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Unformatted text preview: THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MATH1010 University Mathematics Definite Integration 1 Preliminaries Exercise 1. (Level 3) (a) Let f : R\{0} −→ R be the function defined by f (x) = 1 for any x ∈ R\{0}. x2 Consider the functions F1 , F2 , F3 , F4 , F5 , whose ‘formulae of definition’ are given domain R\{0}. Which of them are primitives of f ? Which not? Why? ( 1 −1/x + 3 i. F1 (x) = − for any x ∈ R\{0}. iv. F4 (x) = x −1/x − 2 1 ii. F2 (x) = − + 1 for any x ∈ R\{0}. x −1/x + 3 ( −1/x if x < 0 v. F5 (x) = −1/x − 2 iii. F3 (x) = −1/x + 1 if x > 0 −1/x + 4 (b) Let f : R\{0} −→ R be the function defined by f (x) = below. Each of them has if x < 0 if x > 0 if x < 0 if 0 < x < 1 if x ≥ 1 1 for any x ∈ R\{0}. x Consider the functions F1 , F2 , F3 , F4 , F5 , whose ‘formulae of definition’ are given below. Each of them has domain R\{0}. Which of them are primitives of f ? Which not? Why? ( i. F1 (x) = ln(|x|) for any x ∈ R\{0}. ln(−x) + 3 if x < 0 iv. F4 (x) = ln(x) − 2 if x > 0 ii. F2 (x) = ln(|x|) + 1 for any x ∈ R\{0}. iii. F3 (x) = ( ln(−x) if ln(x) + 1 if x<0 x>0 v. F5 (x) = ( − ln(−x) ln(x) if x < 0 if x > 0 −1 if x < −1 (c) Let f : R\{−1, 1} −→ R be the function defined by f (x) = 0 if −1 < x < 1 1 if x > 1 Consider the functions F1 , F2 , F3 , F4 , whose ‘formulae of definition’ are given below. R\{−1, 1}. Which of them are primitives of f ? Which not? Why? −x if x < −1 −x − 1 if i. F1 (x) = iii. F3 (x) = 0 if −1 < x < 1 0 if x if x > 1 x+1 if −x + 1 if x < −1 −x − 2 if ii. F2 (x) = iv. F4 (x) = 0 if −1 < x < 1 100 if x−1 if x > 1 x+3 if . Each of them has domain x < −1 −1 < x < 1 x>1 x < −1 −1 < x < 1 x>1 2 Solution. (a) F1 , F2 , F3 , F4 are primitives of f , because their respective first derivatives on R\{0} are f . F5 is not even differentiable at 1. Then F5 is not a primitive of f . (b) F1 , F2 , F3 , F4 are primitives of f , because their respective first derivatives on R\{0} are f . F5 is differentiable on R\{0} but F5′ (−1) 6= f (−1). Then F5 is not a primitive of f . (c) F1 , F2 , F3 , F4 are primitives of f , because their respective first derivatives on R\{−1, 1} are f . Exercise 2. (Level 3) Z Find f (x)dx for the following functions f (x). 