Problem Set 5 Solutions: Integration Techniques

SOLUTIONS TO 18.01 EXERCISES Unit 5. Integration techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 a) tan−1 √ π 3= 3 √ b) sin−1 ( 3 π )= 2 3 √ √ √ c)√ tan θ = 5 implies sin θ = 5/ 26, cos θ = 1/ 26, cot θ = 1/5, csc θ = 26/5, sec θ = 26 (from triangle) √ π 3 π d) sin−1 cos( ) = sin−1 ( )= 6 2 3 f) tan−1 tan( π π e) tan−1 tan( ) = 3 3 2π −π −π ) = tan−1 tan( )= 3 3 3 g) lim tan−1 x = x→−∞ 5A-2 � 2 � dx π −1 �2 = tan x = tan−1 2 − 2+1 1 x 4 1 � 2b � 2b dx d(by) b) = (put x = by) 2 2 x +b (by)2 + b2 b b � 2 dy 1 π = = (tan−1 2 − ) 2 b 4 1 b(y + 1) � 1 � dx π −π 1 √ c) = sin−1 x�−1 = − =π 2 2 2 1 − x −1 a) COPYRIGHT DAVID JERISON AND MIT 1996, 2003 1 −π . 2 E. Solutions to 18.01 Exercises 5A-3 a) y = 5. Integration techniques x−1 1 (x + 1) , so 1 − y 2 = 4x/(x + 1)2 , and � = √ . Hence 2 x+1 2 x 1−y dy 2 = dx (x + 1)2 d dy/dx sin−1 y = � dx 1 − y2 2 (x + 1) = · √ (x + 1)2 2 x 1 √ = (x + 1) x b) sech2 x = 1/ cosh2 x = 4/(ex + e−x )2 c) y = x + √ √ x2 + 1, dy/dx = 1 + x/ x2 + 1. √ d dy/dx 1 + x/ x2 + 1 1 √ ln y = = =√ 2 2 dx y x +1 x+ x +1 d) cos y = x =⇒ (− sin y)(dy/dx) = 1 dy −1 −1 = =√ dx sin y 1 − x2 e) Chain rule: d 1 1 1 sin−1 (x/a) = � · =√ dx a2 − x2 1 − (x/a)2 a f) Chain rule: d 1 −a −a sin−1 (a/x) = � · 2 = √ 2 dx x x x2 − a2 1 − (a/x) √ g) y = x/ 1 − x2 , dy/dx = (1 − x2 )−3/2 , 1 + y 2 = 1/(1 − x2 ). Thus d dy/dx 1 tan−1 y = = (1 − x2 )−3/2 (1 − x2 ) = √ dx 1 + y2 1 − x2 Why is this the same as the derivative of sin−1 x? h) y = √ √ x − 1, dy/dx = −1/2 x − 1, 1 − y 2 = x. Thus, dy/dx −1 d sin−1 y = � = � dx 2 x(1 − x) 1 − y2 2 5. Integration techniques E. Solutions to 18.01 Exercises 5A-4 a) y � = sinh x. A tangent line through the origin has the equation y = mx. If it meets the graph at x = a, then ma = cosh(a) and m = sinh(a). Therefore, a sinh(a) = cosh(a) . b) Take the difference: F (a) = a sinh(a) − cosh(a) Newton’s method for finding F (a) = 0, is the iteration an+1 = an − F (an )/F � (an ) = an − tanh(an ) + 1/an With a1 = 1, a2 = 1.2384, a3 = 1.2009, a4 = 1.19968. A serviceable approximation is a ≈ 1.2 (The slope is m = sinh(a) ≈ 1.5.) The functions F and y are even. By symmetry, there is another solution −a with slope − sinh a. 5A-5 a) ex − e−x 2 x e + e−x y � = cosh x = 2 y �� = sinh x y = sinh x = y � is never zero, so no critical points. Inflection point x = 0; slope