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 Post subject: Integral
PostPosted: Mon, 8 Sep 2008 14:02:12 UTC 
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Here's one for y'all:

$\int \frac{dx}{k+\sin x} \qquad k\in \mathbb{R}$

Hint:
Spoiler:
$t=\tan(x/2)$


I *think* I have the answer, let's see if anyone agrees:
Spoiler:
$\int \frac{dx}{k+\sin x} = \frac{2k}{k^{2}-1}\int \frac{dt}{1-\Big(\frac{kt+1}{\sqrt{1-k^{2}}}\Big)^{2}}$

$=\frac{-2}{\sqrt{1-k^{2}}}\tanh ^{-1}\Bigg(\frac{k\tan(x/2)+1}{\sqrt{1-k^{2}}}\Bigg) + C$


Last edited by Tommo on Mon, 8 Sep 2008 19:49:50 UTC, edited 1 time in total.

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 Post subject:
PostPosted: Mon, 8 Sep 2008 15:35:28 UTC 
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Is |k| < 1?


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 Post subject:
PostPosted: Mon, 8 Sep 2008 15:52:27 UTC 
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Assuming my answer is correct (it may very well be wrong), it seems like it will have to be, if it is to be real. I hadn't thought about placing such a restriction initially though. Dunno :roll:


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 Post subject: Re: Integral
PostPosted: Mon, 8 Sep 2008 16:47:34 UTC 
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Hello, Tommo!

My answer's slightly different . . .


Quote:
$\int \frac{dx}{k+\sin x} \qquad k\in \mathbb{R}$

\text{Let }\:t \:=\:\tan\frac{x}{2} \quad\Rightarrow\quad \sin x \:=\:\frac{2t}{1+t^2} \quad\Rightarrow\quad dx \:=\:\frac{2\,dt}{1+t^2}


Substitute: .$\int\frac{\frac{2\,dt}{1+t^2}}{k + \frac{2t}{1+t^2}} \;=\;\int\frac{2\,dt}{kt^2 + 2t + k} \;=\;\frac{2}{k}\int\frac{dt}{t^2 + \frac{2}{k}t + 1}


Complete the square: .$\frac{2}{k}\int\frac{dt}{\left(t^2 + \frac{2}{k}t + \frac{1}{k^2}\right) + \left(1 - \frac{1}{k^2}\right)} \;=\;\frac{2}{k}\int\frac{dt}{\left(t + \frac{1}{k}\right)^2 + \left(\frac{k^2-1}{k^2}\right)}


Integrate: .$\frac{2}{k}\cdot\frac{k}{\sqrt{k^2-1}} \arctan\left(\frac{t+\frac{1}{k}}{\frac{\sqrt{k^2-1}}{k}}\right) + C  \;=\;\frac{2}{\sqrt{k^2-1}}\,\arctan\left(\frac{kt + 1}{\sqrt{k^2-1}}\right)  + C



Back-substitute: . $\boxed{\frac{2}{\sqrt{k^2-1}}\,\arctan\left(\frac{k\tan\frac{x}{2} + 1}{\sqrt{k^2-1}}\right) + C}\qquad |k|\:>\:1



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 Post subject:
PostPosted: Mon, 8 Sep 2008 18:11:57 UTC 
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Ah, nice solution Soroban.

I agree with you up to $\frac{2}{k}\int \frac{dt}{(t + \frac{1}{k})^{2} + (\frac{k^{2}-1}{k^{2}})}$

Then for some reason I divided everything by \frac{k^{2}-1}{k^{2}} as I had an 'arctanh()' in mind for my final answer.


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 Post subject:
PostPosted: Wed, 15 Oct 2008 23:03:50 UTC 
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[url][URL=http://imageshack.us]Image[/url]
Image[/url]


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 Post subject:
PostPosted: Tue, 21 Oct 2008 00:38:57 UTC 
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From my previous post, I find.

(x/k) - ln |csc(x)+cot(x)| + C , k ε R - {0}

Verification:

d/dx [(x/k) - ln |csc(x)+cot(x)| + C ]

= 1 / (k+sinx)


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