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manps
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INT from 0 to 1 : (x^4)*(1-x²)^½.dx = ½*(Γ(5/2)Γ(3/2))/(Γ(4)) = pi/8

INT meaning intergral

thanks
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RichE
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(Original post by manps)
INT from 0 to 1 : (x^4)*(1-x²)^½.dx = ½*(Γ(5/2)Γ(3/2))/(Γ(4)) = pi/8

INT meaning intergral

thanks
Try making a substitution of x = sinθ

The integral then transforms into

 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta

Then use multiple angle formulae or reduction formulae to evaluate this
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amjad123
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http://amjadsa123.blogspot.com/ try this might help
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Jonny W
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The final answer should be pi/32, not pi/8.

--

The beta function is defined for a, b > 0 by

B(a, b) = \int_0^1 t^(a - 1) (1 - t)^(b - 1) dt

and is related to the gamma function by

B(a, b) = G(a)G(b)/G(a + b)

--

Substituting t = x^2, dt/dx = 2x gives

\int_0^1 x^4 (1 - x^2)^(1/2) dx
= \int_0^1 t^2 (1 - t)^(1/2) 1/(2t^(1/2)) dt
= (1/2) \int_0^1 t^(3/2) (1 - t)^(1/2) dt
= (1/2) B(5/2, 3/2)
= (1/2) G(5/2)G(3/2)/G(4)
= (1/2) [(3/2)(1/2)G(1/2)]*[(1/2)G(1/2)]/G(4) . . . . . using G(a) = (a - 1)G(a - 1)
= (3/16)G(1/2)^2/G(4)
= (3/16)pi/6 . . . . . since G(1/2) = sqrt(pi) and G(4) = 3! = 6
= pi/32
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RichE
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(Original post by amjad123)
http://amjadsa123.blogspot.com/ try this might help
how exactly?

Anyway back to the OP's question, to get the middle expression you may recall that



B(m,n) = \Gamma(m)\Gamma(n)/\Gamma(m+n)

where B and \Gamma are Euler's Beta and Gamma functions.

To get the middle form make the substitution u = x2 and you see that what you want is 1/2 B(5/2, 3/2)
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manps
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(Original post by Jonny W)
The final answer should be pi/32, not pi/8.
amazing as the exam paper i took this question from says, show this is equal to pi/8!!
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Jonny W
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(Original post by manps)
amazing as the exam paper i took this question from says, show this is equal to pi/8!!
For all x such that 0 <= x <= 1,

x^4 (1 - x^2)^(1/2) <= x^4

So

\int_0^1 x^4 (1 - x^2)^(1/2) dx
<= \int_0^1 x^4 dx
= 1/5

The answer therefore can't be pi/8 because pi/8 is more than 1/5.
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