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This question was a question in an exam years ago.

Find the values of the following integrals.
(i) $$\int_\Gamma\dfrac{xdy-ydx}{x^2+y^2},$$ where $\Gamma$ is the curve $x=t\cos t$, $y=t^2\sin t$, for $t\in [2\pi,6\pi]$.
(ii) $$\int_A\dfrac{dx}{(1+\|x\|_2^4)^{\frac{1}{4}}},$$ where $A=\{x=(x_1,x_2,x_3)\in \mathbb R^3| x_2\gt0, \|x\|_2\le3\}$.

For $(i)$, I tried substituting the parametrisation of $\Gamma$ into the integral, but got nothing. I thought that this integral might be exact, but found no exact anti-derivatives...
For $(ii)$, I wrote it as $$\frac{4\pi}{2}\int_0^3\dfrac{r^2dr}{(1+r^4)^{\frac{1}{4}}}.$$ The I made the change of variables $s=(1+r^4)$ to re-write it as $$\frac{\pi}{2}\int_1^{82}\dfrac{ds}{(s(s-1))^{1/4}}.$$ Then this becomes an improper integral! Since that integrand is bounded by $(s-1)^{-1/4}$ and $\int (s-1)^{-1/4}ds$ converges in that interval, I can also prove the convergence of this improper integral. But how should I obtain the value? It seems that partial fraction decomposition works not so well here...
Thanks in advance for any hint or help, and edits.

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$$\int \dfrac{xdy-ydx}{x^2+y^2}=\int\dfrac{d\frac yx}{1+\left(\frac yx\right)^2}$$.Put $\frac yx=\tan\theta$ –  lab bhattacharjee Jul 25 '13 at 15:58
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The first one is not exact. –  Integral Jul 25 '13 at 16:01
    
Are you sure that $y=t^2\sin(t)$? –  Mhenni Benghorbal Jul 25 '13 at 16:22
    
Yes, at least that is the version I saw. Is there any inconsistence? –  awllower Jul 25 '13 at 16:23
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The first integral is the winding number of your curve around $0$, so its value can be found without any computation. For the second one, try to use spherical coordinates. –  Etienne Jul 25 '13 at 16:37

2 Answers 2

HINT: $ 1)$Note that $$\frac{xdy-ydx}{x^2+y^2}=d\left(\tan^{-1}\left(\frac{y}{x}\right)\right)=d\left(\tan^{-1}\left(t\tan t\right)\right)$$ on the curve $\Gamma$.

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This is not valid in the entire domain, because you got $x=0$ for some $t$. –  Integral Jul 25 '13 at 16:03
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But there are a finite number of discontinuities at $t=5\pi/2,7\pi/2,9\pi/2,11\pi/2,$ you can write the integral as the sum of the integrals in the intervals in between, right? –  Samrat Mukhopadhyay Jul 25 '13 at 16:09
    
These are discontinuities of second kind, and I dont know you can think that way in this case. –  Integral Jul 25 '13 at 16:14
    
Might I ask do you have any suggestion on the second question? It is funny that I surmised that the second should be solved very quickly, while the first might take a long time... :D –  awllower Jul 25 '13 at 16:16
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Aha, I see, the integral is not quite straightforward as it has this $\|x\|^4$ in it. –  Samrat Mukhopadhyay Jul 25 '13 at 16:44

You can write the integral as

$$ \int_\Gamma\dfrac{xdy-ydx}{x^2+y^2}= \int_\Gamma\dfrac{xdy}{x^2+y^2}+\int_\Gamma\dfrac{ydx}{x^2+y^2}=I_1+I_2 .$$

Using the given parametrization, we have

$$ I_1= \int_{2\pi}^{6\pi}\dfrac{t\cos(t) (2t\sin(t)+t^2\cos(t)) }{t^2\cos^2(t)+t^4\sin^2(t)}dt = \dots.$$

You can do the $I_2$ the same way.

Note:

$$ x=t\cos(t)\implies dx -t\sin(t)+\cos(t) dt$$ $$ y=t^2\sin(t) \implies dy= 2t\sin(t)+t^2\cos(t) dt. $$

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And I do not think that the transformed integrals are obvious to solve?Also, you missed some $d$ in the end. –  awllower Jul 25 '13 at 16:33
    
@awllower: This why I asked you about $y$. Note that, if $y=t\sin(t)$ then the integral will be easy to integrate. You can use Maple or Mathematica to evaluate these integrals. Anyhow, this is the technique. –  Mhenni Benghorbal Jul 25 '13 at 16:34
    
If $y=t\sin t$, then the denominator would be $t^2$. And I would not post here. haha In any case, thanks for the attention. –  awllower Jul 25 '13 at 16:38
    
@awllower: Sorry, it won't make a difference. I think you need to use Maple or Mathematica to evaluate these integrals. –  Mhenni Benghorbal Jul 25 '13 at 16:40

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