Engineering Mathematics 4 Episode 11 pps

METHODS OF DIFFERENTIATION  2xe4x sin x   ln 2t p t  2e4x f C 4x sin x sin2 x ln 2t  p  t3 x cos xg Find the gradient of the curve y D 2x x2 [ 18] at the point (2, 4) dy Evaluate at x D 2.5, correct to dx 2x C significant figures, given y D ln 2x [3.82] 45.4 Function of a function It is often easier to make a substitution before differentiating If y is a function of x then dy dy du = × dx du dx Rewriting u as 5x C gives: dy = −30x sin.5x Y 2/ dx Problem 19 Find the derivative of: y D 4t3 3t Let u D 4t3 Problem 18 Differentiate: y D cos 5x C Let u D 5x C then y D cos u du dy Hence D 10x and D sin u dx du Using the function of a function rule, dy dy du D ð D dx du dx sin u 10x D 30x sin u 3t, then y D u6 du dy D 12t2 and D 6u5 dt dt Using the function of a function rule, Hence dy du dy D ð D 6u5 12t2 dx du dx Rewriting u as (4t3 3t) gives: dy D 4t3 3t 12t2 dt D 18.4t − 1/.4t − 3t /5 Problem 20 Determine the differential p coefficient of: y D 3x C 4x yD This is known as the ‘function of a function’ rule (or sometimes the chain rule) For example, if y D 3x then, by making the substitution u D 3x , y D u9 , which is of the ‘standard’ form du dy Hence D 9u8 and D3 du dx dy du dy D ð D 9u8 D 27u8 Then dx du dx dy Rewriting u as (3x 1) gives: = 27.3x − 1/8 dx Since y is a function of u, and u is a function of x, then y is a function of a function of x 389 p 3x C 4x Let u D 3x C 4x Hence D 3x C 4x 1/2 then y D u1/2 du dy D 6x C and D u dx du 1/2 D p u Using the function of a function rule, dy dy du p D ð D dx du dx u dy 3x Y i.e = p dx 3x Y 4x − Problem 21 6x C D 3x C p u Differentiate: y D tan4 3x Let u D tan 3x then y D 3u4 du Hence D sec2 3x, (from Problem 14), dx dy and D 12u3 du dy du dy D ð D 12u3 sec2 3x Then dx du dx D 12 tan 3x 3 sec2 3x dy = 36 tan3 3x sec2 3x i.e dx 390 ENGINEERING MATHEMATICS Problem 22 Find the differential coefficient of: y D 2t 54 D 2t3 2t3 then y D 2u yD Hence Then du dy D 6t2 and D dt du dy du dy D ð D dt du dt Let u D 2t3 8u D 5, d2 y dy D 12x , D 36x , dx dx u5 6t2 D u5 d2 y (pronounced dee two y by dee x dx00 squared) or f x (pronounced f double–dash x) By successive differentiation further higher derivad3 y d4 y tives such as and may be obtained dx dx Thus if y D 3x , written as −48t 2t − 5/5 d3 y d4 y d5 y D 72x, D 72 and D0 dx dx dx Exercise 156 Further problems on the function of a function f x D 2x f x D 10x In Problems to 8, find the differential coefficients with respect to the variable 2x 5x 2 sin 3 [5 6x x3 2x C 2x 4x C 3x cot 5t2 C tan 3y C 2etan  12x C 24x D 4x 10x − 6/ f x / D 40x Problem 24 sin x, evaluate d2 y x, in the range Ä x Ä , when is zero dx 2] 3x x 2x C If y D cos x Since y D cos x sin x, 5e2tC1 5, [ 10 cos4 ˛ sin ˛] 4x C 3x 5x ] [6 cos 3 cos5 ˛ If f x D 2x Problem 23 find f00 x Now try the following exercise [10e2tC1 ] [ 20t cosec2 5t2 C ] [18 sec2 3y C ] [2 sec2 Âetan  ] with respect Differentiate:  sin  to Â, and evaluate, correct to significant [1.86] figures, when  D 45.5 Successive differentiation When a function y D f x is differentiated with respect to x the differential coefficient is written as dy or f0 x If the expression is differentiated again, dx the second differential coefficient is obtained and is dy D dx dy D dx sin x cos x and cos x C sin x d2 y is zero, cos x C sin x D 0, dx sin x i.e sin x D cos x or D1 cos x When Hence tan x D and x D tan in the range Ä x Ä 1 D 45° or Problem 25 Given y D 2xe d2 y dy C6 C 9y D dx dx y D 2xe 3x (i.e a product) dy Hence D 2x 3e 3x C e dx D 6xe 3x p rads C 2e 3x 3x 3x show that METHODS OF DIFFERENTIATION d2 y D [ 6x dx 3e D 18xe 3x d2 y i.e D 18xe dx 3x 3x 6e 12e C e 3x 3x 6e ]C 6e 3x 3x Exercise 157 C 2xe 12e Evaluate given: y D sec 2 3x d2 y when  D d (b) (b) 72x C 12] p tC1 C t t 2 (a) Given f t D t2 determine f00 t Evaluate f00 t when t D 12 (a) C 3C p 5 t t t3 (b) 4.95 In Problems and 4, find the second differential coefficient with respect to the variable Since y D sec 2Â, then dy D sec 2 tan 2 (from Problem 15) d D sec 2 tan 2 (i.e a product) d2 y D sec 2 sec2 2 d C tan 2 [ sec 2 tan 2Â] D 16 sec 2 C 16 sec 2 tan2 2 When  D 0, d2 y D 16 sec3 C 16 sec tan2 d D 16 C 16 D 16 If y D 3x C 2x 3x C find d2 y d3 y (a) (b) dx dx [(a) 36x C 12x 3x 36xe 3x C 12e 3x C 18xe 3x D d2 y dy Thus when y D 2xe 3x , C 9y D C6 dx dx Problem 26 3x Further problems on successive differentiation 3x d2 y dy C 9y gives: Substituting values into C6 dx dx 18xe 3x 12e 3x C 6xe 3x C 2e 3x D 18xe Now try the following exercise (a) sin 2t C cos t (a) (a) cos2 x (a) sin2 x (b) ln 4 12 sin 2t C cos t (b) 2x cos2 x (b) Â2 (b) 48 2x Evaluate f00  when  D given f  D sec 3 [18] Show that the differential equation d2 y dy C 4y D is satisfied when dx dx y D xe2x 391 46 Some applications of differentiation 46.1 Rates of change If a quantity y depends on and varies with a quantity x then the rate