4x − 1, x < 1 3 (a) f (x) = √ , x≥1 x (b) f (x) = ( e2x , cos2 x, |x−3|(x−3) 2 ( 2x2 − x + 5, x < 1 √ x≥1 6 x, ( e2x 2 , 2x+sin 2x+2 , 4 x<0 x≥0 ( − ln(1 − x), ln(1 + x), x<0 x≥0 +C (d) F (x) + C where F (x) = 3 x < −2 x − 12x, 3 (e) F (x) + C where F (x) = −x + 12x + 32, −2 ≤ x < 2 3 x − 12x + 64, x≥2 (f) F (x) + C where F (x) = 2 1 1 + |x| (f) f (x) = | ln x| Solution. (a) F (x) + C where F (x) = (c) (d) f (x) = (e) f (x) = 3|x2 − 4| x<0 x≥0 (b) F (x) + C where F (x) = (c) f (x) = |x − 3| ( x(1 − ln x), 0<x<1 x(ln x − 1) + 2, x ≥ 1 Integration by substitution Exercise 1. (Level 1) Evaluate the following integrals by trigonometric substitution. 3 (a) Z x2 dx 1 + x2 (b) Z dx (c) Z r (d) Z (1 + x2 ) (e) Z x2 dx √ 9 − x2 (1 − 3 x2 ) 2 1+x dx 1−x dx 3 2 Solution. (a) x − tan−1 x + C x +C (b) √ 1 − x2 √ (c) − 1 − x2 + sin−1 x + C x +C (d) √ 1 + x2 9 x x√ 9 − x2 + C sin−1 − 2 3 2 √ (f) ln |x + 4 + x2 | + C (f) Z dx √ 4 + x2 (g) Z x2 (h) Z x2 (i) Z (j) Z p 16 − x2 dx dx √ x2 + 4 dx (4x2 3/2 + 1) 1 (2x − x2 ) √ (g) 16 − x2 (h) −  3/2  x x3 +C − 2x + 32 sin−1 4 4 √ x2 + 4 +C 4x x (i) √ 4x2 + 1 (e) x−1 (j) √ 2x − x2 Exercise 2. (Level 1) Use a suitable substitution to evaluate the following integral. Z Z 2 dx √ (h) xe−x dx (a) 2 − 5x Z Z 3x ex dx e +1 (i) dx (b) 2 + ex ex + 1 Z Z dx x (j) √ (c) dx x e + e−x 1 − x2 Z Z p 2 3 3 (k) tan xdx 1 + x dx (d) x (e) Z xdx (1 + x2 )2 (f) Z √ (g) Z 1 1 sin dx 2 x x dx x(1 + x) (l) Z dx 1 + ex (m) Z x(x2 + 2)99 dx (n) Z x √ dx 25 − x2 4 (o) Z √ x dx 3x2 + 1 (s) Z √ x x − 1dx (p) Z √ x2 dx 9 − x3 (t) Z √ (x + 2) x − 1dx (q) Z x(x + 2)99 dx (u) Z xdx √ x+9 (r) Z √ (v) Z x3 (1 + 3x2 ) 2 dx xdx 4x + 5 √ Solution. (a) − 52 2 − 5x + C (b) 1 2x 2e − ex + x + C √ (c) − 1 − x2 + C (d) 1 4 (1 4 + x3 ) 3 + C 1 (e) − 2(1+x 2) + C 1 (l) x − ln(1 + ex ) + C (m) 1 2 200 (x + 2)100 + C √ (n) − 25 − x2 + C (o) 1 3 √ 3x2 + 1 + C √ (p) − 23 9 − x3 + C (q) (x+2)101 101 (r) 1 12 (2x (h) − 21 e−x + C (s) 2 15 (x (i) ln(2 + ex ) + C (t) 2 5 (x (j) tan−1 ex + C (u) 2 3 (x (k) − ln | cos x| + C (v) 1 135 (f) 2 tan−1 √ x+C (g) cos x1 + C 2 Exercise 3. (Level 1) Use a suitable substitution to evaluate the following integral. Z Z p dx 3 √ (d) x2 1 + x3 dx (a) 2 − 5x Z Z 3x xdx e +1 (e) (b) dx x (1 + x2 )2 e +1 Z Z x 2 √ dx (c) (f) xe−x dx 1 − x2 Solution. − (x+2)100 50 + C. √ − 5) 4x + 5 − 1)3/2 (3x + 2) − 1)3/2 (x + 4) √ − 18) x + 9 3x2 + 1 3/2 9x2 − 2  (g) Z ex dx 2 + ex (h) Z dx 1 + ex 5 √ (a) − 52 2 − 5x + C (b) 1 2x 2e 1 (e) − 2(1+x 2) + C − ex + x + C 2 (f) − 21 e−x + C √ (c) − 1 − x2 + C (d) 1 4 (1 (g) ln(2 + ex ) + C 4 + x3 ) 3 + C (h) x − ln(1 + ex ) + C Exercise 4. (Level 2) Compute the indefinite integrals below: Z Z √ xdx √ (b) x x − 1dx (a) x+9 (c) Z (x2 + 