N Ji INTEGRATION USING

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2 NTEGRATON TECHNQUES integrate using the trigonometric identities shi (x) = -( - cos(x)), cos (x) = -( + cos(x)) and + taii (x) = sec (x) (ACMSM 6) B use substitution = g(x) to integrate expressions of the form J(g(x) )g'(x) (ACMSM 7) establish and use the formula J.!_dx = n x +c, for x ;,e 0 (ACMSM 8) X find and use the inverse trigonometric functions arcsine, arccosine and arctangent (ACMSMl 9) find and use the derivative of the inverse trigonometric functions, arcsine, arccosine and arctangent (ACMSMl0) integrate expressions of the form and (ACMSM) va -x a +x use partial fractions where necessary for integration in simple cases (ACMSM) integrate by parts. N Ji NTEGRATON USNG TRGONOMETRC DENTTES n this section, you will use your knowledge of the double angle formulas and Pythagorean trigonometric identities to integrate trigonometric expressions. You should remember the following double angle formulas from your work last year. Double angle formulas sin (x) = sin (x) cos (x) cos (x) = cos (x) - sin (x) = cos (x) - = - sin (x) tan(x) tan ( x ) =--- l-tan (x) MPORTANT You can rearrange the formula for cos (x) to eliminate sin (x) in a trigonometric expression as follows. cos (x) = - sin (x) so sin (x) = - cos (x) so sin (x) =.!_ ( - cos (x)] Similarly cos (x) = cos cx) - so cos (x) =.!_ [l + cos (x)] 86 NELSON SENOR MATHS Specialist

3 0 Example Find f cos (x)dx. Rearrange cos (x) = cos (x) -. cos cx) = + cos (x) So cos (x) = ½U +cos(x)] Write the problem and replace cos (x). Simplify by removing the common factor. Separate the trig part. Use the formula f cos(kx)dx = ¼sin(kx)+ c Write the answer. f cos (x)dx = J½[l+cos(x)Jdx =½ f[l +cos(x)jdx =½[fldx+ f cos(x)dx] =½[ x+½sin(x)]+c fcos (x)dx =.!. x +.!. sin (x) + c n Example, each separate step has been shown. You would normally omit some of these steps. 0 Example t Find f 0 3sin (6x)dx. Rearrange cos (x) = - sin (6x). Write the problem and replace sin (6x). Simplify by removing the common factor. Separate the trig part. Use the formula f cos(kx)dx =.!..sin(kx)+c.. k Evaluate. sin (6x) = - cos (x) So sin (6x) = ½[-cos(lx)] t f :3sin (6x)dx = f 0 ¾[ l-cos(lx) ]dx t n = { J;[ l-cos(lx) ]dx =½[f ;dx- f} cos(lx)dx] t = [x-.l..sin(lx )] = [ -_!_sin(t)] =¾( -0) = 8 Write the answer CHAPTER 7: ntegration techniques 87

4 You derived the trigonometric Pythagorean identities last year. You can reverse the derivative of tan (x) to obtain the integral of sec cx). Remember that tan(x) = sec (x). dx Similarly, cot ( x) = -cosec ( x ). dx MPORTANT The Pythagorean identities sin (x) + cos cx) = tan cx) + = sec (x) + cot cx) = cosec cx) MPORTANT f sec (x)dx = tan(x)+ c and so f sec (kx)dx = }tan(kx)+ c f cosec (x)dx =-cot(x)+c and so f cosec (kx)dx =-}cot(kx)+c To make things simpler when integrating, you can rewrite or substitute any trigonometric identities that you already know. 0 Example 3 a Find J tan (x)dx. b Find J sin(x)cos(x)dx. a Rearrange + tan (x) = sec (x). Write the problem and replace tan cx). Separate the trig part. Use the formula J sec ( kx )dx = }tan( kx )+ c. Write the answer. b Rearrange sin (x) = sin (x) cos (x) Write the problem and replace sin (x) cos (x). Simplify by removing the common factor. Use the formula J sin(kx)dx = -}cos(kx)+ c + tan (x) = sec (x) So tan (x) = sec (x) - f tan (x)dx = f[ sec (x)-l]dx = J sec (x)dx-f ldx = tan (x) - x + c J tan (x)dx= tan(x)-x+c sin (x) = sin (x) cos (x) So sin(x)cos(x)=.!.sin(x) f sin(x)cos(x)dx = f ½sin(x)dx = ½ J sin (x )dx = ½[-½cos(x) ]+c =-¼cos(x)+c Write the answer. f sin(x)cos(x)dx =-¼cos(x)+c 88 NELSON SENOR MATHS Specialist

