<s>[INST] Calc $P=\int_C \frac{e^{-x^2y}}{\sqrt{( 1+x^2y^4 )^3}}( xy[ 2( 1+x^2y^4 )-y^3 ]dx+x^2[ e^{-x^2y}\sqrt{(1+x^2y^4)^3}+1+x^2y^4-2y^3]dy)$ Calc the integral:
$$P=\int \limits_C \dfrac{e^{-x^2y}}{\sqrt{\left( 1+x^2y^4 \right)^3}}\Bigg( xy\left[ 
2\left( 1+x^2y^4 \right)-y^3 \right]dx+x^2\left[ e^{-x^2y}\sqrt{\left(1+x^2y^4\right)^3}+1+x^2y^4-2y^3\right]dy\Bigg),$$
C is half circle $x^2+y^2=1, y\ge0$, direction from $A(1,0)$ to $B(-1,0).$

If set $x=\cos t$ and $y=\sin t$ $\left( \text{or} \,y=\sqrt{1-x^2} \right)$, can calc P by single integral. But, it's complex and may be not calc.
I also thought about Green's theorem. But it's easy to go wrong when calc derivative.
I wanna find a beautiful solutions. I hope some hints from you. Thanks.
 [/INST] If you parametrize the half circle as,
$r(t) = (t, \sqrt{1-t^2}), t \in (1,-1)$
$r'(t) = \left(1, - \frac{t}{\sqrt{1-t^2}}\right)$.
So the line integral is, $ \ P = \displaystyle - \int_{-1}^1 \vec F(r(t)) \cdot r'(t) \ dt$
$\vec F(r(t)) \cdot r'(t) = t \ g(t^2)$
The integrand is an odd function and its integral over $t \in (-1, 1)$ will be zero due to symmetry.
where, $ \ g(t^2) = \dfrac{e^{-t^2 \sqrt{1-t^2}}}{\sqrt{\left(1+t^2(1-t^2)^2 \right)^3}} \cdot $
$\Bigg(  \sqrt{1-t^2}\left[ 
2\left( 1+ t^2 (1-t^2)^2 \right) - (1-t^2)^{3/2} \right]$
$- \frac {t^2}{\sqrt{1-t^2}}\left[e^{-2t^2 \sqrt{1-t^2}}\sqrt{\left(1+t^2(1-t^2)^2\right)^3}+1 + t^2 (1-t^2)^2 - 2 (1-t^2)^{3/2}\right]\Bigg)$
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