of change of y with respect to x dy is dx Thus, for example, the rate of change of pressure dp p with height h is dh A rate of change with respect to time is usually just called ‘the rate of change’, the ‘with respect to time’ being assumed Thus, for example, a rate of di and a rate of change of change of current, i, is dt d temperature, Â, is , and so on dt Problem The length l metres of a certain metal rod at temperature  ° C is given by: l D C 0.00005 C 0.0000004 Determine the rate of change of length, in mm/° C, when the temperature is (a) 100 ° C and (b) 400 ° C dl d Since length l D C 0.00005 C 0.0000004 , dl D 0.00005 C 0.0000008 then d (a) When  D 100 ° C, dl D 0.00005 C 0.0000008 100 d D 0.00013 m/° C D 0.13 mm= ° C The rate of change of length means (b) When  D 400 ° C, dl D 0.00005 C 0.0000008 400 d D 0.00037 m/° C D 0.37 mm=° C Problem The luminous intensity I candelas of a lamp at varying voltage V is given by: I D ð 10 V2 Determine the voltage at which the light is increasing at a rate of 0.6 candelas per volt The rate of change of light with respect to voltage dI is given by dV dI D ð 10 V Since I D ð 10 V2 , dV D ð 10 V When the light is increasing at 0.6 candelas per volt then C0.6 D ð 10 V, from which, voltage 0.6 VD D 0.075 ð 10C4 D 750 volts ð 10 Problem Newtons law of cooling is given by:  D Â0 e kt , where the excess of temperature at zero time is Â0 ° C and at time t seconds is  ° C Determine the rate of change of temperature after 40 s, given that Â0 D 16 ° C and k D 0.03 The rate of change of temperature is Since  D Â0 e kt then d dt d k e kt D Â0 dt D kÂ0 e kt 0.03 and t D 40 then When Â0 D 16, k D d D 0.03 16 e dt 0.03 40 D 0.48 e 1.2 D 1.594 ° C=s Problem The displacement s cm of the end of a stiff spring at time t seconds is given by: s D ae kt sin ft Determine the velocity of the end of the spring after s, if a D 2, k D 0.9 and f D ds where s D ae Velocity v D dt product) Using the product rule, ds D ae kt f cos ft dt kt sin ft (i.e a C sin ft ake kt SOME APPLICATIONS OF DIFFERENTIATION When a D 2, k D 0.9, f D and t D 1, velocity, v D 2e 0.9 46.2 Velocity and acceleration cos C sin D 25.5455 cos 10 D 25.5455 0.9 e 393 0.9 0.7318 sin 10 0.7318 When a car moves a distance x metres in a time t seconds along a straight road, if the velocity v is x m/s, i.e the gradient of the constant then v D t distance/time graph shown in Fig 46.1 is constant D 25.55 cm=s (Note that cos 10 means ‘the cosine of 10 radians’, not degrees, and cos 10 Á cos D 1) Now try the following exercise Exercise 158 Further problems on rates of change An alternating current, i amperes, is given by i D 10 sin ft, where f is the frequency in hertz and t the time in seconds Determine the rate of change of current when t D 20 ms, given that f D 150 Hz [3000 A/s] The luminous intensity, I candelas, of a lamp is given by I D ð 10 V2 , where V is the voltage Find (a) the rate of change of luminous intensity with voltage when V D 200 volts, and (b) the voltage at which the light is increasing at a rate of 0.3 candelas per volt [(a) 0.24 cd/V (b) 250 V] The voltage across the plates of a capacitor at any time t seconds is given by v D V e t/CR , where V, C and R are constants Given V D 300 volts, C D 0.12 ð 10 farads and R D ð 106 ohms find (a) the initial rate of change of voltage, and (b) the rate of change of voltage after 0.5 s [(a) 625 V/s (b) 220.5 V/s] The pressure p of the atmosphere at height h above ground level is given by p D p0 e h/c , where p0 is the pressure at ground level and c is a constant Determine the rate of change of pressure with height when p0 D 1.013 ð 105 Pascals and c D 6.05 ð 104 at 1450 metres [ 1.635 Pa/m] Figure 46.1 If, however, the velocity of the car is not constant then the distance/time graph will not be a straight line It may be as shown in Fig 46.2 d d Figure 46.2 The average velocity over a small time υt and distance υx is given by the gradient of the chord υx AB, i.e the average velocity over time υt is As υt υt ! 0, the chord AB becomes a tangent, such that dx at point A, the velocity is given by: v D dt Hence the velocity of the car at any instant is given by the gradient of the distance/time graph If an expression for the distance x is known in terms of time t then the velocity is obtained by differentiating the expression 394 ENGINEERING MATHEMATICS x D 3t3 2t2 C 4t m dx Velocity vD D 9t2 4t C m/s dt d 2x Acceleration a D D 18t m/s2 dx (a) When time t D 0, velocity v D C D m=s and acceleration a D 18 D −4 m=s2 (i.e a deceleration) (b) When time t D 1.5 s, velocity v D 1.5 1.5 C D 18.25 m=s and acceleration a D 18 1.5 D 23 m=s2 Distance d d Figure 46.3 The acceleration a of the car is defined as the rate of change of velocity A velocity/time graph is shown in Fig 46.3 If υv is the change in v and υt υv the corresponding change in time, then a D As υt υt ! 