5 EXERCSE 7.0 ntegration using trigonometric identities Concepts and techniques ;;.,,,;.,. f sin (x)dx= A.!. x +_!_sin (x) + c D.!.x-_!_sin(x) + c 8.!.x- _!_sin(x) + c E x+.!.sin(x) + c C x+.!.sin(x) + c Trigonomelric idenlites J 0 sin (x)dx A D + C 8 3 ;,;,,;.;n J cos (x)dx= A.!. x- _!_sin(x) + c D x-.!. sin(x) + c 8 t J 0 cos (8x)dx= A- D x- _! sin(l6x) + c 3 5 ;@,,nhm J tan (x)dx= A.!. tan(x)-x + c X D tan (x)- + c 6 J sec ( x )dx = A.!. tan(x) -x + c D.!. tan(x) +.!. sec(x) + c t 7 f 0 8 sin(x) cos(x)dx = A_!_ D.!.sin (x)+c 8 8 re 8 E re 8.!. x - _! sin ( x) + c 6 E.!. x+ _! sin(8x) + c C E x+_! sin(l6x) + c 6 8 tan (x) - x + c 8.!.tan(x) -x + c E.!. sec ( x) + c 8 8 E _!_ sin (x) + c C -.!.sin(x) + c 8 C x+ _! sin(l6x) + c 3 C re C -tan(x) + c C CHAPTER 7: ntegration techniques 89

6 Reasoning and communication 8 Show that J sin(x)cos(x)dx =-½ cos(x) + c 9 Show that Jsin(3x )cos(3x )dx =! cos(6x) + c 0 Show that sin (x)= ½ [-cos(x)] Show that sin (3x)=½[l-cos(6x)] Show that J cot (3x)dx = -½cot(3x)-x+c Jfv.. NTEGRATON BY SUBSTTUTON You already know how to use the chain rule to differentiate the function p(x) = (x 3-5x + ) by writing u(x) = x 3-5x + and q(u) = u You get p'(x) = q'(u) x u'(x) = u 3 x (6x - 5) = (x 3-5x + ) 3 (6x - 5) You can integrate to get J p'(x)dx = J u 3 x(6x -5)dx = J u 3 xu'(x)dx = (x 3-5x + ). But (x 3-5x+ ) = u = J u 3 du, so J p'(x)dx = J u 3 x(6x -5)dx = J u 3 du. By writingp'(x) = f(x) and u 3 = g(u), you get J f(x)dx = J g(u) xu'(x)dx = J g(u)du. This is the integration rule that corresponds to the chain rule. ntegration by substitution f a functionf(x) can be written as f(x) = g(u) u'(x), then J f(x)dx = J g(u)u'(x)dx = J g(u)du Alternatively, iff(x) = g[u(x)], then J f(x)dx = J g(u)u'(x)du MPORTANT To use the substitution method, you need to find a function u(x) such that u'(x) is part of the original and the remainder of the expression is a function of u. There is no set method for finding u(x). Even so, it often helps to make u(x) the 'inside' of the most complex function in the integral. 90 NELSON SENOR MATHS Specialist

7 0 Example Choose a suitable substitution. Find u' (x). Write the integral and substitute. Use the substitution rule. Do the integration. Substitute u(x) back in. Let u(x) = x - 3x + u' (x) = 8x -3 f (x -3x+l) (8x-3)dx = f u xu'(x)dx = f u du =.!_ u 5 + C =.!. (x - 3x + ) 5 + c 5 Now do the definite integral, using your result. J:(x -3x+l) (8x-3)dx =[fcx -3x+l) 5 Calculate the answer. =.!.x 5 -.!.x = 3 0 Notice in Example that you could get the same answer using f u du. This would give [.!.u 5 ] =.!. x 5 -.!. x 5 = 3 0. For a definite integral, the substitution rule can be written as follows. ntegration by substitution f a function j(x) can be written as f(x) = g(u) u'(x), then J: f(x)dx = J: g(u)u'(x)dx = f! g(u)du, where a= u(a) and = u(b). Alternatively, iff(x) = g[u(x)], then Jb f p, a f(x)dx= g(u)u (x)du, where a= u(a) and = u(b). a MPORTANT You can manipulate constants to make integration easier CHAPTER 7: ntegration techniques 9

8 0 Example 5 Find f (3x-7) 5 dx Choose a suitable substitution. Find the derivative. Write the integral and manipulate to get u' (x). Substitute in u(x) and u' (x). Let u(x) = 3x - 7 u' (x) = 3 f (3x - 7) 5 dx = ½ J (3x - 7) 5 x 3dx = ½J u 5 xu'(x)dx Use the substitution rule. Do the integration. Simplify and substitute u(x) back in. = lxlu 6 +c 3 6 = (3X - 7) 6 + C 8 You can do this wherever you can write the function to be integrated in the formf (x) = g(u) u'(x). 0 Example 6 7t Evaluate Jo'' sin(x)cos (x)dx. Choose a suitable substitution. Find the derivative. Write the integral and manipulate to get u' (x). Substitute in u(x) and u' (x). Use the substitution rule. Do the integration, including the change of bounds. Do the calculation. Simplify. Express with a rational denominator. Let u(x) = cos (x) u' (x) = -sin (x) fo¾ sin(x )cos (x )dx = -fo¾ cos (x) x [ - sin(x) ]dx 7t = -f 0 u xu'(x)dx = -J u(¾) u du u(o) =-[":t = -½[ ( F J _ 5 ] =½( -) 8-h 0 9 NELSON SENOR MATHS Specialist

