P5 Intergration again - What do you expect its half the bloody book!) Watch

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Rixius
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#1
Report Thread starter 14 years ago
#1
Hello, could anyone help me with question 17 of the review exercise?

INT(between pi/2 and 0) 1/(3+5cosx) dx

Should equal (1/4)ln3

Just a hint would be good
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Mysticmin
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#2
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(Original post by Rixius)
Hello, could anyone ehlp me with question 17 of the review exercise?

INT(between pi/2 and 0) 1/(3+5cosx) dx

Should equal (1/4)ln3

Just a hint would be good
use a t substitution (t = tan x/2) then it comes out nicely.
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It'sPhil...
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(Original post by Rixius)
Hello, could anyone help me with question 17 of the review exercise?

INT(between pi/2 and 0) 1/(3+5cosx) dx

Should equal (1/4)ln3

Just a hint would be good
t = tan(x/2)
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Rixius
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#4
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#4
Thanks - will have a go now
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Jonny W
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(Original post by Rixius)
Just a hint would be good
t = tan(x/2).
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Rixius
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#6
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#6
Still cant get anywhere. end up with lots of root 2's and root 8's and ln's
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Jonny W
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#7
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(Original post by Rixius)
Still cant get anywhere. end up with lots of root 2's and root 8's and ln's
dt/dx = (1/2)sec^2(x/2) = (1/2)(1 + t^2)

(int from 0 to pi/2) 1/(3 + 5cosx) dx
= (int from 0 to 1) 1/[3 + 5(1 - t^2)/(1 + t^2)] * 2/(1 + t^2) dt
= 2(int from 0 to 1) 1/[3(1 + t^2) + 5(1 - t^2)] dt
= 2(int from 0 to 1) 1/(8 - 2t^2) dt
= (int from 0 to 1) 1/(4 - t^2) dt
= (int from 0 to 1) 1/[(2 - t)(2 + t)] dt
= (1/4)(int from 0 to 1) [1/(2 - t) + 1/(2 + t)] dt
= (1/4) [ln(2 + t) - ln(2 - t)] from 0 to 1
= (1/4) ln(3)
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It'sPhil...
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#8
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(Original post by Rixius)
Still cant get anywhere. end up with lots of root 2's and root 8's and ln's
When you get to INT 2/(8-2t^2) dt cancell the 2 on top and bottom
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Rixius
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#9
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#9
Thanks guys

I see where i went wrong - i used dt/dx from book p44
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