0, the chord CD becomes a tangent, such that dv at point C, the acceleration is given by: a D dt Hence the acceleration of the car at any instant is given by the gradient of the velocity/time graph If an expression for velocity is known in terms of time t then the acceleration is obtained by differentiating the expression dv Acceleration a D dt dx vD However, dt d 2x d dx D Hence aD dt dt dx The acceleration is given by the second differential coefficient of distance x with respect to time t Summarising, if a body moves a distance x metres in a time t seconds then: (i) distance x = f t / dx , which is the gradi(ii) velocity v D f t / or dt ent of the distance/time graph d 2x dv = f or , which (iii) acceleration a = dt d t2 is the gradient of the velocity/time graph Problem The distance x metres moved by a car in a time t seconds is given by: x D 3t3 2t2 C 4t Determine the velocity and acceleration when (a) t D 0, and (b) t D 1.5 s Problem Supplies are dropped from a helicopter and the distance fallen in a time t seconds is given by: x D gt2 , where g D 9.8 m/s2 Determine the velocity and acceleration of the supplies after it has fallen for seconds gt D 9.8 t2 D 4.9t2 m 2 dv Velocity vD D 9.8 t m/s dt d 2x and acceleration a D D 9.8 m/s2 dx When time t D s, velocity v D 9.8 D 19.6 m=s and acceleration a = 9.8 m=s2 (which is acceleration due to gravity) Distance xD Problem The distance x metres travelled by a vehicle in time t seconds after the brakes are applied is given by: x D 20t t Determine (a) the speed of the vehicle (in km/h) at the instant the brakes are applied, and (b) the distance the car travels before it stops t 10 dx D 20 t Hence velocity v D dt At the instant the brakes are applied, time D Hence 20 ð 60 ð 60 velocity v D 20 m/s D km/h 1000 D 72 km=h (a) Distance, x D 20t SOME APPLICATIONS OF DIFFERENTIATION D (b) When the car finally stops, the velocity is zero, 10 10 i.e v D 20 t D 0, from which, 20 D t, 3 giving t D s Hence the distance travelled before the car stops is given by: 5 t D 20 6 x D 20t 3 D 120 60 D 60 m (a) Angular displacement  D 9t2 2t3 rad d Angular velocity ω D D 18t 6t2 rad/s dt When time t D s, ! D 18 61 Now try the following exercise Exercise 159 The equation  D 10 C24t 3t2 gives the angle Â, in radians, through which a wheel turns in t seconds Determine (a) the time the wheel takes to come to rest, (b) the angle turned through in the last second of movement [(a) s (b) rads] At any time t seconds the distance x metres of a particle moving in a straight line from a fixed point is given by: x D 4tCln t Determine (a) the initial velocity and acceleration (b) the velocity and acceleration after 1.5 s (c) the time when the velocity is zero dx D 2.2 sin t dt C 3.6 cos t (a) m/s; m/s2 (b) m/s; m/s2 D 11 sin t C 18 cos t cm/s When time t D 30 ms, C 18 cos ð 30 ð 10 (b) 62.5 m] Displacement x D 2.2 cos t C 3.6 sin t 11 sin ð 30 ð 10 The distance s metres travelled by a car in t seconds after the brakes are applied is given by s D 25t 2.5t2 Find (a) the speed of the car (in km/h) when the brakes are applied, (b) the distance the car travels before it stops Problem The displacement x cm of the slide valve of an engine is given by: x D 2.2 cos t C 3.6 sin t Evaluate the velocity (in m/s) when time t D 30 ms velocity D A missile fired from ground level rises x metres vertically upwards in t seconds 25 and x D 100t t Find (a) the initial velocity of the missile, (b) the time when the height of the missile is a maximum, (c) the maximum height reached, (d) the velocity with which the missile strikes the ground [(a) 90 km/h When the angular acceleration is zero, 18 12t D 0, from which, 18 D 12t, giving time, t = 1.5 s Velocity v D Further problems on velocity and acceleration (a) 100 m/s (b) s (c) 200 m (d) 100 m/s Angular acceleration ˛ D (b) 15.69 C 50.39 D 34.7 cm/s D 0.347 m=s D 12 rad=s d 2 D 18 12t rad/s d t2 When time t D s, a D 18 12 D rad=s2 11 sin 27° C 18 cos 27° D Problem The angular displacement  radians of a flywheel varies with time t seconds and follows the equation:  D 9t2 2t3 Determine (a) the angular velocity and acceleration of the flywheel when time, t D s, and (b) the time when the angular acceleration is zero 11 sin 0.4712 C 18 cos 0.4712 D (Note: changing from m/s to km/h merely involves multiplying by 3.6) c s The angular displacement  of a rotating t disc is given by:  D sin , where t is the time in seconds Determine (a) the angular velocity of the disc when t is 395 396 ENGINEERING MATHEMATICS 1.5 s, (b) the angular acceleration when t is 5.5 s, and (c) the first time when the angular velocity is zero   (a) ω D 1.40 rad/s  (b) ˛ D 0.37 rad/s2  (c) t D 6.28 s 23t2 20t3 C 6t C represents x D the distance, x metres, moved by a body in t seconds Determine (a) the velocity and acceleration at the start, (b) the velocity and acceleration when t D s, (c) the values of t when the body