Math 21B - Homework Set 8

Transcription

1 Math B - Homework Set 8 Section 8.:. t cos t dt Let u t, du t dt and v sin t, dv cos t dt Let u t, du dt and v cos t, dv sin t dt t cos t dt u v v du t sin t t sin t dt [ t sin t u v ] v du [ ] t sin t t cos t + cos t dt t sin t + t cos t sin t + C. e x ln x Let u ln x, du x and v 4 x4, dv x. e e x ln x uv e v du e e 4 x4 ln x 4 x4 x e 4 x4 ln x e x 4 e 4 x4 ln x e 6 x4 e 4 e e t e 4t dt

2 Let u t, du t dt and v 4 e4t, dv e 4t dt. Let u t, du dt and v 6 e4t, dv 4 e4t dt. t e 4t dt u v v du 4 t e 4t 4 te4t dt [ ] 4 t e 4t 8 te4t 8 e4t dt 4 t e 4t 8 te4t + e4t + C 4. e θ sin θ dθ Let u sin θ, du cos θ dθ and v e θ, dv e θ dθ. Let u cos θ, du sin θ dθ and v e θ, dv e θ dθ. e θ sin θ dθ u v v du e θ sin θ e θ cos θ dθ [ e θ sin θ u v ] v du [ ] e θ sin θ e θ cos θ + e θ sin θ dθ e θ sin θ e θ cos θ e θ sin θ dθ + C We can use this to solve for e θ sin θ dθ. e θ sin θ dθ e θ sin θ e θ cos θ e θ sin θ dθ + C e θ sin θ dθ e θ sin θ e θ cos θ + C e θ sin θ dθ eθ sin θ e θ cos θ + C

3 5. π/ x tan x Let u x, du and dv tan x, v tan x sec x tan x x. π/ x tan x uv π/ π/ v du x tan x x π/ x tan x x π/ π π 9 π π 8 ln π/ tan x x ln sec x π/ x ln π 8 6. zln z dz Let w ln z, dw z dz. Then we get that z ew. z zln z ln z dz dz w e w dw z Let u w, du w dw and v ew, dv e w dw. Let u w, du dw and v ew, dv e w dw. w e w dw u v v du w e w we w dw [ ] w e w u v v du [ ] w e w wew ew dw w e w wew + 4 ew + C We now need to go ack and sustitute z ack into the solution. zln z dz z ln z z ln z + 4 z + C

4 7. Using integration y parts, with u sin x and dv so that v x sin u, we get sin x x sin x sinu du x sin x + cos u + C x sin x + cos sin x + C x sin x + x + C Section 8.4: x + 4. x + 5x 6 x + 4 x + 6x x + 4 x + 6x A x B x x + 4 A + Bx + A + 6B { A + B 4 A + 6B Therefore, going ack to our integral we get: x + 4 x + 5x 6 7 A 7, B 5 7 x x 7 ln x ln x + C 7. x + x 8x x + xx x + A x + x + xx x + B x + C x + Therefore, going ack to our integral we get: x + x 8x 4 x x + A + B + Cx + 4C 4Bx 4A A + B + C 4C 4B 4A A 4, B 6, C 5 6 x x + 5 ln x + ln x + ln x + + C 6 6 4

5 . x x + x + x + x + x + x + y long division x + x + x + A x + + B x + x + Ax + A + B { A A + B A, Therefore, going ack to our integral we get: x x + x + x B x + x + x + x + ln x ln + ln x + 4. x + x + x + x + A x + + Bx + C x + A + Bx + B + Cx + C + A A + B B + C C + A A, B, C Therefore, going ack to our integral we get: x + x + x + + x + + x + x x + x x + ln x + + tan x ln x + ln + π 4 ln 4 ln + π 8 5

6 5. t + t + 4 t + t t + t + 4 tt + dt A t + Bx + C t + t + t + 4 A + Bt + Ct + A A + B C 4 A A 4, B, C Therefore, going ack to our integral we get: t + t t + t t + t t dt + 4 t + t + t t dt + 4 ln t + tan t ln t + 4 ln + π π ln 4 4 ln ln + π ln π 4 + ln ln ln + π 6. x 4 x x + + x x A x + B x + A + Bx + A B { A + B A B A, B Therefore, going ack to our integral we get: x 4 x x + + x x + x + x + ln x ln x + + C x + x + ln x x + + C 6

7 Section 8.:. π/ sin 5 x π/ π/ sin x cos x sin x cos x sin x + cos 4 x sin x cos x + cos x π/ 5 cos5 x cos 4 πx 8 8 cos4πx + cos 4πx + cos4πx + 4 cos8πx + + cos4πx + [cos8πx + 4 cos4πx + ] 8π sin8πx + sin4πx + x π. π cos x π π π sin x x sin x sin x π cos 4 x sin, x π 7

8 4. π/4 π/4 sec x π/4 π/4 π/4 π/4 π/4 tan x tan x tan x ln sec x π/4 ln ln 5. π/4 sec 4 x π/4 π/4 sec xtan x + sec x tan x + sec x π/4 tan x + tan x π/4 π/4 π/4 6 tan 4 x π/4 π/4 π/4 π/4 π/4 tan xsec tan x sec x tan x tan x sec x sec x + π/4 6 tan x tan x + x π/4 [ 6 + π 4 + π ] 4 π 8 8