of y is 1 there. y is an odd function, like ex /2 for x >> 0. y = sinh 1x y = sinh x b) y = sinh−1 x ⇐⇒ x = sinh y. Domain is the whole x-axis. c) Differentiate x = sinh y implicitly with respect to x: 1 = cosh y · dy dx dy 1 1 = =� dx cosh y sinh2 y + 1 d sinh−1 x 1 =√ 2 dx x +1 3 E. Solutions to 18.01 Exercises 5. Integration techniques d) � dx √ = 2 x + a2 � dx � 2 a x + a2 /a2 � d(x/a) � = (x/a)2 + 1 = sinh−1 (x/a) + c 5A-6 a) 1 π � b) y = π sin θdθ = 2/π 0 √ √ 1 − x2 =⇒ y � = −x/ 1 − x2 =⇒ 1 + (y � )2 = 1/(1 − x2 ). Thus � ds = w(x)dx = dx/ 1 − x2 . Therefore the average is � �� 1 � dx dx 2 √ √ 1−x 2 1−x 1 − x2 −1 −1 � 1 1 The numerator is dx = 2. To see that these integrals are the same as the ones −1 in part (a), take x = cos θ (as in polar coordinates). Then dx = − sin θdθ and the limits of integral are from θ = π to θ = 0. Reversing the limits changes the minus back to plus: � 1� � π dx 2 1−x √ = sin θdθ 1 − x2 −1 0 � 1 � π dx √ = dθ = π 1 − x2 −1 0 (The substitution x = sin t works similarly, but the limits of integration are −π/2 and π/2.) c) (x = sin t, dx = cos tdt) � � � π/2 1 1� 1 π/2 2 2 1 − x dx = cos tdt = cos2 tdt 2 −1 2 −π/2 0 � π/2 1 + cos 2t = dt 2 0 = π/4 5B. Integration by direct substitution Do these by guessing and correcting the factor out front. The substitution used implicitly is given alongside the answer. 4 5. Integration techniques � E. Solutions to 18.01 Exercises � 3 1 x x2 − 1dx = (x2 − 1) 2 + c (u = x2 − 1, du = 2xdx) 3 5B-1 � e8x dx = 5B-2 1 8x e + c (u = 8x, du = 8dx) 8 � ln xdx 1 = (ln x)2 + c (u = ln x, du = dx/x) x 2 � cos xdx ln(2 + 3 sin x) = + c (u = 2 + 3 sin x, du = 3 cos xdx) 2 + 3 sin x 3 � sin2 x cos xdx = 5B-3 5B-4 5B-5 � 5B-6 sin 7xdx = � sin x3 + c (u = sin x, du = cos xdx) 3 − cos 7x + c (u = 7x, du = 7dx) 7 � 6xdx √ = 6 x2 + 4 + c (u = x2 + 4, du = 2xdx) x2 + 4 5B-7 5B-8 Use u = cos(4x), du = −4 sin(4x)dx, � sin(4x)dx −du = cos(4x) 4u ln u ln(cos 4x) =− +c=− +c 4 4 � � tan 4xdx = � ex (1 + ex )−1/3 dx = 5B-9 � 5B-10 sec 9xdx = � 5B-11 � 5B-12 1 ln(sec(9x) + tan(9x)) + c (u = 9x, du = 9dx) 9 sec2 9xdx = tan 9x + c (u = 9x, du = 9dx) 9 2 −x2 xe 3 (1 + ex )2/3 + c (u = 1 + ex , du = ex dx) 2 −e−x dx = + c (u = x2 , du = 2xdx) 2 5 E. Solutions to 18.01 Exercises 5. Integration techniques 5B-13 u = x3 , du = 3x2 dx implies � � x2 dx du tan−1 u = = +c 1 + x6 3(1 + u2 ) 3 tan−1 (x3 ) = +c 3 � π/3 � sin π/3 3 5B-14 sin x cos xdx = u3 du (u = sin x, du = cos xdx) 0 √ � sin 0 3/2 = 0 e � 1 1 � 0 9 64 0 5B-15 = �√3/2 � � u3 du = u4 /4� = � (ln x)3/2 dx = x � ln e u3/2 du (u = ln x, du = dx/x) ln 1 �1 2 � y 3/2 dy = (2/5)y 5/2 � = 5 0 1 � 5B-16 −1 tan−1 xdx = 1 + x2 � tan−1 1 udu (u = tan−1 x, du = dx/(1 + x2 ) tan−1 (−1) �π/4 u2 �� = udu = =0 2 �−π/4 −π/4 π/4 � (tan x is odd and hence tan−1 x is also odd, so the integral had better be 0) 5C. Trigonometric integrals � 5C-1 sin2 xdx = � 5C-2 3 � 1 − cos 2x x sin 2x dx = − +c 2 2 4 sin (x/2)dx = � � 2 (1 − cos (x/2)) sin(x/2)dx = (put u = cos(x/2), du = (−1/2) sin(x/2)dx) 3 = −2u + 2u3 2cos(x/2) + c = −2 cos(x/2) + +c 3 3 6 −2(1 − u2 )du 5. Integration techniques � 5C-3 � sin4 xdx = cos2 (2x) dx = 4 � ( � E. Solutions to 18.01 Exercises 1 − cos 2x 2 ) dx = 2 � 1 − 2 cos 2x + cos2 2x dx 4 1 + cos 4x x sin 4x dx = + +c 8 8 32 Adding together all terms: � � 5C-4 sin4 xdx = cos3 (3x)dx = � 3x 1 1 − sin(2x) + sin(4x) + c 8 4 32 (1 − sin2 (3x)) cos(3x)dx = � 1 − u2 du (u = sin(3x), 3 du = 3 cos(3x)dx) 3 = u u3 sin(3x) sin(3x) − +c= − +c 3 9 3 9 � 5C-5 sin3 x cos2 xdx = � (1 − cos2 x) cos2 x sin xdx = � −(1 − u2 )u2 dy (u = cos x, du = − sin xdx) =− u3 u5 cos x3 cos x5 + +c=− + +c 5 5 3 3 � 5C-6 � 4 sec xdx = 2 � 2 (1 + tan x) sec xdx = (1 + u2 )du (u = tan x, du = sec2 xdx) =u+ � 5C-7 u3 tan3 x + c = tan x + +c 3 3 2 � 2 sin (4x) cos (4x)dx = sin2 8xdx = 4 � (1 − cos 16x)dx 1 sin 16x = − +c 8 8 128 A slower way is to use 2 2 sin (4x) cos (4x) = � 1 − cos(8x) 2 �� 1 + cos(8x) 2 multiply out and use a similar trick to handle cos2 (8x). 7 � E. Solutions to 18.01 Exercises 5. Integration techniques 5C-8 sin2 (ax) dx cos(ax) � 1 − cos2 (ax) = dx cos(ax) � = (sec(ax) − cos(ax))dx � tan2 (ax) cos(ax)dx = = � 1 1 ln(sec(ax) + tan(ax)) − sin(ax) + c a a 5C-9 � 1 − cos2 x sin xdx cos2 x � 1 − u2 du (u = cos x, du = − sin xdx) = − u2 1 = u + + c = cos x + sec x + c u sin3 x sec2 xdx = 5C-10 � 2 � � 2 (tan x + cot x) dx = 2 tan x + 2 + cot xdx = � sec2 x + csc2 xdx = tan x − cot x + c � 5C-11 sin x cos(2x)dx � = 2 � sin x(2 cos x − 1)dx = (1 − 2u2 )du (u = cos x, du − sin xdx) 2 2 = u − u3 + c = cos x − cos3 x + c 3 3 � 5C-12 0 π �π � 2 −2 3 � sin x cos(2x)dx = cos x − cos x� = (See 27.) 