1. initially a particle moves with a velocity of 24m/s along a straight line and its acceleration after t seconds is (8-15t^2) m/s^2. it stops instantaneously and reverses its direction of motion only once.
a)show that this takes place when t=2
b)find the distance travelled in the first 2 seconds
c)find the distance travelled in the first 4 second.
2.find the following indefinite integrals
∫(sin2x)^2 cosx (sinx)^3 dx
3. ∫(sec θ) ^ 6 dθ
4.∫(x^2-1)^1/2 dx
put x =secθ
amaths
2007-11-11 2:38 am
回答 (1)
2007-11-11 7:10 am
✔ 最佳答案
(1) Suppose that x = 0 at t = 0, we have:v = ∫(8 - 15t²)dt
= 8t - 5t³ + C where C is an arbitrary constant.
Sub t = 0, v = 24, then C = 24
So,
v = 8t - 5t³ + 24
When v = 0,
8t - 5t³ + 24 = 0
(-t + 2)(5t² + 10t + 12) = 0
t = 2
(2) ∫sin² 2x cos x sin³ x dx
= 4∫(sin x)^5 cos³ x dx
= 4∫(sin x)^5 cos² x d(sin x)
= 4∫(sin x)^5 (1 - sin² x) d(sin x)
= 4∫(sin x)^5 - (sin x)^7 d(sin x)
= (2/3) (sin x)^6 - (1/2) (sin x)^8 + C
(3) ∫(sec θ) ^ 6 dθ
= ∫(sec² θ)³ dθ
= ∫(sec² θ)² d(tan θ)
= ∫(1 + tan² θ)² d(tan θ)
= ∫[1 + 2 tan² θ + (tan θ)^4] d(tan θ)
= tan θ + (2/3) tan³ θ + (1/5) (tan θ)^5 + C
(4).∫√(x² - 1) dx
Sub x = sec θ, dx = sec θ tan θ dθ
∫√(x² - 1) dx = ∫√(sec²θ - 1) sec θ tan θ dθ
= ∫√(tan²θ) sec θ tan θ dθ
= ∫tan²θ sec θ dθ
= (1/2){ln [cos (x/2) - sin (x/2)] - ln [cos (x/2) + sin (x/2)] + sec θ tan θ}
= (1/2) {x √(x² - 1) - ln [x + √(x² - 1)]}
參考: Myself~
收錄日期: 2021-04-13 14:24:35
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