How to integrate these integrals?

Question

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.

Answer 1

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 $\Gamma$.

Answer 2

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:

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