is at rest, (d) the value of t when the acceleration is 37 m/s2 , and (e) the distance travelled in the third second   (a) m/s, 23 m/s2  (b) 117 m/s, 97 m/s2       (c) s or s  (d) 11 s y R P Negative gradient Positive gradient Q O x Figure 46.4 y Maximum point Maximum point Points of inflexion (e) 75 m Minimum point x Figure 46.5 46.3 Turning points In Fig 46.4, the gradient (or rate of change) of the curve changes from positive between O and P to negative between P and Q, and then positive again between Q and R At point P, the gradient is zero and, as x increases, the gradient of the curve changes from positive just before P to negative just after Such a point is called a maximum point and appears as the ‘crest of a wave’ At point Q, the gradient is also zero and, as x increases, the gradient of the curve changes from negative just before Q to positive just after Such a point is called a minimum point, and appears as the ‘bottom of a valley’ Points such as P and Q are given the general name of turning points It is possible to have a turning point, the gradient on either side of which is the same Such a point is given the special name of a point of inflexion, and examples are shown in Fig 46.5 Maximum and minimum points and points of inflexion are given the general term of stationary points Procedure for finding and distinguishing between stationary points (i) Given y D f x , determine Positive gradient dy (i.e f0 x ) dx (ii) (iii) dy D and solve for the values of x dx Substitute the values of x into the original equation, y D f x , to find the corresponding y-ordinate values This establishes the coordinates of the stationary points Let To determine the nature of the stationary points: Either d 2y and substitute into it the values of (iv) Find d x2 x found in (ii) If the result is: (a) positive — the point is a minimum one, (b) negative — the point is a maximum one, (c) zero — the point is a point of inflexion or (v) Determine the sign of the gradient of the curve just before and just after the stationary points If the sign change for the gradient of the curve is: (a) positive to negative — the point is a maximum one SOME APPLICATIONS OF DIFFERENTIATION (b) (c) negative to positive — the point is a minimum one positive to positive or negative to negative — the point is a point of inflexion Problem 10 Locate the turning point on the curve y D 3x 6x and determine its nature by examining the sign of the gradient on either side Following the above procedure: dy 6x, (i) Since y D 3x D 6x dx dy D 0, hence 6x (ii) At a turning point, dx from which, x D (a) Considering the point (1, 3): If x is slightly less than 1, say 0.9, then dy D 0.9 3, which is negative dx If x is slightly more than 1, say 1.1, then dy D 1.1 3, which is positive dx Since the gradient changes from negative to positive, the point (1, 3) is a minimum point Considering the point ( 1, 7): If x is slightly less than 1, say 1.1, then dy D 1.1 3, which is positive dx If x is slightly more than 1, say 0.9, then dy D 0.9 3, which is negative dx Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point (iii) When x D 1, y D 61 D D 0, Hence the co-ordinates of the turning point is (1, −3) (iv) If x is slightly less than 1, say, 0.9, then dy D 0.9 D 0.6, i.e negative dx If x is slightly greater than 1, say, 1.1, then dy D 1.1 D 0.6, i.e positive dx Since the gradient of the curve is negative just before the turning point and positive just C), (1, −3) is a minimum after (i.e point Problem 11 Find the maximum and minimum values of the curve y D x 3x C by (a) examining the gradient on either side of the turning points, and (b) determining the sign of the second derivative dy D 3x dx dy D0 For a maximum or minimum value dx Hence 3x D 0, Since y D x from which, and 3x C then 3x D x D š1 When x D 1, y D 3 C D When x D 1, y D 3 C D Hence (1, 3) and ( 1, 7) are the co-ordinates of the turning points 397 (b) d 2y dy D 3x 3, then D 6x dx d x2 d 2y is positive, hence (1, 3) is a When x D 1, d x2 minimum value d 2y is negative, hence ( 1, 7) When x D 1, d x2 is a maximum value Thus the maximum value is and the minimum value is Since It can be seen that the second differential method of determining the nature of the turning points is, in this case, quicker than investigating the gradient Problem 12 Locate the turning point on the following curve and determine whether it is a maximum or minimum point: y D 4 C e  dy D4 e d maximum or minimum value Hence D e  , D e  , giving  D ln D 1.3863 (see Chapter 13) When  D 1.3863, Since y D 4 C e yD4  1.3863 C e then 1.3863  D for a D 5.5452 C 4.0000 D 1.5452 398 ENGINEERING MATHEMATICS Thus ( 1.3863, the turning point d 2y D e  d Â2 When  D 1.5452) are the co-ordinates of 1.3863, is a minimum point Knowing ( 2, 9) is a maximum point (i.e crest of a wave), and 3, 11 is a mini6 mum point (i.e bottom of a valley) and that when