1) · √ 3 x − 1dx  Z Z  √ 9 1 2 1 3 (u − 9)du 2 3 xdx √ √ du = u 2 − 18u 2 + C = (x + 9) 2 − 18(x + 9) 2 , = = u− √ 3 3 u u x+9 where C is an arbitrary constant. Solution. (a) Z We have used the substitution u = x + 9. Z Z Z   √ √ 3 2 5 2 3 1 2 2 5 3 u 2 + u 2 du = u 2 + u 2 + C = (x − 1) 2 + (x − 1) 2 + C, where (b) x x − 1dx = (u + 1) udu = 5 3 5 3 C is an arbitrary constant. We have used the substitution u = x − 1. (c) Z = = = = Z Z (x2 + 1) · √ 3 x − 1dx [(u + 1)2 + 1] · 7 4 √ 3 udu 1 (u 3 + 2u 3 + 2u 3 )du 6 7 3 4 3 10 u 3 + u3 + u3 + C 10 7 2 6 3 3 10 7 4 (x − 1) 3 + (x − 1) 3 + (x − 1) 3 + C, where C is an arbitrary constant. 10 7 2 We have used the substitution u = x − 1. Exercise 5. (Level 2) (a) Verify that Z d  x  x + sin(x) 1 tan( ) = . Hence compute the indefinite integral dx. dx 2 1 + cos(x) 1 + cos(x) (b) Hence, or otherwise, compute the indefinite integral Z 2 + 2 cos(x) + 2x sin(x) + x2 dx (1 + cos(x))(x + sin(x)) 6 Solution. (a) Then x  1 x 1 1 d  tan( ) = sec2 ( ) = = . 2 dx 2 2 2 2 cos (x/2) 1 + cos(x) Z = = = = = = x + sin(x) dx 1 + cos(x) x  d  tan( ) dx dx 2 Z x x (x + sin(x))tan( ) − (1 + cos(x)) tan( )dx 2 2 Z x x x (x + sin(x))tan( ) − 2 cos2 ( ) tan( )dx 2 2 2 Z x x x (x + sin(x))tan( ) − 2 cos( ) sin( )dx 2 2 2 Z x (x + sin(x))tan( ) + − sin(x)dx 2 x (x + sin(x))tan( ) + cos(x) + C, where C is an arbitrary constant. 2 Z x + sin(x) (b) = Z 2 + 2 cos(x) + 2x sin(x) + x2 dx (1 + cos(x))(x + sin(x)) Z 1 + 2 cos(x) + cos2 (x) + sin2 (x) + 2x sin(x) + x2 dx (1 + cos(x))(x + sin(x)) (1 + cos(x))2 + (x + sin(x))2 dx (1 + cos(x))(x + sin(x))  Z  1 + cos(x) x + sin(x) dx + = x + sin(x) 1 + cos(x) x = ln(|x + sin(x)|) + (x + sin(x))tan( ) + cos(x) + C, 2 where C is an arbitrary constant. = Z Exercise 6. (Level 2) Let f (x), g(x) be the functions given by f (x) = sin(x) cos(x) , g(x) = . Here a, b are real a cos(x) + b sin(x) a cos(x) + b sin(x) constants satisfying a2 + b2 > 0. Z Z Compute the indefinite integrals (af (x) + bg(x))dx, (bf (x) − ag(x))dx. Hence, or otherwise, compute the Z Z indefinite integrals f (x)dx, g(x)dx. Z Z x Remark. Also use the substitution t = tan( ) for computing the indefinite integrals f (x)dx, g(x)dx re2 spectively. At some stage you have to perform partial fraction decomposition for a rational function whose denominator is a degree-4 polynomial. 7 sin(x) cos(x) , g(x) = . Solution. f (x) = a cos(x) + b sin(x) a cos(x) + b sin(x)  Z Z Z  a cos(x) b sin(x) dx = dx = x+C, where C is an arbitrary + (af (x) + bg(x))dx = a cos(x) + b sin(x) a cos(x) + b sin(x) constant.  