9 You could use your CAS calculator to do the integration from the original expression. However, if the problem says to 'use substitution or otherwise: then you are expected to demonstrate integration techniques like those in the examples. EXERCSE 7.0 ntegration by substitution Concepts and techniques g;.;;,;.,e i3cx -x-) 3 (x-l)dx= x -x- A--- C 56 D 65. E 3 ntegration by substitution A 5 B 5 C g;.;,,;.,a f (x+5)7 dx = A B (x+5) 8 6 C.!.(x+5) 7 7 D (x+5) 8 +c E.!.(x+5) 8 +c 6 8 An antiderivative for J ( - J dx is B - (x -6) 5 05 C -(x - 6) 5 D t inhm The integral f 0 3 -sin(x)cos 3 (x)dx can be expressed as 7t A f -sin(x)cos 3 (x)dx B J 3 cos 3 (x)dx 0 lu D f u 3 du E J -dx 6 The integral J 3tan (x)dx can be expressed as./3 C J; u 3 du A 3 J sec (x)-ldx B J 3sec (x)dx C f 3tan (x)dx D f 3tan (u)du J Find (x -x-) (x-)dx 7t 8 Find J 0 sin (x)cos(x)dx E J 3tan (u)-ldu Reasoning and communication 9 Show that f li n+3.f3 3 sec (x)+ldx=--- o 3 0 Show that an antiderivative of the expression sin (3x) is - sin(6x) CHAPTER 7: ntegration techniques 93

10 d d d ti i NTEGRATON OF x- The natural logarithm is given by log e (x) = ln(x) (for x > 0). The natural logarithm is the inverse of the exponential function. The derivative oflogix) is given b y - log e (x) = - for x >. d X X MPORTANT For y = log e (x), the definition oflogarithms gives r! = x. Now from the chain rule:!! (r!) = e d Y x y d = e>' x y. d dx dx dx x y But d d =, so taking the derivative of both sides of x = e>' gives r! x y =. X But e>' = x so x x d y = and thus d y =.!._ dx dx x This means that f.!..ax= log e (x) for x > 0. What about x < 0? X QED NVESTGATON Definite integrals of - X What do you notice about the shapes of the graph for x > 0 and for x < 0? \ X f you draw an area between the x-axis and the curve for x > 0 and for x < 0 on domains equally distant from the centre, you get the following. X What do you notice about the areas? f- f3l f What can you say about -dx, -dx and -dx? -3 X l X 3 X 9 NELSON SENOR MATHS Specialist

11 From the investigation, it should be clear that for values of x < 0, you can find the integral of_!_ by X using the integral for x > 0. For example, s-l!_dx = f!_dx = [log.(x) ]3 = log. () - log. (3) = X 3 X The sign is negative because the area is below the axis. t is also clear that you cannot perform an integration for a region containing O because!_ is not defined for x = 0. X The integral of!_ = x- is given by f x- dx = log. lxl + c, provided that x "#- 0. X MPORTANT 0 Example 7 Use the method of substitution to find an expression for f- -dx. x+3 Choose a suitable substitution. Find the derivative. Write the integral and manipulate to get u' (x). Let u(x) = x+ 3 u' (x) = f- -dx =lf- -xdx x+3 x+3 Substitute in u(x) and u' (x). Use the substitution rule. Do the integration. Substitute u(x) back in. Write the answer. = lf.!:.xu'(x)dx u = lf.!:.du u =½ log. lul + c = ½ log. lx c f- -dx =.!.. log. lx c x+3 The method of Example 7 can be used to show the following results. MPORTANT f--dx =.!:. log. lax+ bl+ c where a, b are constants and c is the constant of integration. ax+b a j'(x) f f (x) dx=log.j f(x)j+c CHAPTER 7: ntegration techniques 95

12 0 Example 8 f 6x- Find dx. 3x -x+l Write in the form J f' (x) dx. f(x) Use the rule for J f' (x) dx. f(x) f 6x- J f' (x) dx = --dx wheref(x) = 3x - x + 3x -x+l f(x) f 6x- dx = log e l3x -x + ll + c 3x -x+l You can also manipulate expressions involving reciprocals by introducing constants to obtain the derivative of the denominator. 0 Example 9 li sin(x) Evaluate J - -dx 0 cos(x) Choose a suitable substitution. Find the derivative. Write the integral and manipulate to get u' (x). f f ' (x) Use the rule for --dx. j(x) Calculate the value. Manipulate the result. Let u(x) = cos (x) u'(x) = -sin (x) J ¾ sin(x) dx= _ J ¾ [ - sin(x) ] d x o cos(x) 0 cos(x) = [-log e cos(x) l]l = - log e leas( )+ log e lcos (O) = log e ( Jz r + loge (l) = log e () Write the answer. --- x= og e ¾ sin(x) d ( ) o cos(x) EXERCSE 7.03 ntegration of x- ws ntegration of Concepts and techniques.i,,j,jm Use the method of substitution to find an expression for.the following integrals. a J --dx b --dx c J --dx d J dx J 3x+ 9x-7-3x 5-x 96 j NELSON SENOR MATHS Specialist

