And proving integrationWatch

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Thread starter 14 years ago
#1
If I remember exactly the question. It should be proved
Prove I(x) = INT[1/e, tanx](tdt/(t^2 + 1)) + INT[1/e, cotx](dt/(t^2 +1)) = 1.
I hope it's right.
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14 years ago
#2
I think there's a mistake. Check it again.
After integrating, you get
(1/2)ln|t^2 + 1| (from 1/e to tanx) + arctan(t) (from 1/2 to cot x)
This doesn't seem to simplify to 1 with the given limits.
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Thread starter 14 years ago
#3
(Original post by J.F.N)
I think there's a mistake. Check it again.
After integrating, you get
(1/2)ln|t^2 + 1| (from 1/e to tanx) + arctan(t) (from 1/2 to cot x)
This doesn't seem to simplify to 1 with the given limits.
Yeah, right. There's a small mistake here, but make a big difference.
Prove I(x) = INT[1/e, tanx](tdt/(t^2 + 1)) + INT[1/e, cotx](dt/t(t^2 +1)) = 1. with 0<x<pi/2.
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Thread starter 14 years ago
#4
It's not so hard to do ... but need an intelligent move.
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14 years ago
#5
INT[1/e, tanx] ...
Which one's the lower-bound?
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Thread starter 14 years ago
#6
1/e is lower bound.
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Thread starter 14 years ago
#7
Ok, I'll give a hint I(pi/4) = 1.
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14 years ago
#8
Sorry I'm confused.

Have you already done this integral - or are you looking for help with it?
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14 years ago
#9
(Original post by BCHL85)
Yeah, right. There's a small mistake here, but make a big difference.
Prove I(x) = INT[1/e, tanx](tdt/(t^2 + 1)) + INT[1/e, cotx](dt/t(t^2 +1)) = 1. with 0<x<pi/2.
consider second int
let u=1/t
when t=cotx => u=1/cotx=tan x
t=1/e=> u=1/1/e=e
dt=-u^(-2)du
dt/t(t^2 + 1))=-u^(-2)/1/u.(1/u^2+1)=-u^(-1).u^2/(u^2+1)=-u/(u^2+1)
so i(x)=INT[1/e, tanx](tdt/(t^2 + 1)) - INT[e, tanx](dtt/(t^2 +1))
= INT[1/e, tanx](tdt/(t^2 + 1)) - INT[1/e, tanx](tdt/(t^2 +1))
+INT[1/e, e](tdt/(t^2 + 1))

=(1/2){ln|e^2+ 1|-{ln|1+e^2|-ln |e^2|}
=1
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Thread starter 14 years ago
#10
(Original post by evariste)
consider second int
let u=1/t
when t=cotx => u=1/cotx=tan x
t=1/e=> u=1/1/e=e
dt=-u^(-2)du
dt/t(t^2 + 1))=-u^(-2)/1/u.(1/u^2+1)=-u^(-1).u^2/(u^2+1)=-u/(u^2+1)
so i(x)=INT[1/e, tanx](tdt/(t^2 + 1)) - INT[e, tanx](dtt/(t^2 +1))
= INT[1/e, tanx](tdt/(t^2 + 1)) - INT[1/e, tanx](tdt/(t^2 +1))
+INT[1/e, e](tdt/(t^2 + 1))

=(1/2){ln|e^2+ 1|-{ln|1+e^2|-ln |e^2|}
=1
Woa ..good move.
The method I used here is differentiate I(x) w.r.t x.
I found it's interesting question because of that ... using differentiation to prove integration. It's quite strange.
I'(x) = tanx/[(tanx)^2 +1]. 1/cos^2 x + 1/cotx(cot^2x +1).(-1/sin^2x)
= tanx - 1/cotx = 0.
So I(x) is constant.
I(pi/4) = 1 . So I(x) = 1
Is it right?
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