Odd Answers: Chapter Eight Contemporary Calculus 1 { ( 3+2 } = lim { 1. { 2. arctan(a) 2. arctan(3) } = 2( π 2 ) 2. arctan(3)

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1 Odd Answers: Chapter Eight Contemporary Calculus PROBLEM ANSWERS Chapter Eight Section 8.. lim { A 0 } lim { ( A ) ( 00 ) } lim { 00 A } 00.. lim {. arctan() A } lim {. arctan(a). arctan() } ( π ). arctan() Use u ln(). lim { 5. ln( ln() ) A e } lim { 5. ln( ln(a) ) 5. ln( ln(e) ) } lim { 5. ln( ln(a) ) 0 }. DIVERGES.. lim { ln( ) A } lim { ln(a ) ln( ) } lim { ln(a ) 0 }. DIVERGES. 9. lim { A ( ) } lim { (A ) ( ) } lim { (A ) + }.. lim { + A } lim { A+ + } lim { A+ + 5 } 5.. lim A!0 + { 4 A } lim A!0 + { 4 A } lim { 4 6 /4 } lim A!0 + A { 4 (6) /4 4 A/4 } lim { 4 A/4 }. A!0 + A!0 +. lim A! " { arcsin( ) A 0 } lim A! " { arcsin( A ) arcsin( 0 ) } π 0 π. 9. lim { cos() A } lim { cos(a) cos( ) } lim 0.46 cos(a) DNE. DIVERGES.. lim / { ln cos() A 0 } DIVERGES. lim { ln cos(a) ln cos(0) } lim / 0 ln cos(a) DNE. /

2 Odd Answers: Chapter Eight Contemporary Calculus. (a) A R: R A R: R (b) lim R R P d R P( R ) R P( R R R ) RP. R R P d R P( R ) R P( R R R ) RP. R P # d lim R +A R R P( R+A ) R lim R P( R+A R ) R P( R ) RP. 5. ( + ) d d which converges by the p test. + d < Therefore, ( + ) d converges.. Fof > 0, ln() < so + ln() < and Then d + ln() >. d which diverges by the p test. Therefore, + ln() d diverges. 9. cos() so 0 + cos() and 0 + cos(). Then + cos() d d which converges by the p test. Therefore, + cos() d converges.. V 0 π ( 4 d which converges by the p test. + ) d < π 0 Therefore, V 0 π ( + ) d converges.. (a) A d < A k k (b) A d > A k k Section 8.. u +, du d, du 6 d: u du u + C ( + ) + C.. u, du d, du 6 d: u du. u

3 Odd Answers: Chapter Eight Contemporary Calculus 5. u + : 6 ln( + ) + C.. u + : cos( + ) + C. 9. u e +, du e d: sec ( u ) du tan( u ) tan(e + ) 0 tan(e + ) tan(+).9. u ln(), du d: u du u + C ( ln() ) + C.. u sin(), du cos() d: e u du e u + C e sin() + C. 5. u, du d: u du 5 arctan( u ) 5 arctan( ) 5 arctan(9) 5 arctan(). u, du d: cos(u) du sin(u) sin( ) ( sin( ) ( sin() ) u 5 + sin (), du. sin(). cos() d: u du ln u + C ln 5 + sin () + C.. 5 ln C.. ln 5 + ln 48 ln 8 ln 48 8 ln( 6 ) arctan( + ) 0 arctan( ) arctan(.5 ) u e, du e d: + u du arctan( u ) + C arctan( e ) + C. 9. u + ln(), du d: u du ln u ln + ln() e ln.08. u, du d: u du u/ ( ) / 0 (0) / () /. u + sin(), du cos() d: u du 4 u4 + C 4 ( + sin() ) 4 + C. 5. u ln(), du d: u du u/ ( ln() ) / e ( ln(e) ) / ( ln() ) /.