13 3 r --dx can be expressed as 3x- A -du r r u B - r 3 u -du - -dx can be expressed as 3-3x A - -du r u B -- r!o -du 3-7 u r C - 3 u r -du C - -du 3 u D log e ( :) c:) D log e E E r --dx 3x- -0 J --dx -7 3x- hihuhp J -3! dx = A log e lxl + c x +3 B log e lx c C log e lxl + c D log e lx + c E log e lx c 5 r x+l ---dx= x +x A log e lx + xl B flog e lx +xl+c C ½ log e (½) D ½ log e (¾) E ½log e (f).! n... rf sin (x) 6 l: urp J --dx = A log e () o cos(x) B -log e lcos(x)l+c C ½ log e (½) D log e lsin(x) E log e () 7 Find expressions for the following integrals. a f x ---dx 3x + b f sec (x) dx tan(x) C e f- x --dx e X +l.,. d J 6 ; + dx X +x Reasoning and communication 7t 8 Show that f tan(x)dx=llog e () Jo 9 Show that J - X l -dx=-log e () 0 x + 0 Show that f 3 e 3 - l dx = log e e x - ( e- J tl Ji NVERSE TRGONOMETRC FUNCTONS Your calculator has the functions sin- (x), cos- (x) and tan- (x) as the inverses of sin (x), cos (x) and tan (x). However, since sin (x), cos (x) and tan (x) repeat, there are actually many solutions to trigonometric equations like sin (x) = 0.5. You found the general solutions of these equations last year. To make the inverses of trigonometric functions into functions, the domains of sin (x), cos (x) and tan (x) are restricted to the principal values CHAPTER 7: ntegration techniques 97

14 MPORTANT The functions arcsine, arccosine and arctangent (usually abbreviated to arcsin, arccos and arctan) are defined as: arcsin (x) = y if and only if sin (y) = x and- ::; y::; arccos (x) = y if and only if cos (y) = x and O::; y::; n and arctan (x) = y where tan (y) = x and- < y < - The domains of arcsin (x) and arccos (x) are both [-, l]. The domain of arctan (x) is (-oo, oo ). arcsin (x), arccos (x) and arctan (x) are often written as sin- (x), cos- (x) and tan- (x) respectivel y. The graphs of y = arcsin (x), y = arccos (x) and y = arctan (x) are shown below. Each graph is shown with the dashed corresponding trigonometric graph. The graphs are shown as the reflections of sin (x), cos (x) and tan (x) in the line y = x. y -;- ) t , l7 f7 V y= ar,sir tx k :-- -- )_ n Ji / y,= s /i = 7t 7t.Jc': it -f r: c--r--jj : t J -, -[) -tt ---r / V; > V-t-- - 7t-- - /- _J}:J -=, -.. / V V, / <=r, nj Y 7t \ y=.ar.c.o.s X) "' f- ["'-. V /, -(0 r---. _l)_,,, / / V V ol,o) t-t-- V / / / / "-,/ -crc': = l ; cm (x.) x,, - y J Y t' x) V / / " r ( _ J-- -7 /' arc - 7t ' -j) q_.,, e:::::::, v,fa if"('t # l/ X,/, / it.il,,,,,,--- 7t -, l '/ , V / - -, f-- - : The derivatives of arcsin (x), arccos (x) and arctan (x) are of considerable importance. 98 NELSON SENOR MATHS Specialist

15 The rules for the derivatives can all be proven using implicit differentiation and the appropriate Pythagorean trigonometric identities. Given arcsin (x) = y, x = sin (y). Differentiating both sides, you get!!:_ x =!!:_ [ sin (y)]. dx dx Using the chain rule gives = cos (y) x dy, so dy = - -. dx dx cos(y) Using the trigonometric identity sin (x) + cos cx) =. dy gives - = ---;=====. dx J-sin (y) Substitution of x = sin (y) then gives dy = dx hl-x The other derivatives can be proven similarly. MPORTANT h!!:_ [arcsin (x)] = dx l-x d - - [arccos (x)] = and dx \fl-x r-:;- d - [arctan (x)] = -- dx l+x QED 0 Example 0 a Find!!:_ arccos (3x). dx b Find!!:_ tan- (x + ). dx a Write the problem and use the chain rule. Let u = 3x Differentiate, using the rule for arccos. Simplify. b Write the problem and use the chain rule. Let u = x + Differentiate, using the rule for arctan. Substitute for u. d d du - arccos (3x) = - arccos (u) x dx du dx - = ---x3 l-u -3 = l-9x!!:_ tan- (x + ) =!!:_ tan- (u) x du dx du dx =--xx l+u x Simplify. The derivatives of the inverse trigonometric functions allow you to integrate some expressions that would otherwise be difficult to do using other means CHAPTER 7: ntegration techniques 99