Z Z  b cos(x) a sin(x) dx − (bf (x) − ag(x))dx = a cos(x) + b sin(x) a cos(x) + b sin(x) Z −a sin(b) + b cos(x) = dx a cos(x) + b sin(x) = ln(|a cos(x) + b sin(x)|) + C, where C is an arbitrary constant. Then Z f (x)dx =  Z  Z 1 a (af (x) + bg(x))dx + b (bf (x) − ag(x))dx a2 + b 2 = ax + b ln(|a cos(x) + b sin(x)|) + C, where C is an arbitrary constant. a2 + b 2 =  Z  Z 1 b (af (x) + bg(x))dx − a (bf (x) − ag(x))dx a2 + b 2 = bx − a ln(|a cos(x) + b sin(x)|) + C, where C is an arbitrary constant. a2 + b 2 Also, Z g(x)dx Exercise 7. (Level 2) Compute the indefinite integrals below: Z Z sin(x) (a) e cos(x)dx (f) 2sin(x) cos(x)dx Z (3x2 + 1)ex (c) Z x cos(sin(ex )) cos(ex )ex dx (d) Z 2 dx (e) Z 2x xdx (b) (b) +x dx 2 2 2 2 Z (3x + 1)e (g) Z (h) Z (i) Z 2 x Solution. (a) Z 3 (j) Z sin(x) dx (2 + cos(x))2 √ cos( x) √ dx x (k) Z p x x2 + 4dx √ cosh( x) √ dx x (l) Z xdx (x2 + 1)2 (m) Z x3 (x2 + 1)20 dx 1 sec2 x2   1 dx x esin(x) cos(x)dx = esin(x) + C, where C is an arbitrary constant. x3 +x dx = Z ex 3 +x · 3 d 3 (x + x)dx = ex +x + C, where C is an arbitrary constant. dx 8 (c) Z 2 2 2 x cos(sin(ex )) cos(ex )ex dx = = (d) Z x 2 dx = Z e x ln(2) dx = Z Z 2 2 1 d x2 sin′ (sin(ex )) sin′ (ex ) · (e ) · 2xdx 2 dx 2 1 sin(sin(ex )) + C, 2 where C is an arbitrary constant. ex ln(2) 2x 1 · (ex ln(2) · ln(2))dx = +C = + C, where C is an arbitrary ln(2) ln(2) ln(2) constant. Z Z Z x2 ln(2) 2 2 2 2 e · 2x · ln(2) ex ln(2) 2x (e) 2x xdx = ex ln(2) xdx = dx = +C = + C, where C is an arbitrary 2 ln(2) 2 ln(2) 2 ln(2) constant. Z Z sin(x) ln(2) e cos(x) ln(2) esin(x) ln(2) 2sin(x) (f) 2sin(x) cos(x)dx = dx = +C = + C, where C is an arbitrary ln(2) ln(2) ln(2) constant. √ Z Z cos( x) 1 1 1 1 √ (g) dx = 2 cos(x 2 ) · x− 2 dx = 2 sin(x 2 ) + C, where C is an arbitrary constant. 2 x √ Z Z 1 cosh( x) 1 1 1 √ dx = 2 cosh(x 2 ) · x− 2 dx = 2 sinh(x 2 ) + C, where C is an arbitrary constant. (h) 2 x       Z Z 1 1 1 1 1 2 2 dx = − sec · − + C, where C is an arbitrary constant. sec dx = − tan (i) 2 2 x x x x x Z Z sin(x) 1 d 1 (j) dx = − · (2 + cos(x))dx = + C, where C is an arbitrary con(2 + cos(x))2 (2 + cos(x))2 dx 2 + cos(x) stant. Z p Z Z √ 2 d 2 1 3 2 1 1 3 x +4 (k) x x2 + 4dx = · 2xdx = · (x + 4) 2 · (x + 4)dx = (x2 + 4) 2 + C, where C is an 2 3 2 dx 3 arbitrary constant. Z Z Z xdx 1 1 d 2 1 2 −2 (l) = · (x + 1) · 2xdx = − · −(x2 + 1)−2 · (x + 1)dx = − (x2 + 1)−1 + C, where C 2 2 (x + 1) 2 2 dx 2 is an arbitrary constant. (m) Z x3 (x2 + 1)20 dx = = = = = Z Z Z Z x(x2 + 1 − 1)(x2 + 1)20 dx x[(x2 + 1)21 − (x2 + 1)20 ]dx x(x2 + 1)21 dx − Z x(x2 + 1)20 dx 1 · 22(x2 + 1)21 · 2xdx − 44 Z 1 · 21(x2 + 1)20 · 2xdx 42 (x2 + 1)21 (x2 + 1)22 − + C, where C is an arbitrary constant. 