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# C4 integration by substitution question watch

1. From Solomon C integrate 12piSin^2 2t Costt dt limites 0 and pi/2, using substitution u=sint.

I got 4 pi, but the markscheme got 32/5 pi, but i dont understand how they got there
2. *Bumb*
3. You are missing something really important in the question. You can substitute sin(t) but not sin(2t) with u.
4. (Original post by Flo199)
You are missing something really important in the question. You can substitute sin(t) but not sin(2t) with u.
Thank you, stupid mistake
5. (Original post by kf141998)
From Solomon C integrate 12piSin^2 2t Costt dt limites 0 and pi/2, using substitution u=sint.

I got 4 pi, but the markscheme got 32/5 pi, but i dont understand how they got there
New Limits:
t=0, u=0
t=Pi/2 , u=1

u=sintdu/dt=cost
Therefore 1/cost du =dt

12pisin^2(2t)cost dt =
(12pisin^2(2t)cost)/cost=
12pisin^2(2t)
12pi (sin2t)(sin2t)
12pi (2sintcost)(2sintcost)
48pi sin^2(t) cos^2 (t)
48pi sin^2 (t) (1-sin^2 (t))
48pi u^2 (1-u^2)=
48pi u^2 -48pi u^5

Integrate..... = 16pi u^3 - 48/5 u^6

Sub in u values
(u=1) - (u=0)= Just u=1=
80/5 pi -48/5 pi=
32/5 pi
Hope this helped
6. (Original post by 1ncompl)
New Limits:
t=0, u=0
t=Pi/2 , u=1

u=sintdu/dt=cost
Therefore 1/cost du =dt

12pisin^2(2t)cost dt =
(12pisin^2(2t)cost)/cost=
12pisin^2(2t)
12pi (sin2t)(sin2t)
12pi (2sintcost)(2sintcost)
48pi sin^2(t) cos^2 (t)
48pi sin^2 (t) (1-sin^2 (t))
48pi u^2 (1-u^2)=
48pi u^2 -48pi u^5

Integrate..... = 16pi u^3 - 48/5 u^6

Sub in u values
(u=1) - (u=0)= Just u=1=
80/5 pi -48/5 pi=
32/5 pi
Hope this helped
Thank you

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