16 MPORTANT f Jbdx = arcsin (x) + c l-x f - dx = arccos (x) + c "\/, x- f dx = arctan (x) + c l+x 0 Example Find f 6 dx. l-x Choose a substitution. Find the derivative. Write the integral and rearrange to include u' (x). Substitute in u(x) and u' (x). Use the substitution rule. Let u(x) = x u' (x) = f f l-x l-(x) ==dx = - -;==== x dx = ½ f xu'(x)dx -vl-u =½f l u du Do the integration. = ½ arcsin (u) + c Substitute u(x) back in. = ½ arcsin (x) + c Write the answer. f dx =½arcsm (x) + c X You may also be required to integrate expressions of the type 6, Jh and a -x a -x a +x Rearrange f dx into a more suitable form as a +x f dx= f dx a ( +( ) ) a +( ) Choose u(x) = ::..Then u' (x) =.!..Rearrange the integral a a to include u' (x). MPORTANT f h dx = arcsin (.::.) + c a -x f Jh dx = arccos (.::.) + c a -x a f dx =.!:. arctan (.::.) + c a +x a a a 300 NELSON SENOR MATHS Specialist

17 l Then J - - dx=- J a +x a x-dx. +( -f ) a l, l J J a l+u a l+u a Substitute to get - -- x u (x )dx = - -- du= - arctan (u) + c. Substitute u(x) back in so that J dx = _!.. arctan ( ) + c. a +x a a You can prove the other integrations in a similar way. QED 0 Example J 3 - Find dx, correct to 3 decimal places. '\/5-x Write the integral and arrange to use the rule. Use the rule J Jh dx = arccos ( ) + c. a -x Substitute values. Simplify. Make sure that your calculator is set to radians. Round to write the answer. a r =- = l =dx = r = - = l =dx.j5-x.) 5 -X = cos- (0. x 3) - cos- (0. x ) = cos- (0.6) - cos- (0.) = dx,,,-0. '\/'.:J-x- J 3 Notice in Example that cos- was preferred to arccos because it was easier to use with a calculator to obtain an approximate answer. f an exact answer had been required, it would be more usual to leave it as arccos ( 0.6) - arccos ( 0.). You should also check that putting the negative sign out the front and using J b dx gives the same answer. a -x EXERCSE 7.0 nverse trigonometric functions Concepts and techniques i#,i,,j,jj Find the following. d. - ( d - ( 3 a - sm 3 x ) b - tan x - ) dx dx!! cos- (x) = dx A ==.J-x B.J-x d - ( X ) dx C -COS C ½-x d d -tan - ( x +x dx ) - D.J-x E -.====.J-6x CHAPTER 7: ntegration techniques 30

18 3..!!: tan- (x) = dx A - -.!. +x M ihiil J B +x dx = l-6x C +x D + 6 x E + x 5 6 A.!. sin- (x) + c D sin- (x) + c r dx= 0 A.:: - J - dx= o l+x A.:: 7 Find the following. B B t -- a J dx b J --dx h l-x +9x - 8 ah :: HlfJ Find each of the following. t B.!. sin- (6x) + c E.!. cos- (x) + c C sin ( % ) C tan ( % ) C b J - l dx 00-x D sm. - (t) - J3.l. 3 --dx C.!. sin- (x) + c 0 +9x d J ½ C E J ---dx x +9 it d 0 l-9x X 9 Find the value of dx x J 6 e J-/-dx X +6 0 Find J 3 dx correct to decimal places. -t 6-x Reasoning and communication 6. u'(x) 6 Express - - m the form of-- and hence find J - -dx. X + l+u X + d Prove that- [arctan (x)] = - dx l+x - 3 Prove that J Jh dx = arccos ( ) + c. a - x a Prove that J +z-dx =.!. arctan ( ) + c. a +x a a 30 NELSON SENOR MATHS Specialist

19 N - NTEGRATON USNG PARTAL FRACTONS Partial fractions refer to algebraic fractions that have been expressed as a sum of algebraic fractions with simpler denominators. MPORTANT a ax+ b Fractions of the form or can be expressed as partial fractions in (x-c)(x-d) (x-c)(x-d) A B the form x-c ( ) x-d ( ) 0 Example 3.. Express as partla f ractlons. X -x-3 Express. in t h e ti orm. a x -x-3 (x-b)(x-c) Write as partial fractions. Put the RHS over a common denominator. Equate the numerators. Let x = - and x = 3 to find values of A and B. Write the answer.. x -x-3 (x-3)(x+l) A B --+- (x-3)(x+l) (x-3) (x+l) = A(x+l)+B(x-3) (x-3)(x+) A =.!.andb=-.!. =A(x+l)+B(x-3) (x-3) (x+l) ----=--- x -x-3 You can use the method of partial fractions to find integrals where the denominator can be expressed as linear factors CHAPTER 7: ntegration techniques 303