3 3 0 � 1 + (y � )2 dx = 1 + cot2 xdx = csc xdx. �π/2 � π/2 � √ � arclength = csc xdx = − ln(csc x + cot x)� = ln(1 + 2) � π/4 5C-13 ds = � π/4 � 5C-14 0 π/a π sin2 (ax)dx = π π/a � (1/2)(1 − cos(2ax))dx = π 2 /2a 0 8 5. Integration techniques E. Solutions to 18.01 Exercises 5D. Integration by inverse substitution 5D-1 Put x = a sin θ, dx = a cos θdθ: � � dx 1 1 x = 2 sec2 θdθ = 2 tan θ + c = √ +c 2 2 3/2 2 a a (a − x ) a a2 − x2 5D-2 Put x = a sin θ, dx = a cos θdθ: � � � x3 dx 3 3 3 √ =a sin θdθ = a (1 − cos2 θ) sin θdθ a 2 − x2 = a3 (− cos θ + (1/3) cos3 θ) + c � = −a2 a2 − x2 + (a2 − x2 )3/2 /3 + c 5D-3 By direct substitution (u = 4 + x2 ), � xdx = (1/2) ln(4 + x2 ) + c 4 + x2 Put x = 2 tan θ, dx = 2 sec2 θdθ, � � dx 1 = dθ = θ/2 + c 4 + x2 2 In all, � (x + 1)dx = (1/2) ln(4 + x2 ) + (1/2) tan−1 (x/2) + c 4 + x2 5D-4 Put x = a sinh y, dx = a cosh ydy. Since 1 + sinh2 y = cosh2 y, � � � � a2 2 2 2 2 a + x dx = a cosh ydy = (cosh(2y) − 1)dy 2 = (a2 /4) sinh(2y) − a2 y/2 + c = (a2 /2) sinh y cosh y − a2 y/2 + c � = x a2 + x2 /2 − a2 sinh−1 (x/a) + c 5D-5 Put x = a sin θ, dx = a cos θdθ: � √ 2 � a − x2 dx = cot2 θdθ x2 � = (csc2 θ − 1)dθ = − ln(csc θ + cot θ) − θ + c � = − ln(a/x + a2 − x2 /x) − sin−1 (x/a) + c 5D-6 Put x = a sinh y, dx = a cosh ydy. � � � x2 a2 + x2 dx = a4 sinh2 y cosh2 ydy � � 2 4 4 = (a /2) sinh (2y)dy = a /4 (cosh(4y) − 1)dy = (a4 /16) sinh(4y) − a4 y/4 + c = (a4 /8) sinh(2y) cosh(2y) − a4 y/4 + c = (a4 /4) sinh y cosh y(cosh2 y + sinh2 y) − a4 y/4 + c � = (1/4)x a2 + x2 (2x2 + a2 ) − (a4 /4) sinh−1 (x/a) + c 9 E. Solutions to 18.01 Exercises 5. Integration techniques 5D-7 Put x = a sec θ, dx = a sec θ tan θdθ: � √ 2 � x − a2 dx tan2 θdθ = x2 sec θ � � (sec2 θ − 1)dθ = = (sec θ − cos θ)dθ sec θ = ln(sec θ + tan θ) − sin θ + c � � = ln(x/a + x2 − a2 /a) − x2 − a2 /x + c � � = ln(x + x2 − a2 ) − x2 − a2 /x + c1 (c1 = c − ln a) 5D-8 Short way: u = x2 − 9, du = 2xdx, � � x x2 − 9dx = (1/3)(x2 − 9)3/2 + c direct substitution Long way (method of this section): Put x = 3 sec θ, dx = 3 sec θ tan θdθ. � � � x x2 − 9dx = 27 sec2 θ tan2 θdθ � = 27 tan2 θd(tan θ) = 9 tan3 θ + c = (1/3)(x2 − 9)3/2 + c √ (tan θ = x2 − 9/3). The trig substitution method does not lead to a dead end, but it’s not always fastest. 