44 42 9 Exercise 8. (Level 3) Compute the indefinite integrals below: Z Z (a) i. sin3 (x) cos(x)dx iv. cos (x) sin(x)dx iii. Z sin(x) dx cos2 (x) i. Z ii. Z cosh (x) sinh(x)dx iii. Z sinh(x) dx cosh2 (x) i. Z ii. ii. (b) (c) v. Z 5 vi. 3 sinh (x) cosh(x)dx Z Z viii. Z cos3 (x)dx cot(x)dx ix. Z cos4 (x) sin2 (x)dx tan(x)dx x. Z sin4 (x) cos2 (x)dx viii. Z cosh3 (x)dx coth(x)dx ix. Z cosh4 (x) sinh2 (x)dx tanh(x)dx x. Z sinh4 (x) cosh2 (x)dx csc3 (x) cot(x)dx sin(x) dx cos4 (x) vii. Z sin4 (x) cos3 (x)dx iv. Z sinh(x) dx cosh4 (x) v. Z 5 vi. Z vii. Z sec (x) tan (x)dx vi. Z sec (x) tan (x)dx xi. Z Z sec(x) tan3 (x)dx vii. Z sec4 (x) tan5 (x)dx xii. Z sech2 (x) tanh2 (x)dx iii. Z tan3 (x)dx viii. Z csc2 (x) cot2 (x)dx xiii. Z sech(x) tanh3 (x)dx iv. Z sec (x) tan(x)dx ix. Z csc(x) cot (x)dx xiv. Z tanh3 (x)dx v. Z sec4 (x) tan6 (x)dx x. Z cot3 (x)dx xv. Z sech3 (x) tanh(x)dx 2 2 3 Solution. (a) i. Z sin3 (x) cos(x)dx = Z sinh4 (x) cosh3 (x)dx 5 7 3 1 1 · 4 sin3 (x) sin′ (x)dx = sin4 (x) + C, where C is an arbitrary con4 4 stant. Z Z 1 1 5 ii. cos (x) sin(x)dx = − · 6 cos5 (x) cos′ (x)dx = − cos6 (x) + C, where C is an arbitrary constant. 6 6 Z Z Z sin(x) cos′ (x) 1 iii. dx = − dx = −(cos(x))−2 cos′ (x)dx = + C, where C is an arbitrary concos2 (x) cos2 (x) cos(x) stant. Z Z 1 1 sin(x) dx = · −3(cos(x))−4 cos′ (x)dx = + C, where C is an arbitrary constant. iv. cos4 (x) 3 3 cos3 (x) Z Z Z cos(x) sin′ (x) v. cot(x)dx = dx = dx = ln(| sin(x)|) + C, where C is an arbitrary constant. sin(x) sin(x) 10 vi. Z tan(x)dx = Z sin(x) dx = cos(x) Z − cos′ (x) dx = − ln(| cos(x)|) + C, where C is an arbitrary constant. cos(x) vii. Z sin4 (x) cos3 (x)dx = = = = Z Z Z sin4 (x)(1 − sin2 (x)) cos(x)dx 4 sin (x) cos(x)dx − Z sin6 (x) cos(x)dx 1 · 5 sin4 (x) sin′ (x)dx − 5 Z 1 · 7 sin6 (x) sin′ (x)dx 7 sin5 (x) sin7 (x) − + C, where C is an arbitrary constant. 5 7 viii. Z 3 cos (x)dx = = Z Z cos2 (x) cos(x)dx (1 − sin2 (x)) cos(x)dx 1 cos(x)dx − 3 = Z = sin(x) − Z 3 sin2 (x) cos(x)dx sin3 (x) + C, where C is an arbitrary constant. 3 ix. Z = Z cos4 (x) sin2 (x)dx sin2 (x) cos2 (x) · cos2 (x)dx sin2 (2x) cos(2x) + 1 · dx 4 2 Z Z sin2 (2x) sin2 (2x) cos(2x) = dx + dx 8 8 Z Z sin2 (2x) · 2 cos(2x) 1 − cos(4x) dx + dx = 16 16 = Z = x sin(4x) sin3 (2x) − + + C, where C is an arbitrary constant. 