I am currently trying to verify the computation of following integral using two different substitutions.
The area below $f$ on the unit the circle with $$f(x,y) = \frac{x^2 + y^2}{4} + \frac{xy}{2} $$
The calculation follows the basic formular:
$$ \int_C f(x,y) ds = \int_a^b f(g(t), h(t)) * \sqrt{g'(t)^2 + h'(t)^2}$$ for suitable $g, h, a \text{ and } b $
The easy parametrization uses $x = 2 \sin(t)$ and $y = 2 \cos(t)$ for $t \in [0, 2 \pi] $
Plugging everything in the integral immediately simplifies to:
$$ \int_0^{2 \pi} 1+ 2\sin(2t)2 dt = 4 \pi$$
However, one can parameterize a circle differently:
Using $x = \frac{1-t^2}{1+t^2}$ and $\frac{2t}{1+t^2}$ with $t \in (-\infty, +\infty) $
One may then create the following abomination:
$$p(t) = \frac{(\frac{1-t^2}{1+t^2})^2 + (\frac{2t}{1+t^2})^2}{4} + \frac{\frac{1-t^2}{1+t^2} \frac{2t}{1+t^2}}{2} = \frac{(-1 - 2 t + t^2)^2}{4 (1 + t^2)^2} $$
with: $g'(t) = \frac{-(4 t)}{(1 + t^2 (2 + t^2))} $ and $h'(t) = \frac{-(2 (-1 + t) (1 + t))}{(1 + t^2)^2}$
Fortunately: $\sqrt{g'(t)^2 + h'(t)^2}$ simplifies to: $\sqrt{\frac{4}{(1+t^2)^2}} = \frac{2}{1+t^2}$
Putting now these terms together we get:
$$\frac{(-1 - 2 t + t^2)^2}{4 (1 + t^2)^2} * \frac{2}{1+t^2} = \frac{(-1 - 2 t + t^2)^2}{(2 (1 + t^2)^3)}$$
This can then be integrated $$\int_{-\infty}^{+\infty} \frac{(-1 - 2 t + t^2)^2}{(2 (1 + t^2)^3)} = \frac{\pi}{2}$$ which is pretty close. However, this begs the questions: Where did the calculation go wrong?