x D 0, y D , a sketch may be drawn as shown in Fig 46.6 Hence d 2y D e C1.3863 D 4.0, which d Â2 is positive, hence (−1.3863, −1.5452) is a minimum point 3, −11 Problem 13 Determine the co-ordinates of the maximum and minimum values of the x3 x2 graph y D 6x C and distinguish 3 between them Sketch the graph Following the given procedure: (i) Since y D x3 dy D x2 dx (ii) x2 dy D dx At a turning point, x2 i.e from which yD x xC2 x When x D then x Hence (iii) 6x C D 0, D 0, or x D xD Figure 46.6 2, 2 When x D 3, y D 33 Problem 14 Determine the turning points on the curve y D sin x cos x in the range x D to x D radians, and distinguish between them Sketch the curve over one cycle C D9 32 63 C Thus the co-ordinates of the turning points are (−2, 9) and 3, −11 D (iv) Since dy D x2 dx x 11 then d 2y D 2x d x2 d 2y D 2 D When x D 2, d x2 which is negative Hence (−2, 9) is a maximum point d 2y D23 When x D 3, d x2 positive 5, D 5, which is Since y D sin x cos x then dy D cos x C sin x D 0, for a turning point, dx sin x D from which, cos x D sin x and cos x D tan x Hence x D tan D 126.87° or 306.87° , since tangent is negative in the second and fourth quadrants When x D 126.87° , y D sin 126.87° When cos 126.87° D x D 306.87° , y D sin 306.87° cos 306.87° D 48 Integration using algebraic substitutions 48.1 Introduction Functions that require integrating are not always in the ‘standard form’ shown in Chapter 47 However, it is often possible to change a function into a form which can be integrated by using either: du Let u D 3x C then D and rearranging gives dx du dx D Hence cos 3x C d x D D trigonometric substitutions (see Chapters 49 and 51), (iii) du cos u d u, which is a standard integral (i) an algebraic substitution (see Section 48.2), (ii) cos u partial fractions (see Chapter 50), or D (iv) integration by parts (see Chapter 52) sin u C c Rewriting u as 3x C gives: 48.2 Algebraic substitutions With algebraic substitutions, the substitution usually made is to let u be equal to f x such that f u d u is a standard integral It is found that integrals of the forms: k [f x ]n f0 x d x and k f0 x n dx [f x ] (where k and n are constants) can both be integrated by substituting u for f x 48.3 Worked problems on integration using algebraic substitutions Problem Determine: cos 3x C d x D which may be checked by differentiating it Problem Find: dx 2x (2x 5) may be multiplied by itself times and then each term of the result integrated However, this would be a lengthy process, and thus an algebraic substitution is made du du Let u D 2x then D and d x D dx Hence 2x dx D cos 3x C d x cos 3xC7 d x is not a standard integral of the form shown in Table 47.1, page 408, thus an algebraic substitution is made sin.3x Y 7/ Y c, D Rewriting u as (2x u7 du D 2 u8 CcD u7 d u u Cc 16 5) gives: 2x − 5/7 dx = 2x − 5/8 Y c 16 INTEGRATION USING ALGEBRAIC SUBSTITUTIONS Problem Find: 5x dx 3 Hence u5 d u D D Let u D 5x du du D and d x D dx then Hence 5x du D u 5 dx D du u ln u C c ln.5x − 3/ Y c D Evaluate: 2e 6x 1 then Hence 2e 6x Thus 2e 6x Let u D sin  then dx D D D 24 u5 d u, by cancelling u6 C c D 4u6 C c D sin  C c /6 24 sin5  cos  d  Thus 6x 1 [e ]0 e D [4 sin6 Â]0 du cos  D sin6  C c 2e u [e 24 sin5  cos  d  du du D cos  and d  D d cos  24u5 cos  D d x, correct du D eu d u 1 D e u C c D e 6x C c 3 4x Y 3/6 Y c 16 24 sin5  cos  d  Hence du du D and d x D dx dx D u Cc 16 D 24 Let u D 6x Cc D Evaluate: to significant figures u6 /6 Problem D Problem ] D 49.35, D4 /6 D4 sin D sin or 0.0625 16 6 correct to significant figures Now try the following exercise Problem Determine: Let u D 4x C then 3x 4x C d x du du D 8x and d x D dx 8x Hence 3x 4x C d x D D 3x u du 8x u5 d u, by cancelling The original variable ‘x’ has been completely removed and the integral is now only in terms of u and is a standard integral Exercise 166 Further problems on integration using algebraic substitutions In Problems to 6, integrate with respect to the variable sin 4x C cos 4x C C c 2 cos 2 sin 2 sec2 3t C 5 Cc tan 3t C C c 415 416 ENGINEERING MATHEMATICS 5x 5x 70 2x 7Cc ln 2x Cc 3e 3ÂC5 Hence e 3ÂC5 C c D 4x dx u 1/2 p d u, by cancelling u du    1/2 C1 3x C d x 1u  u   Cc D  Cc 4 C1 2 1p 1p D uCcD 4x − Y c 2 x 2x C d x  1/2 D [227.5] 2x 2x d u p D u 8x D In Problems to 10, evaluate the definite integrals correct to significant figures p [4.333] /3 sin 3t C dt [0.9428] Problem Show that: 10 cos 4x dx [0.7369] tan  d  D ln sec  C c tan  d  D 48.4 Further worked problems on integration using algebraic substitutions Problem Find: du D d Hence then x dx C 3x sin  d  Let u D cos  cos  du sin  and d  D sin  sin  d D cos  D du du D 6x and d x D Let u D C 3x then dx 6x x dx Hence C 3x D x du D u 6x d u, by cancelling, u Let u D 4x Determine: then du sin  d u D ln u C c u ln cos  C c D ln cos  C c, by the laws of logarithms p 2x 4x dx du du D 8x and d x D dx 8x Hence tan q d