4 Odd Answers: Chapter Eight Contemporary Calculus 4. u 5 + tan(), du sec () d: u du ln u + C ln 5 + tan() + C. 9. u 5, du d: tan(u) du ln sec(u) + C ln sec( 5) + C. 4. u 5, du 5 d: 5 eu du 5 eu 5 e5 0 5 e5 5 e (+) + d. arctan( + ) + C. 45. (+5) + 4 d. arctan( +5 ) + C. ( ) + 49 d. arctan( ) + C. 49. ln arctan( + ) + C. 5.. ln arctan( ) + C d + ( ) + 9 d. ln arctan( ) + C. Section 8... ln() d u ln(). Then dv d, du d, and v 6. v du ln() d 6 ln() 6 d 6 ln() +C.. 4 ln() d dv 4 d. Then u ln(), du d, and v 5 5. v du ln() d 5 5. ln() 5 4 d 5.. arctan() d dv d. Then u arctan(), du v du arctan().. + d, and v. + d. arctan() 5 5. ln() C. + d. arctan() { arctan() } + C. arctan(). +. arctan() } + C.. 0 e d 0. e d. Put u. Then dv e d, du d, and v e. v du. e e d. e 9 e 0 {. e () 9 e () } { 0. e (0) 9 e (0) } e.

5 Odd Answers: Chapter Eight Contemporary Calculus sec(). tan() d. Put u. Then dv sec(). tan() d, du d, and v sec(). v du. sec() sec() d. sec() ln sec() + tan() + C. π/. π/. cos() d Put u. Then dv cos() d, du d, and v sin(). v du. sin() sin(). d. sin() + 9 cos() π/ {.π. sin(.π ) + 9 cos(.π } {.π. sin(.π ) + 9 cos(.π ) }.88. π/.. cos( ) d. Use u substitution! Put u. Then du 6 d and du d. cos( u ). du. sin( u ) + C. sin( ) + C. 5. ln( + 5) d. Put u ln( + 5). Then dv d, du + 5 d, and v. v du ln( + 5) d. ln( + 5) d. ln( + 5) { 5. ln + 5 } {. ln() + 5. ln } {. ln() + 5. ln }. ln(). ln() 4.8. e. ( ln() ) d. Put u ( ln() ). Then dv d, du. ln(). d, and v. v du ( ln() ).. ln(). d. ( ln() ). ln() d. ( ln() ) {. ln() } e { e( ln(e) ) e. ln(e) + e } { ( ln() ). ln() + } e arcsin() d. Put u arcsin(). Then dv d, du v du arcsin()... arcsin() d, and v. d (use u sub with u ) d. arcsin() u ( ) du. arcsin() + u + C. arcsin() + + C.

6 Odd Answers: Chapter Eight Contemporary Calculus 6.. arctan() d. Put u arctan(). Then dv d, du v du arctan() d, and v. + 9 d. arctan() / + 9 d. arctan() { 9. arctan() } + C. arctan() arctan() } + C.. ln() d. Use u substitution! Put u ln(). Then du d. u du u ( ln() ) ( ln() ) ( ln() ) ( ln() ) (a) sin () d { S. C + S d } { S. C C } + K { sin (). cos(). cos() } + K. (b) sin 4 () d 4 { S. C + S d } 4 { S4. C + [ ( SC + ) ] } + K 4 { S. C SC + } + K 4 { sin (). cos() sin(). cos() + } + K (c) on your own.. (a) sec () d { sec()tan() + sec() d } { sec(). tan() + ln sec() + tan() } + K (b) and (c) on your own. 9. cos ( + ) d. First do a substitution: u +. Then du d and d du. cos ( u ) du { ( C. S + C du ) } 6 { C. S + S } + K 6 { cos (u). sin(u) + sin(u) } + K 6 { cos ( + ). sin( + ) + sin( + ) } + K.. ( + 5) 9 d. (a) By parts: put u. Then dv ( + 5) 9 d, du d, and v 40 ( + 5) 0. v du. 40 ( + 5) ( + 5) ( + 5) + C. 40 ( + 5) 0 d (b) Substitution: put u + 5. Then du d and d du. Also, ( u 5 ). ( u 5 ). u 9 du 4 u 0 5u 9 du 4 { u 5 0 u0 } + C 84 ( + 5) 5 80 ( + 5) 0 + C. The answers (antiderivatives) in parts (a) and (b) look different, but you can check that the derivative of each answer is. ( + 5) 9.