20 0 Example x-3 Find f dx. X -x-6 x-3.. E xpress as parha f ract0ns. X -x-6 Put the RHS over a common denominator. Simplify and equate denominators. Equate coefficients. Solve for A and B. Write the integral and express it using partial fractions. ntegrate using logs. Write the answer. x-3 x-3 x -x-6 (x-3)(x+) A B =--+- x-3 x+ x-3 A(x+)+B(x-3) x -x-6 (x-3)(x+) x - 3 = Ax + A + Bx - 3B A + B = and A - 3B = -3 A = i = 0.6 and B =?.. =. 5 5 f ; x-3 dx = f dx + f!i_dx x -x-6 x-3 x+ = 0.6 log. x log. x+ + c f : x-3 dx X -x-6 = 0.6 log. x log. x+ + c You could use the properties oflogarithms to write the answer in Example as log e l(x - 3) 0 6(x + )'- + c or log e, (x-3) 3 (x + ) 7 + c. Partial fractions of more complicated forms can be used in integration. f the denominator of an algebraic fraction can be expressed with only linear factors, with some linear factors repeated, then it can still be expressed as a sum of partial fractions. MPORTANT a ax+b F ractons o f t h e fi orm or can b e expresse d as parha f rachons m (x-c) (x-d) (x-c) (x-d) A B C the form (x-c) (x-c) (x-d) 30 NELSON SENOR MATHS Specialist

21 0 Example 5 Find f x dx. (x-) (x+3) Express as partial fractions. Put the RHS over a common denominator. Equate the numerators. Use x =, -3, and O to find A, B and C. Write the integral and express it using partial fractions. Take out any common factors. x A B C -----= (x-) (x+3) (x-) (x-) (x+3) A =, B =. and C = A(x+3)+ B(x-)(x+3)+C(x-) = (x-) (x+3) x = A(x + 3) + B(x - ) (x + 3) + C(x - ) X..]_.]_ f----- dx =f s + - _s s_ dx (x-) (x+3) (x-) (x-) (x+3) _ f 3 f ,-dx+- --dx s (x-) 5 (x-) f (x+3 /x ntegrate each term. Simplify terms. Write the answer. = +.l log e x - S(x-) log e x+ 3 5 = }s log e -:-:-: - -5 ( - x ) f (x-) (x+3) dx = - f(x ) + is loge l::! CHAPTER 7: ntegration techniques 305

22 You can use a CAS calculator to change an algebraic fraction into partial fractions. Tl Nspire CAS & expand ( x -S x-6 ) 7 (x-6) - 7 (x-+) ClassPad Tap Action, Transformation and expand and enter the expression and the variable of interest. Q Edit Action nteractive 'l½ ½ W:;i Shnp!5? -W n expand(- - -, x) x -5x ( ) + 7-(;-6) Math Line vii " ;} Math fl" o n i 00 a Meth3 a f.-o, o re- Jim (D] [ ] [ 8 o Ver ebc o!lo sin cos tan 0 t ens EXE Alg Standard Real Rad <ii! EXERCSE 7.05 ntegration using partial fractions Partial froclions Concepts and techniques ijlj,.j,jiij Express the following in partial fraction form. 3x x b -3x+ x -x- x -x-6 -iuh/m Find the following. a f f ----dx x -3x+ C ----dx x -x-6 3 f dx = (x+l)(x-5) A.!.tn() 6 d (x-)(x -x-6) b 3x ----dx x -x- f l d f X dx o (x-)(x -x-6) C ¼ log e (¾) D 6 log e (½) E -6 log e(½) f (x-l) x+) dx = A ¼log e l(x-l)(x+) +c B log e l(x-l)(x+) l+c C log e l(x-l)(x+)i+c D i0 g x+ + c E.!.0 g J (x+) J+c 5 e x- 5 e x- 306 NELSON SENOR MATHS Specialist

23 5 x can be written in partial fraction form as x -x-3 A---- A+B A Bx x -x-3 x -x-3 x -x-3 A B A Bx x+l x D --+- x- x+3 x -x -7x- A+B+C A----, ,--- x 3 -x -7x- A+B C D ---x-- (x+l) x- 3x- 7 Find f l+ dx X -3x+ B z E --+- can be written in partial fraction form as A B C B (x+l) x+l x- AB C (x+l) x- E x R(x) f x 8 Express- - as and hence find - -dx. X - X - X - 9 Find f -f--dx. OX - C - A -+ - B - x+l x-3 C A + B + f (x+l) x- 0 Findf½-l-dx. OX - N : NTEGRATON BY PARTS Previously you used the product rule to calculate the derivative of products. For y = u(x) v(x) this is dy = v du + u dv. The corresponding dx dx dx rule for integration is called integration by parts. You use integration by parts to integrate expressions of the formf(x) i (x). This method allows you to integrate some products of functions where other simpler methods are not possible. n this case, it is generally not possible to remove products completely from an integral. The product is altered using integration by parts to a form that makes it easier to integrate.. dy du dv dv dy du Rearrangmg the product rule - = v-+ u-, you get u - = -- v-. dx dx dx dx dx dx Take the integral of both sides, giving fu dv dx = f dy dx-f v du dx where y = u(x) v(x). dx dx dx CHAPTER 7: ntegration techniques 307

