 9 2 → = 2 √4 2 − 9 ❶ = 3 2 , ❷ 2 = 9 4 2 ❸ = 3 2 ➍ √4 2 −9 = √4 ( 9 4 2) −9 = √9 2 −9 = √9(2 − 1) = √9 2 = 3 = = = (4 2 − 9) 3 2 ⁄ = ((4 2 − 9) 1 2 ⁄ ) 3 = (√4 2 − 9) 3 To find the value of a (if the coefficient of x2 ≠ 1) : √4 2 − 9 = √4 ( 2 − 9 4 ) = 2√ 2 − ( 3 2 ) 2 √ 2 − 2, take = , where = 3 2
Chapter 3 : Integration Techniques 189 More Examples (from previous Semester papers) Example 5/ page 119/ MAT238/ SEP 2013/ Q3a (10 marks) Using trigonometric substitution x sin 2 3 = , solve dx x x − 2 2 3 4 . Solution : Substitute into the given question ∫ 2 √3 − 4 2 = ∫ 3 4 2 √3 ∙ √3 2 = 3 8 ∫ 2 = 3 8 ∫ 1 2 (1 − 2) = 3 16 ∫(1 − 2) = 3 16 ( − 1 2 2) + = 3 16 − 3 32 2 + = 3 16 (−1 ( 2 √3 )) − 3 32 ( 4√3 − 4 2 3 ) + = 3 16 −1 ( 2 √3 ) − 1 8 √3 − 4 2 + # √3 2 √3 − 4 2 ❶ = √3 2 , ❷ 2 = 3 4 2 ❸ = √3 2 ➍ √3 − 4 2 = √3 − 4 ( 3 4 2) = √3 − 3 2 = √3(1 − 2) = √3 2 = √3 ❶ = √3 2 = 2 √3 → = −1 ( 2 √3 ) = √3− 4 2 √3 sin 2 = 2 sin = 2 ( 2 √3 ) ( √3− 4 2 √3 ) = 4√3 − 4 2 3
MAT238/ MAT438 : Foundation of Applied Mathematics 190 Example 5/ page 70/ MAT238/ APR 2010/ Q3b (11 marks) Use the substitution in the form of x = 3 tan to solve dx x x + 4 2 9 . Solution : Substitute into the given question ∫ √9 + 2 4 = ∫ 3 81 4 ∙ 3 2 = 1 9 ∫ 3 4 = 1 9 ∫ 3 4 = 1 9 ∫ 1 3 ∙ 4 4 = 1 9 ∫ 4 = 1 9 ∫ 4 ∙ = 1 9 ∫ 1 4 = 1 9 ∫ −4 = 1 9 ( −3 −3 ) + = − 1 27 3 + = − 1 27 3 + = − 1 27 3 + = − 1 27 ( √9 + 2 ) 3 + = − (√9 + 2) 3 27 3 + # 3 ❶ = 3 = 3 = √9 + 2 → = √9 + 2 ❶ = 3 , ❷ 2 = 9 2 , 4 = 81 4 ❸ = 3 2 ➍ √ 9 + 2 = √9 + 9 2 = √9(1 + 2) = √9 2 = 3 = = =
Chapter 3 : Integration Techniques 191 Example 6/ page 43/ MAT238/ DEC 2008/ Q3b (10 marks) Evaluate − x dx 2 4 25 by using the trigonometric substitution 5x = 2sin . Solution : Substitute into the given question ∫ √4 − 25 2 = ∫ 2 ∙ 2 5 = 4 5 ∫ 2 = 4 5 ∫ 1 2 (1 + 2) = 2 5 ∫(1 + 2) = 2 5 ( + 1 2 2) + = 2 5 + 1 5 2 + = 2 5 (−1 ( 5 2 )) + 1 5 ( 5√4 − 25 2 2 ) + = 2 5 −1 ( 5 2 ) + 1 2 √4 − 25 2 + # 2 5 √4 − 25 2 ❶ 5 = 2 → = 2 5 , ❷ 2 = 4 25 2 ❸ = 2 5 ➍ √4 − 25 2 = √4 − 25 ( 4 25 2) = √4 − 4 2 = √4(1 − 2) = √4 2 = 2 ❶ = 2 5 = 5 2 → = −1 ( 5 2 ) = √4− 25 2 2 sin 2 = 2 sin = 2 ( 5 2 ) ( √4− 25 2 2 ) = 5√4 −25 2 2
MAT238/ MAT438 : Foundation of Applied Mathematics 192 TUTORIAL 3.2 : Integration by Trigonometric substitution Evaluate the following integrals using Trigonometric substitution. 1. ∫ √ 2 − 9 3 . ∶ √ 2 − 9 3 − −1 ( 3 ) + 2. ∫ 4 ( 2 + 2) 3 2 . ∶ 2 √ 2 + 2 + 3. ∫(4 − 2 ) − 3 2 . ∶ 4√4 − 2 + 4. ∫ √4 2 − 9 2 . ∶ 2 | 2 + √4 2 − 9 3 | − √4 2 − 9 + 5. ∫ √16 − 4 2 . = 2 ∶ − 2 3 (√4 − 2) 3 + 6. ∫ √9 − 2 2 0 . ∶ ∫ √9 − 2 = −√9 − 2 + ∫ √9 − 2 2 0 = 0.7639 7. ∫ √3 + 2 . = √3 ∶ (√3 + 2) 3 3 + 8. ∫ √1 − 2 . ∶ − 1 3 (√1 − 2) 3 + 9. ∫ 2√1 − 2 . ∶ 1 8 −1 − 1 32 (4−1) + 10. ∫ 1 2√1 + 2 . ∶ − √1 + 2 + 11. ∫ 3 √4 2 − 1 . ∶ 1 16 √4 2 − 1 + 1 48 (√4 2 − 1) 3 + 12. ∫ √16 + ( − 1) 2 . ∶ √ 2 − 2 + 17 + | 1 4 ( − 1 + √ 2 − 2 + 17)| +
Chapter 3 : Integration Techniques 193 3.4 Integration by Parts (using ‘I LATE’ rule) u dv = uv − v du An acronym that is very helpful to remember when using integration by parts is I LATE. Whichever function comes first in the following list should be u : I Inverse Trigonometric functions tan−1x, sin−1x, etc NR L Logarithmic functions ln x NR A Algebraic functions x, 3x 2 , x 3 , etc NR T Trigonometric functions sinx, cos3x, tan2x, etc R E Exponential functions e x , e2x , e3x , etc R Note : Priority to choose u and dv : u (easy to differentiate), dv(easy to integrate) NR = Non Repeatable Integral R = Repeatable Integral Example 1 Evaluate ln x dx . Solution : ∫ = ∫ = − ∫ = ∙ − ∫ ∙ 1 = − ∫ 1 = − + # = = = 1 =