9 7. π sinx cosx π sin x + sin5x see p465 cos x cos5x π Section 8.:. dy 9 + y! " # Let y tan θ, dy sec θ dθ for π < θ < π. dy 9 + y sec θ tan θ dθ sec θ dθ ln sec θ + tan θ + C ln sec tan y + tan tan y + C 9 + y ln + y + C or ln 9 + y + y + C. 8 + x 9

10 Let x tan θ, sec θ dθ for π < θ < π. tan 8 + x sec θ tan θ dθ tan π/4 π/4 θ 4 π 6 π/4 sec θ 8 sec θ dθ 4 dθ OR 8 + x 8 + x 8 tan x π 6. / 9 x Let x sin θ, cos θ dθ for π θ π. / π/6 cos θ 9 x 9 9 sin θ dθ π/6 π/6 θ π/6 π 6 cos θ cos θ dθ dθ x 9 for x > 5

11 Let x 5 sec θ, 5 sec θ tan θ dθ for θ < π 5 5x 9 5 /5 sec θ tan θ 5 9/5 sec θ 9 dθ sec θ tan θ sec θ dθ sec θ tan θ tan θ dθ sec θ dθ ln sec θ + tan θ + C ln 5x 5x sec sec + tan sec + C 5 ln x + 5x 9 5x + C or ln + 5x 9 + C 5. x for x > x

12 Let x sec θ, sec θ tan θ dθ for θ < π. x x sec θ tan θ sec θ sec θ dθ sec θ tan θ sec θ tan θ dθ sec θ tan θ sec θ tan θ dθ tan θ, θ < π cos θ dθ sin θ + C sin sec x + C x + C x 6. x x +

13 Let x tan θ, sec θ dθ for π < θ < π x x + sec θ tan θ tan θ + dθ sec θ tan θ sec θ dθ sec θ sec tan θ sec θ dθ θ >, π < θ < π sec θ tan θ dθ cos θ sin θ dθ sin θ + C sin tan x + C x + + C x 7. ln 4 e t e t + 9 dt Firstly, let x e t, e t dt. ln 4 e t 4 e t + 9 dt x + 9

14 Now, let x tan θ, sec θ dθ for tan < θ < tan 4. 4 tan 4/ x + 9 sec θ 9 tan θ + 9 dθ tan / tan 4/ tan / tan 4/ tan / sec θ sec θ dθ sec θ dθ 4 sec θ >, tan θ tan ln sec θ + tan θ tan 4/ tan / ln sec tan 4/ + tan tan 4/ ln sec tan / + tan tan / 5 ln + 4 ln + ln ln + + ln ln ln + 8. x x Let x sec θ, sec θ tan θ dθ for < θ < π x x sec θ tan θ sec θ sec θ dθ sec θ tan θ sec θ tan θ dθ dθ 9. x θ + C sec x + C 4

15 Let x sin θ, cos θ dθ for π < θ < π cos θ x sin θ dθ cos θ cos cos θ dθ θ >, π < θ < π dθ θ + C sin x + C Section 8.7:. lim x + x lim x + lim +. lim x x lim sin x lim sin sin π. x x lim + lim + x x sec x + lim c + lim c sec x lim + [ sec sec ] + lim c π + π π c x x [ sec c sec ] π 5

16 4. e θ sin θ dθ We will first consider the indefinite integral e θ sin θ dθ. Let u sin θ, du cos θ dθ and v e θ, dv e θ dθ. Let u cos θ, du sin θ dθ and v e θ, dv e θ dθ. e θ sin θ dθ u v v du e θ sin θ + e θ sin θ dθ e θ sin θ + u v v du e θ sin θ e θ cos θ e θ sin θ dθ Let I e θ sin θ dθ. Solving for I we get: I e θ sin θ e θ cos θ I I e θ sin θ e θ cos θ I e θ sin θ e θ cos θ So we have that e θ sin θ dθ e θ sin θ e θ cos θ + C. Now let s consider the improper integral e θ sin θ dθ lim lim e θ sin θ dθ. e θ sin θ dθ e θ sin θ e θ cos θ lim [ e sin e cos ] 5. xe x lim lim lim xe x + lim c e x + lim c + e + lim c c e x c xe x e c + 6

17 6. x ln x lim + x ln x [ ] lim + x ln x x lim + x ln x 4 x [ lim + 4 ln ] 4 [ ] ln 4 [ [ 4 lim + lim + lim + ] ] Integration y Parts l Hopital s Rule lim x lim 4 + lim x c + + lim x c + lim x + lim lim + 6 c + c 4 c x x x + lim c + 4 c 4 c Section.: 7

18 . x cos t, y + sin t, t π, x-axis Area π π π π π + sin t sin t + cos t dt π + sin t sin t + cos t dt + sin t dt π t cos t π π[4π ] 8π. x lnsec t + tan t sin t, y cos t, t π ; x-axis Area π/ π π π π π π π/ sec t tan t + sec t π cos t cos t + sin t sec t + tan t dt sec ttan t + sec t cos t cos t + sin t dt sec t + tan t cos t sec t cos t + sin t dt π/ π/ π/ π/ π/ cos t sec t + cos t + sin t dt cos t sec t dt cos t tan t dt sin t dt π cos t π/ π π 8