7 Odd Answers: Chapter Eight Contemporary Calculus. (a) Make an informed prediction. (b) 0 sin() d cos() 0 ( cos()) ( cos(0)) cos(0) cos() sin() d. cos() + sin() 0 {. cos() + sin()} { 0. cos(0) + sin(0)} sin() cos() (a) Make an informed prediction. (b) See problem : V ais V y ais. On your own. y0 e π. y. e y dy π π( ln() ) d π(e ).5. y. e y dy (use integration by parts with u y, dv e y dy : see Eample ) y0 π( y. e y e y ) 0 π(. e e ) π(0. e 0 e 0 ) π(0) π( ) π arctan() d. Put u arctan(). Then dv d, du + d, and v. v du arctan() On your own.. arctan() d + + d (dividing by + ). arctan() { ln( + ) } + C. arctan() ln( + ) } + C. Section 8.4. A + B A + + B Divide first: ( + 5) ( + 5) ( + 5)( ) A + B +5 + C A( + 4 5) + B( ) + C( + 5) ( + 5)( ). Solving : A + B + C 6 : 4A B + 5C 9 k: 5A 5 we get A, B, and C 0 so

8 Odd Answers: Chapter Eight Contemporary Calculus ( + 5)( ) ( + ) A + B + C + A( + ) + (B + C) ( + ). Solving : A + B 8 : C k: A we get A, B 5, and C so 8 + ( + ) ( + ) A + B + C + A()( + ) + B( + ) +C( ) ( + ). Solving : A + C : A + B k: B 6 we get A 0, B, and C so ( + ) ( + )( 5) d d ln ln 5 + C d + + d ln + + ln ln + 5 ln( 4 6 ) ln( ) () d. ln + 5. ln + + C () d. ln ln + C d + 5 d + 5. ln + C d d +. ln + ln C.. d. Use u substitution with u. Then du so u du ln u + C ln + C. A partial fraction decomposition also works but takes longer.

9 Odd Answers: Chapter Eight Contemporary Calculus d + 6 d + 6ln + C d d. ln ln + + C d + + d. ln +. arctan( ) + C.. (a) + + d (+) + d arctan( + ) + C. (b) + + d ( + ) d ( + ) + C + + C. (c) d ( + ) d / + / + d ln. Prob. : f() () + 5. ( + ). Then Prob. : g() f '(). () 5. ( + ) and f ''() 4. () + 0. ( + ) ( + ) + 4. ( ). Then ln + + C. g '(). ( + ) 4. ( ) and g ''() 6. ( + ) + 8. ( ) Prob. 5: f() + 5 Prob. 6: On your own. + 5 so f '() 5 and f ''() 0. d. (a) Solve dt (00 ). Separate the variables: (00 ) d dt. Use partial fractions: (00 ) A + B (solving A+B 0 and 00A ) so { } d dt and, integrating, d dt. Then 0.0. ln 0.0. ln 00 t + C so ln Using the initial condition (0) 50: ln 00 00t + K. ( K 00C is a constant ) ln( ) 00. (0) + K so K ln( ). Finally, ln t + ln() so 00 e00t. e ln(). e 00t and 00. e 00t. e 00t. Graph this on your own.

10 Odd Answers: Chapter Eight Contemporary Calculus 0 (b) In this part it is easier to use the form Put 0 and solve Put 0 and solve Put 00. Then e 00t e 00t : 6. e 00t so t e 00t :. e 00t so t ln( ) ln( ) 0.0. is undefined (division by 0) so (t) is never equal to 00. (c) limit (as t becomes arbitrarily large) of 00. e 00t. e 00t is (d) The population (t) is decling to the "carrying capacity" M 00 of the enviroment. 9. (a) Solve d dt ( )(5 ) by separating the variables and using partial fractions to rewrite the fraction. (b) 5 5 ekt and (t). e kt 5. e kt. (As t gets big, approaches 5.) Solve d dt (6 )(6 ) (6 ) by separating the variables: (6 ) d dt and integrating. 6 t + C (C 6 ) so 6t + t + 6 6t + 6 6t +. (As t gets big, approaches 6.) Section sin(θ) (a) 9 9 9sin (θ) 9( sin (θ) ) 9cos (θ) so (b) d cos(θ) dθ 9 cos(θ)... sec(θ) (a) 9 9sec (θ) 9 9( sec (θ) ) 9tan (θ) so (b) d sec(θ)tan(θ) dθ 9 tan(θ). 5. tan(θ) (a) + + tan (θ) ( + tan (θ) ) sec (θ) so (b) d sec (θ) dθ + sec(θ)... sin(θ) (a) θ arcsin( / ) (b) & (c) f(θ) cos(θ). tan(θ) cos( arcsin(/) ). tan( arcsin(/) ) sec(θ) (a) θ arcsec( / ) (b) & (c) f(θ) + sin (θ) + sin ( arcsec(/) ) + ( 9 )