5D-9 y � = 1/x, ds = � 1 + 1/x2 dx, so � b arclength = � 1 + 1/x2 dx 1 Put x = tan θ, dx = sec2 θdθ, � √ 2 � x + 1dx = x � = � = sec θ sec2 θdθ tan θ sec θ(1 + tan2 θ) dθ tan θ (csc θ + sec θ tan θ)dθ = − ln(csc θ + cot θ) + sec θ + c � � = − ln( x2 + 1/x + 1/x) + x2 + 1 + c � � = − ln( x2 + 1 + 1) + ln x + x2 + 1 + c � � √ √ arclength = − ln( b2 + 1 + 1) + ln b + b2 + 1 + ln( 2 + 1) − 2 10 5. Integration techniques E. Solutions to 18.01 Exercises Completing the square � 5D-10 dx = 2 (x + 4x + 13)3/2 � dx (x + 2 = 3 tan θ, dx = ((x + 2)2 + 32 )3/2 3 sec2 θdθ) = 1 9 � cos θdθ = 1 (x + 2) sin θ + c = √ +c 9 9 x2 + 4x + 13 5D-11 � � � � x −8 + 6x − x2 dx = x 1 − (x − 3)2 dx � = (sin θ + 3) cos2 θdθ = (−1/3) cos3 θ + (3/2) (x − 3 = sin θ, dx = cos θdθ) � (cos 2θ + 1)dθ = −(1/3) cos3 θ + (3/4) sin 2θ + (3/2)θ + c = −(1/3) cos3 θ + (3/2) sin θ cos θ + (3/2)θ + c = −(1/3)(−8 + 6x − x2 )3/2 � + (3/2)(x − 3) −8 + 6x − x2 + (3/2) sin−1 (x − 3) + c 5D-12 � � � � −8 + 6x − x2 dx = 1 − (x − 3)2 dx (x − 3 = sin θ, dx = cos θdθ) � = cos2 θdθ � 1 = (cos 2θ + 1)dθ 2 1 θ = sin 2θ + + c 4 2 1 θ = sin θ cos θ + + c 2 2 √ (x − 3) −8 + 6x − x2 sin−1 (x − 3) = + +c 2 2 � � dx dx √ � 5D-13 = . Put x − 1 = sin θ, dx = cos θdθ. 2 2x − x 1 − (x − 1)2 � = dθ = θ + c = sin−1 (x − 1) + c � � xdx xdx √ � 5D-14 = . Put x + 2 = 3 tan θ, dx = 3 sec2 θ. 2 x + 4x + 13 (x + 2)2 + 32 � = (3 tan θ − 2) sec θdθ = 3 sec θ − 2 ln(sec θ + tan θ) + c � � = x2 + 4x + 13 − 2 ln( x2 + 4x + 13/3 + (x + 2)/3) + c � � = x2 + 4x + 13 − 2 ln( x2 + 4x + 13 + (x + 2)) + c1 (c1 = c − ln 3) 11 E. Solutions to 18.01 Exercises 5. Integration techniques � √ 2 � � (2x − 1)2 + 42 dx 4x − 4x + 17dx 5D-15 = 2x − 1 2x − 1 (put 2x − 1 = 4 tan θ, dx = 2 sec2 θdθ as in Problem 9) � sec θ =2 sec2 θdθ tan θ � sec θ(1 + tan2 θ) =2 dθ tan θ � = 2 (csc θ + sec θ tan θ)dθ = −2 ln(csc θ + cot θ) + 2 sec θ + c � � = −2 ln( 4x2 − 4x + 17/(2x − 1) + 4/(2x − 1)) + 4x2 − 4x + 17/2 + c � � = −2 ln( 4x2 − 4x + 17 + 4) + 2 ln(2x − 1) + 4x2 − 4x + 17/2 + c 5E. Integration by partial fractions 5E-1 5E-2 5E-3 5E-4 1 1/5 −1/5 = + (cover up) (x − 2)(x + 3) x−2 x+3 � dx = (1/5) ln(x − 2) − (1/5) ln(x + 3) + c (x − 2)(x + 3) x 2/5 3/5 = + (cover up) (x − 2)(x + 3) x−2 x+3 � xdx = (2/5) ln(x − 2) + (3/5) ln(x + 3) + c (x − 2)(x + 3) x 1/10 1/2 −3/5 = + + (cover up) (x − 2)(x + 2)(x + 3) x−2 x+2 x+3 � xdx = (1/10) ln(x − 2) + (1/2) ln(x + 2) − (3/5) ln(x + 3) (x2 − 4)(x + 3) 3x2 + 4x − 11 2 −2 3 dx = + + (cover-up) 2 (x − 1)(x − 2) x−1 x+1 x−2 � 2dx −2dx 3dx + + = 2 ln(x − 1) − 2 ln(x + 1) + 3 ln(x − 2) + c x−1 x+1 x−2 . 