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# P5 Integral - t = tan(x/2) sub. watch

1. Hi, I get zero for the answer of this integral but I think the answer is PiSqrt3/3.

I = Integral (0 to Pi) dx/(2 + cosx)

t = tan(x/2) => cosx = (1 - t^2)/(1 + t^2), dx = 2dt/(1 + t^2)
x = Pi => t = Infinity, x = 0 => t = 0

:. I = 2 . Integral (0 to Infinity) 1/(3 + t^2) dt
Let t = Sqrt3.sinhu => dt = Sqrt3.coshu du
t = Infinity => u = Infinity
t = 0 => u = 0

:. I = (2Sqrt3/3) Integral (0 to Infinity) sechu du
= (2Sqrt3)[tanhu/coshu] (0 to Infinity)
= 0

I think my mistake is evaluating the limit of tanhu/coshu as u -> Infinity, seems like I need Pi/2 as the answer; I get 1/Infinity = 0 though!

Help appreciated.
2. (Original post by Nima)
:. I = (2Sqrt3/3) Integral (0 to Infinity) sechu du
= (2Sqrt3)[tanhu/coshu] (0 to Infinity)
Hmm?

Also why did you use t=sqrt(3)sinh(u)? It seems to me that t=sqrt(3)tan(u) works better.
3. (Original post by dvs)
Hmm?

Also why did you use t=sqrt(3)sinh(u)? It seems to me that t=sqrt(3)tan(u) works better.
True perhaps the tan sub would be better.

But: I = 2 . Integral (0 to infinity) dt/(3 + t^2)
= 2 . Integral (0 to infinity) 1/(3cosh^2u) . Sqrt3 cosh u du
= 2Sqrt3/3 . Integral (0 to infinity) 1/coshu du
= 2Sqrt3/3 . Integral (0 to infinity) sechu du
= (2Sqrt3/3)[sechutanhu] (0 to infinity)
= (2Sqrt3/3)[tanhu/coshu] (0 to infinity)

and so on...
4. (Original post by Nima)
= 2Sqrt3/3 . Integral (0 to infinity) sechu du
= (2Sqrt3/3)[sechutanhu] (0 to infinity)
You're not integrating sech(u) properly.
5. (Original post by dvs)
You're not integrating sech(u) properly.
Ooops I differentiated it instead Thanks for spotting that silly error...
And yes tan sub works better.

Thanks!

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Updated: April 10, 2006
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