q = ln.sec q/ Y c, since D ln u C x D ln.2 Y 3x / Y c Problem D sin  u cos  D D sec  cos  48.5 Change of limits When evaluating definite integrals involving substitutions it is sometimes more convenient to change the limits of the integral as shown in Problems 10 and 11 INTEGRATION USING ALGEBRAIC SUBSTITUTIONS Problem 10 Evaluate: 2x 5x C d x, taking positive values of square roots only du du D 4x and d x D dx 4x It is possible in this case to change the limits of integration Thus when x D 3, u D C D 25 and when x D 1, u D 2 C D Let u D 2x C 7, then xD3 Hence taking positive values of square roots only Now try the following exercise Exercise 167 2x C d x 5x i.e the limits have been changed  9 1/2 p u  p D  D [ 1] D 3,  2 Further problems on integration using algebraic substitutions xD1 uD25 uD9 D 25 25 p du 5x u D 4x D p In Problems to 7, integrate with respect to the variable u du u1/2 d u 2x 2x 2 cos5 t sin t Thus the limits have been changed, and it is unnecessary to change the integral back in terms of x sec2 3x tan 3x xD3 2x C d x 5x Thus xD1 3/2 25 u 3/2 p D [ 253 D 2 27 D 81 Evaluate: p du du Let u D 2x C then D 4x and d x D dx 4x Hence p xD2 3x 2x C dx D xD0 D 4 p 2t 3t2 6Cc cos6 t C c ln   tan 2t p 3x d x, 2x C taking positive values of square roots only Problem 11 1 sec2 3x C c or tan2 3x C c 2 p 25 D u p 93 ] D 125 2x 12 3Cc ln  C c ln sec 2t C c 2e t et C xD2 u 3t2 p [4 e t C C c] In Problems to 10, evaluate the definite integrals correct to significant figures 3x d u p u 4x 1/2 du 3xe 2x2 dx [1.763] xD0 /2 Since u D 2x C 1, when x D 2, u D and when x D 0, u D Thus xD2 u xD0 1/2 du D uD9 u uD1 1/2 d u, sin4  cos  d  [0.6000] 10 3x 4x dx [0.09259] 417 49 Integration using trigonometric substitutions 49.1 Introduction 49.2 Worked problems on integration of sin2 x , cos2 x , tan2 x and cot2 x Table 49.1 gives a summary of the integrals that require the use of trigonometric substitutions, and their application is demonstrated in Problems to 19 Problem Evaluate: cos2 4t d t Table 49.1 Integrals using trigonometric substitutions f x dx fx Method cos2 x xC sin 2x Cc Use cos 2x D cos2 x sin2 x x sin 2x Cc Use cos 2x D tan2 x tan x See problem cot x cosm x sinn x sin2 x Use C tan2 x D sec2 x xCc cot x 2 Use cot x C D cosec x xCc cos A cos B sin A sin B 10 11 12 p p a2 x2 a2 x2 a2 C x 2 sin2 x [sin A C B C sin A B ] sin A B ] Use [sin A C B Use [cos A C B C cos A B ] [cos A C B cos A B ] Use cos A sin B 7, sin A cos B 5, (b) If both m and n are even, use either cos 2x D cos2 x or cos 2x D (a) If either m or n is odd (but not both), use cos2 x C sin2 x D Use x Cc a xp a2 x sin C a a sin x Cc a 10 11 12     Use x D a sin   1 tan a x2 C c 13, 14  substitution    15, 16 Use x D a tan  substitution 17–19 INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS Since cos 2t D cos2 t (from Chapter 26), Problem 1 C cos 2t and cos2 4t D C cos 8t then cos2 t D Hence cos 4t d t Hence D tC   D sin 8t C D   0C  cot  2 D   [ sin p D or 0.7854 cot 2 cosec2 2 d  D D  1 cot2 2 d   sin 1 C cos 8t d t cot2 2 d  Since cot2  C D cosec2 Â, then cot2  D cosec2  and cot2 2 D cosec2 2 Evaluate D2 419 0.2887       cot 6 Determine: 0.2887 sin2 3x d x 0.5236 ] Now try the following exercise Exercise 168 sin2 x (from Chapter 26), Since cos 2x D 1 sin2 3x D then sin2 x D cos 2x and Further problems on integration of sin2 x , cos2 x , tan2 x and cot2 x In Problems to 4, integrate with respect to the variable cos 6x sin 3x d x D Hence Find: Problem 3 tan 4x d x D D3 cos2 t sin 2t Cc tan2 3 Since C tan x D sec x, then tan x D sec x and tan2 4x D sec2 4x Hence Cc cot2 2t Yc 2 sin 4x tan2 4x d x x cos 6x d x sin 6x x− D sin2 2x 1 2 sec 4x dx tan 4x −x tC tan 3 [  Cc cot 2t C 2t C c] In Problems to 8, evaluate the definite integrals, correct to significant figures /3 sin2 3x d x or 1.571 /4 Yc cos2 4x d x   1.0472 D 0.0269 Problem or 0.3927 420 ENGINEERING MATHEMATICS tan2 2t d t [ 4.185] sin2 x D sin2 x cos x sin2 x cos x d x /3 cot2  d  [0.3156] D /6 sin3 x D Problem  D sin5  d  Determine: D Since cos2  C sin2  D then sin2  D D cos  C cos  d  sin  cos2  C sin  cos4  d  2cos3 q cos5 q − Yc Evaluate: 0 4 D D D n sin  cos  d  d C cos 2 C cos2 2 d  C cos 2 C Alternatively, an algebraic substitution may be used as shown in Problem 6, chapter 50, page 415] cos  sin  d  cos2  d  D 1 C cos 4 C cos 2 C cos 4 2 3 sin 4 C sin 2 C C sin Evaluate: sin2 x cos3 x d x sin /4 C sin x cos x d x D sin x cos x cos x d x [0] C1 D 2.178, correct to significant figures Problem d 0 d D 0] cos4  d Â, 1 C cos 2 D4 sinnC1  D C c] nC1 Problem [0 D n 4 cos4  d  D cosnC1  Cc nC1 and   D or 0.1333 15 D [Whenever a power of a cosine is multiplied by a sine of power 1, or vice-versa, the integral may be determined by inspection as shown In general, cos2  d  sin  D −cosq Y 5 sin sin  sin correct to significant figures sin  sin2  d  D D Problem sin  d  sin  cos2  D sin5 