I need help with creating suitable parameters to solve the integral
thanks
Since you are integrating along the path $\displaystyle xy^2=1$ you can use it in your integral (where ever possible). So your integral becomes
$\displaystyle \int_1^4 \left(x^2 - e^{-x}\right) dx + \int_1^{1/2} \left( \frac{2}{y^3} + \cos y\right) dy$
Wow, that's very clever! jonnyl, if it had not worked out that nicely (or, like me, you didn't see to the core as quickly as Danny), here's the hard way:
From $\displaystyle xy^2= 1$, you have $\displaystyle x= \frac{1}{y^2}= y^{-2}$ so $\displaystyle dx= -2y^{-3}dy= \frac{-2}{y^3}dy$. Now, $\displaystyle x^4y^4= \frac{y^4}{y^8}= y^{-4}$, $\displaystyle x^2y^2= \frac{y^2}{y^4}= y^{-2}$, $\displaystyle \frac{2x}{y}= \frac{2}{(y^2)(y)}= \frac{2}{y^3}$, and $\displaystyle xy^3= \frac{y^3}{y^2}= y$. That makes the path integral
$\displaystyle \int_1^{1/2}\left[\left(y^{-4}- e^{-y^{-2}}\right)\left(-2y^{-3}dy\right)+ (2y^{-3}+ cos(y))dy\right]$