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

• Homework Help
• 65
• 60% (5) 3 out of 5 people found this document helpful

This preview shows page 57 - 60 out of 65 pages.

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.
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Chapter 5 / Exercise 80
Applied Calculus
Berresford/Rockett
Expert Verified
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.

Course Hero member to access this document

Course Hero member to access this document

End of preview. Want to read all 65 pages?

Course Hero member to access this document

Term
Summer
Professor
Meisel
Tags
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
The document you are viewing contains questions related to this textbook.
Chapter 5 / Exercise 80
Applied Calculus
Berresford/Rockett
Expert Verified