16 64 48 11 x. Z sin4 (x) cos2 (x)dx Z = sin2 (x) cos2 (x) · sin2 (x)dx sin2 (2x) − cos(2x) + 1 · dx 4 2 Z Z sin2 (2x) cos(2x) sin2 (2x) dx − dx = 8 8 Z Z sin2 (2x) · 2 cos(2x) 1 − cos(4x) dx − dx = 16 16 (b) i. ii. iii. iv. v. vi. Z = Z = x sin(4x) sin3 (2x) − − + C, where C is an arbitrary constant. 16 64 48 sinh3 (x) cosh(x)dx = Z 1 1 · 4 sinh3 (x) sinh′ (x)dx = sinh4 (x) + C, where C is an arbitrary constant. 4 4 1 1 · 6 cosh5 (x) cosh′ (x)dx = cosh6 (x) + C, where C is an arbitrary constant. 6 6 Z Z Z cosh′ (x) 1 sinh(x) dx = dx = (−1) · [−(cosh(x))−2 ] cosh′ (x)dx = − + C, where C is an 2 2 cosh(x) cosh (x) cosh (x) arbitrary constant. Z Z 1 sinh(x) 1 · 3(cosh(x))−4 cosh′ (x)dx = − dx = + C, where C is an arbitrary constant. 3 cosh4 (x) 3 cosh3 (x) Z Z Z sinh′ (x) cosh(x) dx = dx = ln(| sinh(x)|) + C, where C is an arbitrary constant. coth(x)dx = sinh(x) sinh(x) Z Z Z sinh(x) cosh′ (x) tanh(x)dx = dx = dx = ln(| cosh(x)|) + C, where C is an arbitrary constant. cosh(x) cosh(x) Z 5 cosh (x) sinh(x)dx = Z vii. Z = = = = Z Z Z sinh4 (x) cosh3 (x)dx sinh4 (x)(1 + sinh2 (x)) cosh(x)dx sinh4 (x) cosh(x)dx + Z sinh6 (x) cosh(x)dx 1 · 5 sinh4 (x) sinh′ (x)dx + 5 Z 1 · 7 sinh6 (x) sinh′ (x)dx 7 sinh5 (x) sinh7 (x) + + C, where C is an arbitrary constant. 5 7 12 viii. Z cosh3 (x)dx = = Z Z cosh2 (x) cosh(x)dx (1 + sinh2 (x)) cosh(x)dx = Z = sinh(x) + cosh(x)dx + 1 3 Z 3 sinh2 (x) cosh(x)dx sinh3 (x) + C, where C is an arbitrary constant. 3 ix. Z = Z = Z = = = cosh4 (x) sinh2 (x)dx sinh2 (x) cosh2 (x) · cosh2 (x)dx sinh2 (2x) cosh(2x) + 1 · dx 4 2 Z Z sinh2 (2x) sinh2 (2x) cosh(2x) dx + dx 8 8 Z Z sinh2 (2x) · 2 cosh(2x) −1 + cosh(4x) dx + dx 16 16 − x sinh(4x) sinh3 (2x) + + + C, where C is an arbitrary constant. 16 64 48 x. Z = sinh4 (x) cosh2 (x)dx Z sinh2 (x) cosh2 (x) · sinh2 (x)dx sinh2 (2x) cosh(2x) − 1 · dx 4 2 Z Z sinh2 (2x) cosh(2x) sinh2 (2x) = − dx + dx 8 8 Z Z sinh2 (2x) · 2 cosh(2x) 1 − cosh(4x) dx + dx = 16 16 (c) i. Z = Z = x sinh(4x) sinh3 (2x) − + + C, where C is an arbitrary constant. 16 64 48 sec2 (x) tan2 (x)dx = Z 1 tan3 (x) · 3 tan2 (x) tan′ (x)dx = + C, where C is an arbitrary constant. 3 3 13 ii. Z 3 sec(x) tan (x)dx = = = Z Z tan2 (x) sec′ (x)dx (sec2 (x) − 1) sec′ (x)dx sec3 (x) − sec(x) + C, where C is an arbitrary constant. 3 iii. Z 3 tan (x)dx Z = Z = Z = tan(x) tan2 (x)dx tan(x)(sec2 (x) − 1)dx tan(x) sec2 (x)dx − Z tan(x)dx tan2 (x) + ln(| cos(x)|) + C, where C is an arbitrary constant. 