24 This gives u dv dx=uv-jv du dx. ntegration by parts can also be written as f f(x) g'(x)dx = f(x) g(x) - J f'(x) g(x)dx. dx dx ntegration by parts f a function can be written as f(x) g' (x), then f f(x) g'(x)dx= f(x) g(x) -f f'(x) g(x)dx MPORTANT n practice, you select f (x) when f '(x) is easier to find and g' (x) when g(x) is no harder to find. t may take several tries to find the choice that makes the integral easier. t is not necessary to include the constant of integration because it will be part of the new integral at the end. C) Example 6 Find J xcos(x)dx Write x cos (x) in the form f (x) g' (x). Substitute f(x) g'(x) into the integral. Find f'(x) and g(x). Write the problem and set it using the formula f f(x) g'(x)dx = f(x) g(x)-j J'(x) g(x)dx ntegrate where necessary. Write the answer. Setf(x) = x and g' (x) = cos (x) J xcos(x)dx = J f(x)g'(x)dx f(x) = x so f'(x) = and g'(x) = cos (x) so g(x) = sin (x) J x cos(x )dx = x sin(x )- J sin(x )dx = x sin(x) + cos(x) + c J xcos(x)dx =xsin(x)+cos(x)+c n Example 6, it would have been much harder if f (x) and g(x) were chosen the other way around. ntegrals involving products with x, x, x 3, etc. can often be found in a similar way to Example 6. t may be necessary to apply integration by parts several times to simplify the integral. n some cases, you can use integration by parts to obtain an expression where a power is reduced. Repetition of the procedure eventually leads to an integral that can be evaluated. This is an example of a recursive procedure. 308 NELSON SENOR MATHS Specialist

25 0 Example 7 Find J x 3 e x dx. Step Write x 3 e x in the formf(x) g'(x). Substitute f(x) g'(x) into the integral. Findf'(x) andg(x). Write the problem and set it using the formula f f(x) g'(x)dx == f(x) g(x)-j f'(x) g(x)dx Set f(x) = x 3 and g'(x) = e x f x 3 e x dx= f f(x)g'(x)dx f(x) = x 3 so f'(x) = 3x and g'(x) = e x so g(x) = e x J x 3 e x dx==x 3 e x -J 3x e x dx Repeat the procedure for the integral J 3x e x dx. Step Write 3xV in the formf(x) g'(x). Substitute f(x) g'(x) into the integral. Findf'(x) andg(x). Write the problem and set it using the formula ff (x) g'(x )dx == f (x ) g(x)-j f'(x) g(x )dx Write the answer to the original problem. Set f(x) = 3x and g'(x) = e x J 3x e x dx== J f(x)g'(x)dx f(x) = 3x so f'(x) = 6x and g' (x) = e x so g(x) = e x J 3x e x dx == 3x e x -J 6xe x dx f x 3 e x dx==x 3 e x -(3x e x -f 6xe x dx) = x 3 e x -3x e x + J 6xe x dx Repeat the procedure for the integral J 6xe x dx. Step 3 Write 6xe x in the formf(x) g'(x). Substitute f(x) g'(x) into the integral. Find f'(x) and g(x). Write the problem and set it using the formula f f(x) g'(x)dx == f(x) g(x)-j f'(x) g(x)dx ntegrate where possible. Write the answer to the original problem. Write the answer. Setf(x) = 6x andg'(x) = e x J 6xe x dx == J f(x)g'(x)dx f(x) = 6x so f'(x) = 6 and g'(x) = e x so g(x) = e x J 6xe x dx == 6xe x - J 6e x dx = 6xe x - 6e x + c J x 3 e x dx = x 3 e x -( 3x e x - J 6xe x dx) ==x 3 e x -3x e x +6xe x -6e x +c CHAPTER 7: ntegration techniques 309