x  49.3 Worked problems on powers of sines and cosines Hence sin4 x cos x d x Find: sin2 t cos4 t d t INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS sin2 t cos4 t d t D sin2 t cos2 t d t 49.4 Worked problems on integration of products of sines and cosines 1 D D cos 2t D C cos 2t cos 2t C cos 2t C cos2 2t d t C cos 2t C cos2 2t 2 C cos 2t Problem cos 2t d t D sin2 2t cos 4t C cos 2t sin2 2t 2 t sin 4t sin3 2t − Y D dt dt sin 5t C sin t d t cos3  Cc cos  C sin 2x sin3 x cos4 x cos5 x cos7 x C Cc sin 2 sin2 t cos2 t 1 sin 4 C sin 8 C c 32 t sin 4t C c 32 sin 3x d x −cos 7x cos 3x Y Yc Evaluate: cos 6 cos  d Â, correct to decimal places cos 6 cos  d  3 2x ] d x, D2 sin 7x Problem 11 2 cos3 t C cos5 t C c sin t cos t sin 5x sin3 2x Cc 3 cos 5x sin 2x d x from of Table 49.1 D D Integrate the following with respect to the variable: 2 cos3 2x Find: Yc cos 5x sin 2x d x 1 D [sin 5x C 2x Exercise 169 Further problems on integration of powers of sines and cosines a − cos 5t − cos t Problem 10 Yc Now try the following exercise sin3  2t ] d t, from of Table 49.1, which follows from Section 26.4, page 221, C cos 4t D D sin 3t cos 2t d t [sin 3t C 2t C sin 3t D cos 2t d t cos 2t Determine: sin 3t cos 2t d t 3 cos 2t C cos 2t dt cos 2t cos 2t D D 421 [cos 6 C  C cos 6  ] d Â, from of Table 49.1 D sin 7 sin 5 C sin sin C cos 7 C cos 5 d  D D sin sin C 422 ENGINEERING MATHEMATICS ‘sin 7’ means ‘the sine of radians’ (Á 401.07° ) and sin Á 286.48° In Problems to 8, evaluate the definite integrals /2 cos 6 cos  d  Hence D 0.09386 C 0.19178 or 0.4286 cos 4x cos 3x d x 0 D −0.0979, correct to decimal places sin 7t cos 3t d t [0.5973] /3 Problem 12 Find: sin 5x sin 3x d x sin 5 sin 2 d  [0.2474] 3 cos 8t sin 3t d t [cos 5x C 3x D3 cos 5x 3x ] d x, from of Table 49.1 D 3 D− cos 8x 49.5 Worked problems on integration using the sin q substitution cos 2x d x sin 8x sin 2x − Problem 13 Yc Now try the following exercise Let x D a sin Â, then d x D a cos  d  Hence Exercise 170 Further problems on integration of products of sines and cosines D In Problems to 4, integrate with respect to the variable D sin 5t cos 2t cos 7t cos 3t C sin 2x 2 sin 3x sin x D Cc sin 4x Cc 3 cos 6x cos x Determine: p or sin 2x − sin 8x / Y c 16 [ 0.1999] sin 5x sin 3x d x sin 7x sin 5x C Cc cos 4 sin 2 D p a2 x2 cos 2 cos 6 Cc dx dx a cos  d  a cos  d  a2 sin2  a cos  d  p , since sin2  C cos2  D a2 cos2  a cos  d  D d D  C c a cos  x Since x D a sin Â, then sin  D and  D sin a x p Hence d x D sin−1 Y c a a2 x x2 dx D a cos  and d a2 sin2  a2 a2 Problem 14 Evaluate p x2 dx x a INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS From Problem 13, p dx x2 x since a D D sin p D sin 1 sin D or 1.5708 a2 Thus Problem 15 a2 Find: a2 [ C sin  cos Â] p x x a2 x Cc C a a a x2 d x D D a2 sin D a2 x xp sin−1 Y a − x2 Y c a x2 d x Problem 16 Evaluate: x2 d x 16 Let x D a sin  then d x D a cos  d  D a2 D a2 D x2 d x 16 16 sin x x 16 x C p D sin 1 C [8 sin x2 d x D p From Problem 15, a2 Hence dx D a cos  and d a2 sin  a cos  d  sin2  a cos  d  D sin 1D8 C 0] D 4p or 12.57 a2 cos2  a cos  d  Now try the following exercise a cos  a cos  d  D Da 2 cos  d  D a C cos 2 2 Exercise 171 d (since cos 2 D cos2  a D sin 2 ÂC 2 D a ÂC Further problems on integration using the sine q substitution 1 Determine: p t2 dt Cc sin sin  cos  2 Determine: p x2 x and  D sin a x a Determine: Also, cos  C sin  D 1, from which, cos  D D sin2  D a2 x2 a2 p D a2 x a x a Determine: sin 16 1 x Cc x2 d x sin 2 t Cc dx a2 [ C sin  cos Â] C c Since x D a sin Â, then sin  D sin Cc since from Chapter 26, sin 2 D sin  cos  D x xp C 2 x2 C c 9t2 d t 3t 16 C 9t2 C c 423 424 ENGINEERING MATHEMATICS Evaluate: p x2 16 tan 1 tan p D or 0.3927 D dx or 1.571 D Evaluate: 4x d x [2.760] Problem 19 d x, C 2x Evaluate: correct to decimal places 49.6 Worked problems on integration using the tan q substitution Problem 17 1 dx a2 C x Determine: D D Let x D a tan  then d x D a sec2 Âd  Hence dx D a sec2  and d D dx C x2 a sec2  d  a2 C a2 tan2  a sec2  d  a2 C tan2  a sec2  d  , since C tan2  D sec2  a2 sec2  1 d D  C c a a D D D Since x D a tan Â,  D tan Hence a 1 dx [ 3/2]2 C x p p tan 3/2 2 tan Evaluate: Exercise 172 x 3/2 tan 0] Further problems on integration using the tan q substitution dt C t2 Determine: tan 2 d 16 C 9 tan 12 Determine: dx C x2 From Problem 17, dx C x2 x tan since a D D 2 p Now try the following exercise D 1.3976, correct to decimal places x a 1 x dx = tan−1 Y c 2/ Yx a a Problem 18 5 dx 2[ 3/2 C x ] D 2.0412 [0.6847 a2 D dx D C 2x Evaluate: Evaluate: 1 x Cc 3 Cc dt C t2 [2.356] dx C x2 [2.457] INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS ln x dx x p d (9) (c) 2 Evaluate the following definite integrals: (b) Assignment 13 This assignment covers the material in Chapters 47 to 49 The marks for each question are shown in brackets at the end of each question /2 (a) sin 2t C