2 Z Z Z sec3 (x) iv. sec3 (x) tan(x)dx = sec2 (x) · sec(x) tan(x)dx = sec2 (x) · sec′ (x)dx = + C, where C is an 3 arbitrary constant. v. Z sec4 (x) tan6 (x)dx = Z = Z = Z = sec2 (x) tan6 (x) tan′ (x)dx (1 + tan2 (x)) tan6 (x) tan′ (x)dx (tan6 (x) + tan8 (x)) tan′ (x)dx tan7 (x) tan9 (x) + + C, where C is an arbitrary constant. 7 9 = vi. Z = = = = Z Z Z sec5 (x) tan7 (x)dx sec4 (x) tan6 (x) sec′ (x)dx sec4 (x)(sec2 (x) − 1)3 sec′ (x)dx (sec10 (x) − 3 sec8 (x) + 3 sec6 (x) − sec4 (x)) sec′ (x)dx sec11 (x) sec9 (x) 3 sec7 (x) sec5 (x) − + − + C, 11 3 7 5 where C is an arbitrary constant. 14 vii. Z = = = sec4 (x) tan5 (x)dx Z sec3 (x) tan4 (x) sec′ (x)dx Z (sec2 (x) − 1)2 sec3 (x) sec′ (x)dx Z = (sec7 (x) − 2 sec5 (x) + sec3 (x)) sec′ (x)dx sec8 (x) tan6 (x) sec4 (x) − + + C, where C is an arbitrary constant. 8 3 4 1 cot3 (x) − · 3 cot2 (x) cot′ (x)dx = − + C, where C is an arbitrary constant. 3 3 Z Z Z csc3 (x) 3 2 ′ + csc(x) + C, ix. csc(x) cot (x)dx = − cot (x) csc (x)dx = (− csc2 (x) + 1) csc′ (x)dx = − 3 where C is an arbitrary constant. viii. Z 2 2 csc (x) cot (x)dx = Z x. Z 3 cot (x)dx = = = Z Z Z = − xi. Z csc3 (x) cot(x)dx = Z cot(x) cot2 (x)dx cot(x)(csc2 (x) − 1)dx 2 cot(x) csc (x)dx − Z cot(x)dx cot2 (x) − ln(| sin(x)|) + C, where C is an arbitrary constant. 2 − csc2 (x) · (− csc(x) cot(x))dx = Z − csc2 (x) · csc′ (x)dx = − csc3 (x) + C, where 3 C is an arbitrary constant. Z Z 1 tanh3 (x) 2 2 xii. sech (x) tanh (x)dx = · 3 tanh2 (x) tanh′ (x)dx = + C, where C is an arbitrary constant. 3 3 xiii. Z = = = = Z Z Z sech(x) tanh3 (x)dx − tanh2 (x) · (− tanh(x)sech(x))dx − tanh2 (x)sech′ (x)dx (sech2 (x) − 1)sech′ (x)dx sech3 (x) − sech(x) + C, where C is an arbitrary constant. 3 15 xiv. Z = = Z Z = − = − tanh3 (x)dx tanh(x) tanh2 (x)dx tanh(x)(−sech2 (x) + 1)dx Z 2 tanh(x)sech (x)dx + Z tanh(x)dx tanh2 (x) + ln(| cosh(x)|) + C, where C is an arbitrary constant. 2 xv. Z sech3 (x) tanh(x)dx = Z = Z = − −sech2 (x) · (−sech(x) tanh(x))dx −sech2 (x) · sech′ (x)dx sech3 (x) + C, where C is an arbitrary constant. 3 Exercise 9. (Level 3) Compute the indefinite integrals below: Z Z dx (ln(x))3 ii. (a) i. dx x ln(x) x Z Z 2ex dx ex dx iii. (b) i. x e +2 e2x + 4 Z Z dx 2ex dx ii. iv. ex + 1 e2x − 4 iii. (b) i. ii. Z v. Z sech(x)dx vi. Z csch(x)dx Z Z (ln(x))3 1 1 1 (ln(x))4 dx = · 4(ln(x))3 · dx = · 4(ln(x))3 · ln′ (x)dx = + C, where C is x 4 x 4 4 an arbitrary constant. Z Z Z 1 1 dx 1 = · dx = · ln′ (x)dx = ln(| ln(x)|) + C, where C is an arbitrary constant. x ln(x) ln(x) x ln(x) √ Z Z Z