12 5. Integration techniques 5E-5 E. Solutions to 18.01 Exercises 3x + 2 2 B 1 = + + (coverup); to get B, put say x = 1: 2 x(x + 1) x x + 1 (x + 1)2 5 B 1 =2+ + =⇒ B = −2 4 2 4 � 3x + 2 1 dx = 2 ln x − 2 ln(x + 1) − +c 2 x(x + 1) x+1 2x − 9 Ax + B C 5E-6 = 2 + (x2 + 9)(x + 2) x +9 x+2 By cover-up, C = −1. To get B and A, x = 0 =⇒ −9 B 1 = − =⇒ B = 0 9 9·2 2 −7 A 1 = − =⇒ A = 1 10 3 10 · 3 � 2x − 9 1 dx = ln(x2 + 9) − ln(x + 2) + c 2 (x + 9)(x + 2) 2 x = 1 =⇒ 5E-7 Instead of thinking of (4) as arising from (1) by multiplication by x−1, think of it as arising from x − 7 = A(x + 2) + B(x − 1) by division by x + 2; since this new equation is valid for all x, the line (4) will be � −2, in particular it will be valid for x = 1 . valid for x = 5E-8 Long division: a) x2 1 =1+ 2 −1 x −1 x2 b) x3 x =x+ 2 2 x −1 x −1 c) x2 1/9 = x/3 + 1/9 + 3x − 1 3x − 1 d) x+2 1 7/3 = + 3x − 1 3 3x − 1 e) x8 B3 x3 + B2 x2 + B1 x + B0 4 3 2 = A x +A x +A x +A x+A + 4 3 2 1 0 (x + 2)2 (x − 2)2 (x + 2)2 (x − 2)2 5E-9 a) Cover-up gives 1/2 −1/2 1 1 = = + x2 − 1 (x − 1)(x + 1) x−1 x+1 13 E. Solutions to 18.01 Exercises 5. Integration techniques From 8a, 1/2 −1/2 x2 =1+ + 2 x −1 x−1 x+1 and x2 dx = x + (1/2) ln(x − 1) − (1/2) ln(x + 1) + c x2 − 1 b) Cover-up gives � x2 x x 1/2 1/2 = = + −1 (x − 1)(x + 1) x−1 x+1 From 8b, 1/2 1/2 x3 =x+ + 2 x −1 x−1 x+1 � and x3 dx = x2 /2 + (1/2) ln(x − 1) + (1/2) ln(x + 1) + c x2 − 1 c) From 8c, x2 dx = x2 /6 + x/9 + (1/27) ln(3x − 1) + c 3x − 1 � d) From 8d, � x+2 dx = x/3 + (7/9) ln(3x − 1) 3x − 1 e) Cover-up says that the proper rational function will be written as a1 a2 b1 b2 + + + x − 2 (x − 2)2 x + 2 (x + 2)2 where the coefficients a2 and b2 can be evaluted from the B’s using cover-up and the coefficients a1 and b1 can then be evaluated using x = 0 and x = 1, say. Therefore, the integral has the form A4 x5 /5 + A3 x4 /4 + A2 x3 /3 + A1 x2 /2 + A0 x + c + a1 ln(x − 2) − a2 b2 + b1 ln(x + 2) − x−2 x+2 5E-10 a) By cover-up, x3 � 1 −1 1/2 1/2 1 = = + + −x x(x − 1)(x + 1) x x−1 x+1 dx 1 1 = − ln x + ln(x − 1) + ln(x + 1) + c x3 − x 2 2 b) By cover-up, � (x + 1) −3 4 = + . Therefore, (x − 2)(x − 3) x−2 x−3 (x + 1) dx = −3 ln(x − 2) + 4 ln(x − 3) + c (x − 2)(x − 3) 14 5. Integration techniques c) E. Solutions to 18.01 Exercises (x2 + x + 1) −7x + 1 =1+ 2 . By cover-up, x2 + 8x x + 8x −7x + 1 −7x + 1 1/8 −57/8 = = + 2 x + 8x x(x + 8) x x+8 � and (x2 + x + 1) = x + (1/8) ln x − (57/8) ln(x + 8) + c