Determine: p (a) t5 d t (b) C  d p 4x dx (8) /2 sin2 t d t dx (10) /3 (b) cos 5 sin 3 d  6t C d t Evaluate the following definite integrals, correct to significant figures: Determine the following integrals: (a) (10) cos3 x sin2 x d x (a) C C x x dx Determine the following integrals: sin 2t d t (b) (9) /3 (b) (a) Evaluate the following integrals, each correct to significant figures: (a) 3xe 4x dt p dx x2 (c) (b) (c) dx C x2 (14) 425 50 Integration using partial fractions 50.1 Introduction The process of expressing a fraction in terms of simpler fractions — called partial fractions — is discussed in Chapter 7, with the forms of partial fractions used being summarised in Table 7.1, page 51 Certain functions have to be resolved into partial fractions before they can be integrated, as demonstrated in the following worked problems 50.2 Worked problems on integration using partial fractions with linear factors Problem Determine: 11 3x dx C 2x x2 As shown in Problem 1, page 51: 11 3x Á C 2x x x xC3 11 3x Hence dx x C 2x dx D x xC3 Á or ln or x Y 1/4 x Y 3/ x − 2/3 ln Problem 2x 9x 35 dx xC1 x xC3 It was shown in Problem 2, page 52: 2x 9x 35 Á xC1 x xC3 xC1 x C xC3 x2 C dx x 3x C Determine: x2 C Á1 3x C x2 Hence x2 Á x C x2 C dx 3x C C x x x dx D x − ln.x − 1/ Y ln.x − 2/ Y c or x − 2/5 x − 1/2 x Y ln Y c by the laws of logarithms Find: Yc By dividing out (since the numerator and denominator are of the same degree) and resolving into partial fractions it was shown in Problem 3, page 52: Problem Problem dx D ln.x Y 1/ − ln.x − 2/ Y ln.x Y 3/ Y c D ln.x − 1/ − ln.x Y 3/ Y c (by algebraic substitutions — see chapter 50) x − 1/2 x Y 3/5 2x 9x 35 dx xC1 x xC3 C xC1 x xC3 Hence x3 Yc Evaluate: 2x 4x x2 C x d x, correct to significant figures By dividing out and resolving into partial fractions, it was shown in Problem 4, page 53: x3 2x 4x Cx x2 Áx 3C xC2 x INTEGRATION USING PARTIAL FRACTIONS Hence x3 2x 4x x2 C x 2 3C x Á dx xC2 3x dx x x2 D D 3x C ln x C ln x 9 C ln 6 C ln 3 ln D −1.687, correct to significant figures 2x 16x C 20 dx x xC2   3x 2x C ln x   ln x C C c In Problems and 7, evaluate the definite integrals correct to significant figures ln 2 427 x 3x C dx xx x [0.6275] x2 x2 [0.8122] x 14 dx 2x Now try the following exercise Exercise 173 Further problems on integration using partial fractions with linear factors In Problems to 5, integrate with respect to x 12 x2    Problem dx ln x or ln 4x x2 2x ln x C C c x xC3 Cc   xC1 x Cc  Cc 2x 8x dx x C x C 2x  ln x C ln x C ln 2x    xC4 Cc C c or ln x C 2x It was shown in Problem 5, page 54: 2x C Á C x x x x C 9x C dx x2 C x x C ln x C C ln x C c or x C ln x C x C c 2x C dx x 22 Thus ln x 2x C dx x 22 Determine:  dx  ln x C  or ln 50.3 Worked problems on integration using partial fractions with repeated linear factors Á x C D ln.x − 2/ − x 2 dx Yc x − 2/ d x is determined using the algex 22 braic substitution u D x , see Chapter 48 Problem Find: 5x 2x xC3 x 19 dx 12 It was shown in Problem 6, page 54: 5x 2x xC3 x 19 Á C xC3 x x 428 ENGINEERING MATHEMATICS Hence 5x 2x 19 dx xC3 x C Á xC3 x x D ln.x Y 3/ Y ln.x − 1/ Y ln x Y 3/2 x − 1/3 Y or 5x 2 dx Yc x − 1/ ln x Problem Evaluate: 3x C 16x C 15 d x, correct to xC3 significant figures It was shown in Problem 7, page 55: 3x C 16x C 15 Á xC3 xC3 xC3 xC3 Á D ln x C C D ln C 2 x 2 Cc [1.663] 18 C 21x x dx x xC2 [1.089] 50.4 Worked problems on integration using partial fractions with quadratic factors xC3 xC3 10 x x C 7x C x2 x C Problem C 6x C 4x 2x dx x2 x2 C 3 4x C 6x C 4x 2x 2 Á C 2C x2 x2 C x x x C3 xC3 C xC3 xC3 C 16 Find: It was shown in Problem 9, page 56: 3x C 16x C 15 dx xC3 Hence C In Problems and 4, evaluate the definite integrals correct to significant figures Yc x − 1/ 30x C 44 dx x 23 ln C dx Thus Á 2 C 1 D −0.1536, correct to significant figures D C 6x C 4x 2x dx x2 x2 C 3 4x dx C 2C x x x C3 C 2C x x x C3 4x x2 C dx p dx D dx C3 x2 C 3 x D p tan p , from 12, Table 49.1, page 418 3 4x d x is determined using the algebraic subx2 C stitution u D x C x2 Now try the following exercise Exercise 174 Further problems on integration using partial fractions with repeated linear factors In Problems and 2, integrate with respect to x 4x dx xC1 ln x C C Cc xC1 4x dx C 2C 2C3 x x x C3 x x ln x C C c D ln x C p tan p x 3 Hence D ln x 2Y3 x − p x Y tan−1 p Y c x ... (b) (a) 41 1   D   Â2 Â2 C d Â2 d ? ?4 C1 C1  C 2 2 ? ?4 C1    C1 p 2p   C4   D 3 2 p p 2 D 3C4 3C4 3 16 D C8 C4 3 2 D5 C8 4D8 3 D C 41 2 ENGINEERING MATHEMATICS /2 Problem 14 Evaluate:... 144 Substituting into equation (1) gives: VD 144 h2 h D 144 h h3 dV h2 D 144 D 0, for a maximum or dh minimum value Hence 144 D h2 , from which, 144 D 13.86 cm h d 2V D dh d V When h D 13.86,... 3x (b) 4p x Cc (b) Cc 4x 1p x dx (b 1p x Cc (a) (a) p t3 (b) p dx x4 10 (a) p C c t dx 4x (b) (a) (b) x Cc 24 (b)  3x 5x C c  (a)       (b) 4? ? C 2 C Cc 3x e Yc 4t e dt e 4t C c 4t e C