Ln7 1 5 2 ln7 11 2 ln11 7 2 ln7 2 438 17 1 e lnx 2 dx

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ln(7)1 +52.ln|7| }=112.ln(11) –72.ln(7) – 24.38.17.1e( ln(x) )2dx .Putu = ( ln(x) )2.Thendv = dx,du = 2.ln(x).1xdx ,andv = x.= uv –v du=( ln(x) )2xx.2.ln(x).1xdx= x.( ln(x) )22.ln(x) dx = x.( ln(x) )2– 2{ x.ln(x) – x }|e1= { e( ln(e) )2– 2e.ln(e) + 2e } – {1( ln(1) )2– 2.ln(1) + 2 } =e – 20.718.19.arcsin(x) dx .Putu = arcsin(x).Thendv = dx,du =11 – x2dx ,andv = x .= uv –v du=arcsin(x).x –x.11 – x2dx= x.arcsin(x) –x1 – x2dx(use u–sub withu = 1 – x2)=x.arcsin(x) –1u(–12)du=x.arcsin(x) +u+ C=x.arcsin(x) +1 – x2+C.
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Chapter 5 / Exercise 80
Applied Calculus
Berresford/Rockett
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Odd Answers: Chapter EightContemporaryCalculus6621.x.arctan(3x) dx.Putu = arctan(3x).Thendv = x dx,du =31 + 9x2dx,andv =12x2.= uv –v du=arctan(x).12x212x2.31 + 9x2dx=12x2.arctan(3x) –12131/31 + 9x2dx=12x2.arctan(3x) –12{x319.arctan(3x) } + C=12x2.arctan(3x) –16x +118arctan(3x) } + C.23.12ln(x)xdx.Useu–substitution!Putu = ln(x).Thendu =1xdx.=u du =12u2=12( ln(x) )2|21=12( ln(2) )212( ln(1) )2=12( ln(2) )20.240.25. (a)sin3(x) dx=13{ –S2.C + 2S dx } =13{ –S2.C – 2C } + K=13{ –sin2(x).cos(x) – 2.cos(x) } + K.(b)sin4(x) dx =14{ –S3.C + 3S2dx }=14{ –S4.C + 3[12( –SC + x ) ] }+ K=14{ –S3.C –32SC +32x } + K=14{ –sin3(x).cos(x) –32sin(x).cos(x) +32x } + K(c) on your own.27. (a)sec3(x) dx=12{ sec(x)tan(x) +sec(x) dx } =12{ sec(x).tan(x) + ln| sec(x) + tan(x) | } + K(b) and (c)on your own.29.cos3(2x + 3) dx . First do a substitution:u = 2x + 3.Thendu = 2 dxanddx =12du .=12cos3( u ) du =12{13( C2.S + 2C du )}=16{C2.S + 2S}+ K=16{cos2(u).sin(u) + 2sin(u)}+ K=16{cos2(2x + 3).sin(2x + 3) + 2sin(2x + 3)}+ K31.x.(2x + 5)19dx.(a) By parts:putu = x.Thendv = (2x + 5)19dx , du = dx ,andv =140(2x + 5)20.= uv –v du=x.140(2x + 5)20140(2x + 5)20dx=x.140(2x + 5)20140142(2x + 5)21+ C.(b) Substitution:putu = 2x + 5.Thendu = 2 dxanddx =12du.Also,x =12( u – 5 ) .=12( u – 5 ).u1912du =14u20– 5u19du=14{121u21520u20} + C=184(2x + 5)21580(2x + 5)20+ C.The answers (antiderivatives) in parts (a) and (b) look different, but you can check that the derivative ofeach answer isx.(2x + 5)19.
Odd Answers: Chapter EightContemporaryCalculus7733. (a) Make an informed prediction.(b)01sin(x) dx=–cos(x)|10=(–cos(1)) – (–cos(0)) =cos(0) – cos(1)1 – 0.54 =0.46.01x.sin(x) dx = –x.cos(x) + sin(x)|10= {–1.cos(1) + sin(1)} – {–0.cos(0) + sin(0)} = sin(1) – cos(1)0.30.35. (a) Make an informed prediction.(b) See problem 17: Vx–axis=x=1eπ( ln(x) )2dx=π(e – 2)2.257.Vy–axis=y=012π.y.eydy = 2πy=01y.eydy(use integration by parts with u = y, dv = eydy : see Example 2)=2π( y.ey– ey)|10= 2π(1.e1– e1) – 2π(0.e0– e0) = 2π(0) – 2π(–1) = 2π6.283.

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Applied Calculus
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Chapter 5 / Exercise 80
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Berresford/Rockett
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