x2 + 8x d) Seeing double? It must be late. e) x3 1 1 A B C = 2 = + 2+ 2 +x x (x + 1) x x x+1 Use the cover-up method to get B = 1 and C = 1. For A, x = 1 =⇒ 1 1 =A+1+ =⇒ A = −1 2 2 In all, � f) dx = x3 + x2 x3 � � � 1 1 1 1 − + 2+ dx = − ln x + ln(x + 1) − + c x+1 x x x x2 + 1 x2 + 1 A B C = = + + 2 + 2x + x x(x + 1)2 x x + 1 (x + 1)2 By cover-up, A = 1 and C = −2. For B, 2 B 2 =1+ − =⇒ B = 0 and 2 4 4 � � � � x2 + 1 1 2 2 dx = − dx = ln x + +c x3 + 2x2 + x x (x + 1)2 x+1 x = 1 =⇒ g) Multiply out denominator: (x + 1)2 (x − 1) = x3 + x2 − x − 1. Divide into numerator: x3 −x2 + x + 1 =1+ 3 3 2 x +x −x−1 x + x2 − x − 1 Write the proper rational function as −x2 + x + 1 A B C = + + (x + 1)2 (x − 1) x + 1 (x + 1)2 x−1 By cover-up, B = 1/2 and C = 1/4. For A, x = 0 =⇒ −1 = A + 1 1 5 − =⇒ A = − 2 4 4 15 and E. Solutions to 18.01 Exercises � 5. Integration techniques x3 dx = (x + 1)2 (x − 1) � � � −5/4 1/2 1/4 1+ + + dx x + 1 (x + 1)2 x−1 1 = x − (5/4) ln(x + 1) − + (1/4) ln(x − 1) + c 2(x + 1) � h) (x2 + 1)dx = x2 + 2x + 2 � (1 − 1 + 2x )dx = x − 2 x + 2x + 2 � (2y − 1)dy (put y = y2 + 1 x + 1) = x − ln(y 2 + 1) + tan−1 y + c = x − ln(x2 + 2x + 2) + tan−1 (x + 1) + c 5E-11 Separate: dy = dx y(1 − y) Expand using partial fractions and integrate � � 1 1 ( − )dy = dx y y−1 Hence, ln y − ln(y − 1) = x + c Exponentiate: y = ex+c = Aex y−1 (A = ec ) Aex Aex − 1 (If you integrated 1/(1 − y) to get − ln(1 − y) then you arrive at y= y= Aex Aex + 1 This is the same family of answers with A and −A traded.) 5E-12 a) 1 + z 2 = 1 + tan2 (θ/2) = sec2 (θ/2). Therefore, cos2 (θ/2) = 1 1 + z2 and sin2 (θ/2) = 1 − 1 z2 = 2 1+z 1 + z2 Next, 1 − z2 1 z2 − = and 2 2 1+z 1 + z2 1+z � � 1 z2 2z sin θ = 2 sin(θ/2) cos(θ/2) = 2 = 2 1+z 1 + z2 1 + z2 cos θ = cos2 (θ/2) − sin2 (θ/2) = Finally, dz = (1/2) sec2 (θ/2)dθ = (1/2)(1 + z 2 )dθ =⇒ dθ = 16 2dz 1 + z2 5. Integration techniques E. Solutions to 18.01 Exercises b) π � 0 tan π/2 2dz/(1 + z 2 ) 1 + 2z/(1 + z 2 ) 0 �tan � ∞ ∞ 2dz 2dz = = 2 z + 1 + 2z (z + 1)2 0 0 �∞ −2 �� = =2 1 + z� dθ = 1 + sin θ � 0 c) π � 0 � ∞ 2dz/(1 + z 2 ) 2(1 + z 2 )dz = (1 + 2z/(1 + z 2 ))2 (1 + z)4 tan 0 0 � ∞ 2 2(1 + (y − 1) )dy = (put y = z + 1) y4 1 � ∞ � ∞ (2y 2 − 4y + 4)dy = = (2y −2 − 4y −3 + 4y −4 )dy y4 1 1 �∞ = −2y −1 + 2y −2 − (4/3)y −3 � = 4/3 dθ = (1 + sin θ)2 � tan π/2 1 � () π � ∞ 2z 2dz = 2 1 + z2 1 + z 0 �∞ −2 �� = =2 1 + z 2 �0 � sin θdθ = 0 0 ∞ 4zdz (1 + z 2 )2 d 5E-13 a) z = tan(θ/2) =⇒ 1 + cos θ = 2/(1 + z 2 ) and 0 ≤ θ ≤ π/2 corresponds to 0 ≤ z ≤ 1. � π/2 � 1 dθ 2dz/(1 + z 2 ) A= = 2 2(1 + cos θ) 8/(1 + z 2 )2 0 0 � 1 �1 = (1/4)(1 + z 2 )dz = (1/4)(z + z 3 /3)�0 = 1/3 0 b) The curve r = 1/(1 + cos θ) is a parabola: r + r cos θ = 1 =⇒ r + x = 1 =⇒ r2 = (1 − x)2 =⇒ y 2 = 1 − 2x √ This is the region under y = 1 − 2x in the first quadrant: � 1/2 �1/2 √ � 1 − 2xdx = −(1/3)(1 − 2x)3/2 � A= = 1/3 0 0 5F. Integration by parts. Reduction formulas � 5F-1 a) a x ln xdx = � xa+1 xa+1 ln xd( ) = ln x · − a+1 a+1 17 � xa+1 1 · dx a+1 x E. Solutions to 18.01 Exercises = xa+1 ln x − a+1 � b) � 5F-2 a) � b) c) xex dx = x2 ex dx = � � � � x2 d(ex ) = x2 · ex − � 3 x 3 x x d(e ) = x · e − x e dx = � xd(ex ) = x · ex − ex dx = x · ex − ex + c � ex · 2xdx xex dx = x2 · ex − 2x · ex + 2ex + c 3 x = x3 · ex − 3 d) xa xa+1 ln x xa+1 − dx = + c (a = � −1) a+1 a+1 (a + 1)2 x−1 ln xdx = (ln x)2 /2 + c (u = ln x, du = dx/x) = x2 · ex − 2 � � 5. Integration techniques � ex · 3x2 dx x2 ex dx = x3 · ex − 3x2 · ex + 6x · ex − 6ex + c � � n ax x e dx = eax n n ·x − = a a � eax eax n ·x − x d( )= a a n � eax · nxn−1 dx a xn−1 eax dx 5F-3 � � � 4dx sin−1 (4x)dx = x · sin−1 (4x) − xd(sin−1 (4x)) = x · sin−1 (4x) − x · � 1 − (4x)2 � du √ = x · sin−1 (4x) + (put u = 1 − 16x2 , du = −32xdx) 8 u 1√ = x · sin−1 (4x) + u+c 4 � 1 = x · sin−1 (4x) + 1 − 16x2 + c 4 5F-4 � x � x � e d(sin x) = e sin x − ex sin xdx � = ex sin x − ex d(− cos x) � x x = e sin x + e cos x − ex cos xdx e cos xdx = x 18 5. Integration techniques � Add E. Solutions to 18.01 Exercises ex cos xdx to both sides to get � 2 ex cos xdx = ex sin x + ex cos x + c Divide by 2 and replace the arbitrary constant c by c/2: � ex cos xdx = (ex sin x + ex cos x)/2 + c 5F-5 � � cos(ln x)dx = x · cos(ln x) − xd(cos(ln x)) � = x · cos(ln x) + sin(ln x)dx � = x · cos(ln x) + x · sin(ln x) − xd(sin(ln x)) � = x · cos(ln x) + x · sin(ln x) − cos(ln x)dx � Add cos(ln x)dx to both sides to get � 2 cos(ln x)dx = x cos(ln x) + x sin(ln x) + c Divide by 2 and replace the arbitrary constant c by c/2: � cos(ln x)dx = (x cos(ln x) + x sin(ln x))/2 + c 5F-6 Put t = ex =⇒ dt = ex dx and x = ln t. Therefore � � xn ex dx = (ln t)n dt Integrate by parts: � � � (ln t)n dt = t · (ln t)n − td(ln t)n = t(ln t)n − n (ln t)n−1 dt because d(ln t)n = n(ln t)n−1 t−1 dt. 19