26 ntegration by parts is very useful in finding integrals involving functions with e X, log,(x) and arcsine, arccosine and arctangent. Find J:og e (x)dx. Write log e (x) in the form f(x) g'(x). Substitute f(x) g'(x) into the integral. Find f'(x) and g(x). Write the problem and set it using the formula f f(x) g'(x)dx= f(x) g(x)- f f'(x) g(x)dx Separate and simplify terms. ntegrate where necessary. Evaluate terms. Simplify terms. Collect terms. Setf(x) = log e (x) andg'(x) = s:og e (x)dx =ff (x)g'(x)dx f(x) = log,(x) so f'(x) = - and x g'(x) = so g(x) = x Ji log e (x)dx= xlog,(x)- f X -dx [ ] X l =[xlog,(x)]- J:dx n n = [ X log,(x) - [ X = [ log e ( )-½ log e ( ½)] -(-½) = [ log e ()-½log e ( ½) ]-½ = log e ( )+ ½log e ( )-½ Write the answer. J _ 5 ( ) 3 log e (x) dx - log e - 30 NELSON SENOR MATHS Specialist

27 EXERCSE 7.06 ntegration by parts Concepts and techniques ;;.;;.;.;5r,j J x sin ( x) dx = A x sin (x) + cos (x) + c D x cos (x) + sin (x) + c J 0 t xcos(x)dx= A D x sin (x) + cos (x) + c 8 -x cos (x) + sin (x) + c E COS (x) + C 8-7[ E x sin (x) + c C x sin (x) + c C - ntegral calculus 3 J 0 tx cos(x)dx= 5 A T 8-7[ D x sin (x) + x cos (x) - sin (x) + c M MHH J x e x dx = A x e x + c D e- J x 3 e x dx = A x e x + c C ¾xe x (x -3x+ 3x- )+c E ±e x (x 3-6x +6x-3)+c 8 x e x - xe x + e x + c E e x (-x +)+c 8 x e x - xe x + e x + c C -T E x cos (x) + c D ¾xe x (x -3x+3x)+c 6 Find the following. a J xe x dx b J xcos(x)dx c J xlog,(x)dx d J:Xe x dx e J0 if-xsin(x)dx 7,j,ij,jif: Find the following. a J)og e (x)dx 3 l d fi" cos- (x)dx e J cos- (3x)dx Reasoning and communication 8 Use a recursive procedure to find J x 3 sin(x)dx. Jl 9 Use a recursive procedure to find O x e x dx CHAPTER 7: ntegration techniques 3

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30 CHAPTER REVEW NTEGRATON TECHNQUES Multiple choice l#h,,i,jp f cos (x)dx = A x+ sin(x) + c D x+ sin(x) + c f 0.!f sin (3x)dx A n D =- -!. sin ( 6x) + c 6 3 i#huhftii f tan ( x )dx = A!. x-!. tan(x) + c D -x+!. tan(x) + c r.!f sin (8x)dx= Jo A..!::. D!. x sin(l6x) + c 8 8 n 5 f 0 sin(x)cos(x)dx= B x- sin(x) + c E x+ sin(x) + c B E n 6 7t 8 B!. x- tan(x) + c 6 E -x-!.tan(x)+c B..!::. + -/3 56 E!. x - - -cos(l6x) + c 8 8 C x+ sin(x) + c C -_! sin(6x) + c C x +!. sec (x) + c C 6 A!_ B C D --cos(x) + c E!. sin(x) + c 7 l#huhr f (x - 7) 6 dx = A D (x-7) 8 + c 6 B (x -x) B C E (x-7) 7 + c D 0 E 30 C!_ (x-7)7 + c 7 8 iihi,hp f- x -dx = x + A log.lxl + c D log.lx + c B ½ log. lx + + c E log.lx + + c C log.jx + J + c 3 NELSON SENOR MATHS Specialist

31 !!l n- J ¾cos(x) 9 l uu-h. dx = J sm(x) A log e (¾) B log e lsin (x) + c D ½ log e!sin (x ) E log e ( %) 0 BJ.li,i,ji J x cos (x )dx == A x sin (x) + cos (x) + c D.!.x sin(x) +.!. cos(x) + c Short answer 6 i-lr,hiil J 0 f sin(x)cos (x)dx == B.!.x sin(x) +.!. cos(x) + c E cos (x) + c C.!.x sin(x) + c J.l.,l ld Use the method of substitution to find an expression for the following integrals. a J--dx b f-_dx -3x 7x+ 3 Find an expression for the following integrals. a J-- x -dx 3x +3 M l,ihi Find the following. b J - e x --dx e +l d -(X) b d, -( ) a cos dx dx sm x - 5 Sketch the graph of y == ½ sin -l ( J stating the domain. 6 i#h HEi Express the following in partial fraction form. a z X -3x-8 7 ii li iill Find the following. a J 3x dx l X -x- 8 i@,j,, j,jiij Find J x e xdx. 9 fo lj!jij:j Find f log e (x)dx. Application 0 Show that½ J sin(x )cos(x )dx == ¼sin (x) + c Show that J sec (x)-ldx==tan(x)-x + c b b x (x+) (x-6) X J d 3x -x- x Express sin- ( ;) in the form of sin- ( )and hence find J p-dx. 3 Use a recursive procedure to find J x cos(x)dx. tlr Practice quiz CHAPTER 7: ntegration techniques 35

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