11 Odd Answers: Chapter Eight Contemporary Calculus. 5. tan(θ) (a) θ arctan( /5 ) (b) & (c) f(θ) cos (θ) + cot(θ) ( ) + ( 5 ) Same as Practice. Sorry. 5.. tan(θ). d sec (θ) dθ tan (θ) ( tan (θ) + ) 49 sec (θ) d 49 sec (θ) sec (θ) dθ sec(θ) dθ ln sec(θ) + tan(θ) + C ln sec( arctan(/) ) + tan( arctan(/) ) + C ln C.. 6. sin(θ). d 6 cos(θ) dθ. 6 6 cos (θ). 6 d 6 cos (θ) 6 cos(θ) dθ 6 cos (θ) dθ (use Table #4) 6 { θ + sin(θ). cos(θ) } + C 6 { arcsin( /6 ) + ( 6 ). ( 6 6 ) } + C tan(θ). d 6. sec (θ) dθ sec (θ). 6 + d sec (θ) dθ sec(θ) dθ (use Table #) ln sec(θ) + tan(θ) + C ln C or ln K.. Similar to 9:. tan(θ) d ln C.. 5. sin(θ). 5 cos( arcsin(/5) ) + C C 5 + C. 5.. tan(θ). ln cos( arctan(/) ) + C ln + C (now some algebra) 49 + ln + ln C ln K. (A u substitution with u 49 + is much easier.)

12 Odd Answers: Chapter Eight Contemporary Calculus.. sec(θ). 9 cos(θ) sin (θ) dθ (put u sin(θ) ) 9 sin(θ) + C csc(θ) + C C sec(θ). θ + C arcsec( 5 ) + C.. 5. sin(θ). 5 ln sec(θ) + tan(θ) + C 5 ln C (after lots of algebra) 0 ln C.. Similar to 9. ln a + a + a + C ln a K. 5. a. tan(θ). a a + + C.. + u. Then u. tan(θ). ln sec(θ) + tan(θ) + C ln u u + C ln u +9 + u ln() + C ln u +9 + u + K ln (+) (+) + K. 9. arctan( +5 ) + C 4. ln (+) C. Section 8.6. sin ( ) d sin( 6 ) + C sin( ). cos( ) 6 + C.. Put u sin( e ). sin ( e ) + C. ( If you put w cos( e ) then cos ( e ) + C.) π. 4 4 sin ( ) + C or 4 cos ( ) + C. Put u cos( ). 8 cos4 ( ) + C.

13 Odd Answers: Chapter Eight Contemporary Calculus. sin ( ) cos ( ) d ( cos(6) ) ( + cos(6) ) d 4 cos ( 6 ) d 4 4 cos ( 6 ) d 4 4 { + sin( ) 4 } + C tan ( 5 ) + C.. 9 sec ( ) + C. π 9. m n. 0 sin( m ) sin( n ) d { sin( (m n)π ) m n { sin( (m n) ) m n sin( (m+n)π ) m + n } sin( (m+n) ) m + n } π 0 sin( (m n)0 ) { m n sin( (m+n)0 ) m + n } { 0 0 } { 0 0 } 0. π. 0 π sin( m ) sin( m ) d 0 { π sin ( m ) d sin( mπ ) cos( mπ ) m } { sin( m ) cos( m ) m π 0 sin( 0 ) cos( 0 ) m } π. 0. On your own.