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Thread starter 14 years ago
#1

1. Find ∫ (1-cost)^4 dt.

2. Show that I_n = ∫ x^(n/2) (1-x)^(1/2) dx can be written as: I_n = n/(n+3) I_(n-2).
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14 years ago
#2
1. Expand this to give: (note: i am leaving out the 't' in cost for simplicity)
cos^4 - 4cos^3 + 6cos^2 - 4cos + 1
Integrate each component on its own..
cos^4 = (cos^2 t)^2 = ((1+cos2t)/2)^2 = 1/4(1+cos2t)^2 = 1/4(1+2cos2t+(cos^2)2t) = 1/4(1+2cos2t+(1+cos4t/2))
Integrating, = 1/4(3t/2 + sin2t + sin4t/4)
Now, 4cos^3 = 4(cos^2)(cos t)=4(1-sin^2)(cos)=4(cos-cos.sin^2)
Integrating, = 4 (sin t - (sin^3)t)/3)
Now, 6cos^2 = 6((1+cos2t)/2)=3(1+cos 2t)
Integrating, = 3(t+(sin2t)/2)
Now you're left with 4cos, which is easily integrated = 4sint, and 1, which is t.
1/4(3t/2 + sin2t + (sin4t)/4) - 4(sin t - (sin^3)t)/3) + 3(t+(sin2t)/2) -4sint + t

I might have gone wrong somewhere, but you get the idea. You also have the unenviable task of simplifying that.
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14 years ago
#3
Alternatively for (1), use

(1 - cos(t))^4 = (2 sin^2(t/2))^4 = 16 sin^8(t/2)

and the reduction formula for sin^(2n).

--

I think the integrals in (2) should have limits 0 and 1.

I(n)
= (int from 0 to 1) x^(n/2) (1 - x)^(1/2) dx
= [ x^(n/2) * -(2/3)(1 - x)^(3/2) ] (from 0 to 1)
= + (int) (n/2)x^(n/2 - 1) * (2/3)(1 - x)^(3/2) dx
= 0 + (n/3) (int) x^(n/2 - 1) (1 - x) (1 - x)^(1/2) dx
= (n/3) [ (int) x^(n/2 - 1) (1 - x)^(1/2) dx - (int) x^(n/2) (1 - x)^(1/2) dx ]
= (n/3) [I(n - 2) - I(n)]

(1 + n/3)I(n) = (n/3)I(n - 2)

I(n) = [n/(n + 3)] I(n - 2)
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Thread starter 14 years ago
#4
Thanks guys.
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Thread starter 14 years ago
#5
quick question

(Original post by Jonny W)
= (n/3) [ (int) x^(n/2 - 1) (1 - x)^(1/2) dx - (int) x^(n/2) (1 - x)^(1/2) dx ]
= (n/3) [I(n - 2) - I(n)]
How is that I(n-2)? Shouldn't it be I(n-1)?

...wait, is it because n/2 - 1 = (n-2)/2?
0
14 years ago
#6
(Original post by shift3)
...wait, is it because n/2 - 1 = (n-2)/2?
Yes!
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