ln( x) 1 1 1 1 dx = ln(x) · dx = · 2 ln(x) · ln′ (x)dx = (ln(x))2 + C, where C is an arbitrary x 2 x 4 4 constant. Z Z 1 d x exp(x)dx = · (e + 2)dx = ln(ex + 2) + C, where C is an arbitrary constant. x x e +2 e + 2 dx Z Z Z dx 1 1 d −x · (−e )dx = − · =− (1 + e−x )dx = − ln(1 + e−x ) + C, where C is ex + 1 1 + e−x 1 + e−x dx an arbitrary constant. Solution. (a) ii. √ ln( x) dx x iii. i. Z 16 iii. Z 2ex dx e2x + 4 = Z 2ex dx 4 + (ex )2 = Z 2du 4 + u2 = Z 2 · 2 sec2 (v)dv 4 + 4 tan2 (v) = Z u dv = v + C = arctan( ) + C 2 = arctan( ex ) + C, where C is an arbitrary constant. 2 We have used the substitutions u = ex , u = 2 tan(v) successively. iv. Z 2ex dx e2x − 4 = = = = = 2du −4   Z 1 1 1 du − 2 u−2 u+2 Z u2 1 (ln(|u − 2|) − ln(|u + 2|)) + C 2 1 u − 2 ln( )+C 2 u + 2 1 ex − 2 ln( x ) + C, where C is an arbitrary constant. 2 e + 2 We have used the substitution u = ex . Z Z Z Z 2dx 2ex dx dx = = = 2 arctan(exp(x)) + C, where C is an arbitrary v. sech(x)dx = cosh(x) ex + e−x 1 + e2x constant. vi. Z Z Z Z dx 2ex dx 2dx csch(x)dx = = = x −x sinh(x) e −e e2x − 1  Z  1 1 dx + = ex + 1 ex − 1  Z  −x e−x e dx + = 1 + e−x 1 − e−x = = = = −(ln(|1 + e−x |) + ln(|1 − e−x |)) + C   1 − e−x +C ln 1 + e−x   sinh(x/2) +C ln cosh(x/2)   x   ln tanh + C, where C is an arbitrary constant. 2 17 Exercise 10. (Level 3) Compute the indefinite integrals below. Leave your answer in terms of sec(x), tan(x), csc(x) and cot(x), where appropriate. Z Z (a) sec(x)dx (b) csc(x)dx (Hint: Write cos(x) 1 = .) cos(x) 1 − sin2 (x) Solution. (a) Z sec(x)dx Z Z dx cos(x)dx cos(x)dx = = = cos(x) cos2 (x) 1 − sin2 (x)   Z 1 1 1 sin′ (x)dx + = 2 1 + sin(x) 1 − sin(x) Z Z 1 1 d 1 1 d = · · (1 + sin(x))dx + − · · (1 − sin(x))dx 2 1 + sin(x) dx 2 1 − sin(x) dx Z = = = = = 1 1 ln(|1 + sin(x)|) − ln(|1 − sin(x)|) + C 2 2   1 + sin(x) 1 +C ln 2 1 − sin(x)   (1 + sin(x))2 1 +C ln 2 1 − sin2 (x)   (1 + sin(x))2 1 +C ln 2 cos2 (x)   1 + sin(x) +C ln cos(x) = ln(| sec(x) + tan(x)|) + C, where C is an arbitrary constant. 18 (b) Z = = = = = = = = = csc(x)dx Z Z sin(x)dx dx sin(x)dx = = sin(x) 1 − cos2 (x) sin2 (x)   Z 1 1 1 cos′ (x)dx + − 2 1 + cos(x) 1 − cos(x) Z Z 1 d d 1 1 1 − · (1 + cos(x))dx + · (1 − cos(x))dx 2 1 + cos(x) dx 2 1 − cos(x) dx Z 1 1 − ln(|1 + cos(x)|) + ln(|1 − cos(x)|) + C 2 2   1 − cos(x) 1 +C ln 2 1 + cos(x)   (1 − cos(x))2 1 +C ln 2 1 − cos2 (x)   (1 − cos(x))2 1 +C ln 2 sin2 (x)   1 − cos(x) +C ln sin(x) ln(| csc(x) − cot(x)|) + C, where ...
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