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# sorry, yet another question watch

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1. by using the fact that sinx >= (2/pi)x for 0<= x <= pi/2, show that:

integral between pi/2 and 0 of: (x^2)/(1+sin^2) < (pi^3 / 8)(1-pi/4)
2. i dont understand that last step:

< pi^3/8 < (pi^3 / 8)(1-pi/4) QED

that's surely wrong....pi^3/8 > (pi^3 / 8)(1-pi/4)
3. Sorry, I integrated the wrong way round. Initially I integrated between 0 and pi/2 but I just read in the original question that you asked for between pi/2 and 0. In that case:

Start off by saying:

sinx >= (2x/pi)
(sinx)^2 >= (2x/pi)^2
1 + (sinx)^2 > (2x/pi)^2

1/[1 + (sinx)^2] < 1/(2x/pi)^2
x^2/[1 + (sinx)^2] < pi^2/4

Integrate the RHS:

< pi^2.x/4

Putting in the limits (pi/2 ... 0):

< -[pi^3]/8 < (pi^3 / 8)(1-pi/4) QED

Euclid.
4. no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

so it's still not right!
5. (Original post by Willa)
no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

so it's still not right!
Why dont you try to find the integral of x^2/(1+ (2x/pi)^2) from 0 ->pi/2. It maybe work
6. (Original post by Willa)
no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

so it's still not right!
Arrghh! Ok this one is definately right!:

sinx >= (2x/pi)
(sinx)^2 >= (2x/pi)^2
1 + (sinx)^2 >= 1 + (2x/pi)^2

1/[1 + (sinx)^2] <= 1/[1+ (2x/pi)^2]
x^2/[1 + (sinx)^2] <= [x^2]/[1+ (2x/pi)^2]

x^2/[1 + (sinx)^2] <= (pi^2).(x^2)/(pi^2 + 4x^2)

Use the substitution u = 2x/pi, and then integrating the RHS will get:

(x.pi^2)/4 - (arctan(2x/pi).pi^3)/8

Putting in the limits (0 ... pi/2):

<= (pi^3)/8 - (pi^4)/32
<= [(pi^3)/8](1 - pi/4) ..........

Euclid.
7. (Original post by Euclid)
Arrghh! Ok this one is definately right!:

sinx >= (2x/pi)
(sinx)^2 >= (2x/pi)^2
1 + (sinx)^2 >= 1 + (2x/pi)^2

1/[1 + (sinx)^2] <= 1/[1+ (2x/pi)^2]
x^2/[1 + (sinx)^2] <= [x^2]/[1+ (2x/pi)^2]

x^2/[1 + (sinx)^2] <= (pi^2).(x^2)/(pi^2 + 4x^2)

Use the substitution u = 2x/pi, and then integrating the RHS will get:

(x.pi^2)/4 - (arctan(2x/pi).pi^3)/8

Putting in the limits (0 ... pi/2):

<= (pi^3)/8 - (pi^4)/32
<= [(pi^3)/8](1 - pi/4) ..........

Euclid.
Yeah I know it works . But I'm lazy. Why didn't u just give him a clue?
8. (Original post by BCHL85)
Yeah I know it works . But I'm lazy. Why didn't u just give him a clue?
Ah well! I think the first one didn't work because of the loss of accuracy when adding 1

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