We first write the integrand into partial fractions and then integrate. By EXAMPLE 27 we have the identity

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1 Solutios 8 Complete solutios to Miscellaeous Eercise 8. We ave v v v m KE m vdv m v. We ave l l EA EA EAl W d. We ave W k d k k. Multiplyig bot sides by μ gives ( ) T dt T T μθ l T l ( T) l ( T) l T T T by (8.) by (5.) T Takig epoetials of bot sides: μθ T l e ( Because e ) T Hece T Te μθ ( Multiplyig by T ) 5. pq W ( ) d 6EI pq pq pq 6EI + 5 6EI + 5 EI 6. Takig out te gives J r D r dr Di D Di D Di D D i D D i We first write te itegrad ito partial fractios ad te itegrate. By EXAMPE 7 we ave te idetity Hece l ( A) l ( B) l A (5.) B

2 Solutios 8 d d d Itegratig gives: 9. + l l C + l ( )( ) C l C l( ) C Δ.8T + by (5.) ( ) T kj kg T T Δ R a+ bt + ct + dt + et dt b c d e R at + T + T + T + T 5 5 b c d e R a T T + T T + T T + T T + T T 5. We ca use te trigoometric idetity for cos cos si (): cos cos si d d cos si cos si T T 5 5 ( ) ( ) ( ) ( ) ( ) d si d cos si d cos si cos si d cos. r q q r dr V dr z ε r ε z r + cosec d sec d cot ta C q q ε ε r l ( r) l ( r) l ( z) z by (8.) q q z l ( z) l ( r) l ε ε r by (5.) (5.) (5.) (8.) l( A)+ l( B) l( AB) l( A) l( B) l AB dr r l r

3 Solutios 8. From PV C we ave P C. Substitutig tis ito te itegral gives V V dv W CV V V Cl V V Cl ( V) l ( V) V V Cl PV l V C V. We ave P C CV.. Substitutig tis ito W gives: V. V W CV. V dv C V. dv. V by (8.) V V C V.+ C V.. +. V V V C V.. V. C. V.. V [ ] 99 (i) Area of rectagles ( 99 ) (ii) d.5 (iii) Differece By cosiderig more rectagles. 6 7 r k r 5. P k ( + r ) d k (i) P wsi d w si d (8.9) by (8.9) by it cos w w w cos( ) cos si( k + m) d cos( k + m) k

4 Solutios 8 (ii) R w si d. How do we itegrate tis? Use itegratio by parts formula(8.5) u v si cos u v si d cos R w cos + () cos d w w cos + si w + 7. Itegratig gives: v v trc l ( w v) v w trc l ( w v) l ( w) w v trc l by (5.) w Takig epoetial gives: w v trc e w trc w v we 8. We ave: trc ( ) v w e t ( ) ( 8 ) i 6e + e d 6 ( ) ( 8 ) e e 8 by (8.) t ( ) ( 8 ).e.5e ( ) t ( ) t.e.5e.e.5e ( ) t ( 8 ) i.e.5e 8 t.75 t (5.) (8.) l( A) l( B) l( AB) e kt + m dt e kt +m k d (8.5) u v d uv u v

5 Solutios (i) ( t) ( t) ( ) t cos i si ( t) dt 5 5 cos + cos ( t) cos( t ) 5.5 (ii) Usig te formula give w ( 5 ) cos( t) ( 5 ) cos( t) ( t) cos w. μin b μin b μin μin b Φ dr l ( r) l ( b) l ( a) l a a r a by (8.) Hece d μ in l b μ N N N l b μ N l b di a a a. We ave d ta ta ta + by (8.6). m I ( y) y dy m ( y y ) dy t my y m m m 6. We ave: y D cos dy Dsi d d y D cos d (8.) (8.6) dr r l r d ta + a a a

6 Solutios 8 6 EI V D cos d EI D cos d ( ) cos d + cos d + cos d Substitutig tis ito (). Need to evaluate by (.68) ( ) si ( ) si EID EID EID V 6 6 ηi p λ si * ( θ) si ( θ ) si ( θ) ( cos ( θ) ) d θ by (.6) et u cos( θ) si ( θ) si ( θ ) Te limits are u cos( ) ad u cos Substitutig tese values si( θ ) cos ( () θ ) si( θ) u Substitutig tis ito (*) gives 5. (.6) y e si( θ ) ( u ) u u ηi ηi. p λ λ y e e d e l l ( e) l si ( θ)+ cos ( θ) cos A + cos ( A ) [ ] (.68) 6. MAPE commads are sow o te web site.

7 Solutios Use (8.) wit a ad u : d + si + si [ ] si () ( ) 8. We ave u acos( θ). Differetiatig asi ( θ ) gives asi ( θ) ( 6.6) Substitutig u acos( θ) ad asi( θ) ito te itegral gives: asi ( θ) (*) u a a cos θ a Te deomiator, a cos ( θ) a, simplifies: a θ a a θ a θ a si ( θ ) ( θ ) cos cos si si Substitutig tis ito (*) gives a si ( θ ) d d C ** θ θ θ + u a a si ( θ ) Wat is θ? u θ cos We kow a cos( θ ) u so cos() θ u wic gives a a. u Replacig θ wit cos a ito (**) displays our result: u cos C + u a a 9. We ave u ata( θ ) asec () θ gives asec ( θ) Substitutig u ata θ ( 6.7) ad sec θ asec ( θ) a u a a ta ( θ ) we ave (*) Te deomiator simplifies to: a a ta ( θ ) a ( ta ( θ) ) a sec ( θ) sec ( θ ) (6.6) (8.) ( θ ) si ( θ ) cos u a u a u a u + si

8 Solutios 8 8 Substitutig tis ito (*) gives asec ( θ) θ a u a sec θ a a We ca obtai θ from te substitutio ata θ u Tus + u u ta ( θ) wic gives θ ta a a u ta + C a u a a. How do we evaluate ωt cos( ωt)d( ωt)? Use itegratio by parts u ωt v cos ωt Usig (8.5) gives ( ω ) si v cos( ωt) d( ωt) d t ωt ( ωt) si ( ωt) si ωtcos( ωt) d( ωt) ωt d ωt ωt ( ) because si ( ω ) si si ωt d t ( ω ) C ( ωt) si t d ( ω t ) ωt cos( ωt) cos ( ) cos [ ] ωt (6.7) (8.5) ( θ ) sec ( θ ) ta uvdt u vdt vdt dt