curve integral and reparameterization of X = (exp(y)+y*exp(x)...) I'm given a vector field
$$X = \left(\begin{array}{c} {\mathrm{e}}^y+y\,{\mathrm{e}}^x\\ {\mathrm{e}}^x+x\,{\mathrm{e}}^y\\ 1 \end{array}\right) \quad\text{with the curve} \quad \gamma(t) = \left(\begin{array}{c} \cos\left(t\right)\\ \sin\left(t\right)\\ t \end{array}\right)$$
Now the curve integral for $t \in[0,\pi]$ should be computed: $\int\langle X(\gamma(t)),\gamma'(t)\rangle\,\mathrm{dt}$:
$$\begin{align}\begin{array}\\ \\=\displaystyle{\int_0^{\pi}}\biggl\langle \left(\begin{array}{c} {\mathrm{e}}^{\sin\left(t\right)}+{\mathrm{e}}^{\cos\left(t\right)}\,\sin\left(t\right)\\ {\mathrm{e}}^{\cos\left(t\right)}+{\mathrm{e}}^{\sin\left(t\right)}\,\cos\left(t\right)\\ 1 \end{array}\right),\left(\begin{array}{c} -\sin\left(t\right)\\ \cos\left(t\right)\\ 1 \end{array}\right)\biggr\rangle\, \mathrm{dt} \\ \\= \int_0^{\pi}\cos\left(t\right)\,\left({\mathrm{e}}^{\cos\left(t\right)}+{\mathrm{e}}^{\sin\left(t\right)}\,\cos\left(t\right)\right)-\sin\left(t\right)\,\left({\mathrm{e}}^{\sin\left(t\right)}+{\mathrm{e}}^{\cos\left(t\right)}\,\sin\left(t\right)\right)+1 \\ \\ = \left[t+{\mathrm{e}}^{\cos\left(t\right)}\,\sin\left(t\right)+{\mathrm{e}}^{\sin\left(t\right)}\,\cos\left(t\right)\right]^\pi_0 = \pi-2\end{array}\end{align}$$
However, now I'm told to change the parameterization: $\gamma(t) = \gamma(g(t))$ where $g(t) = t^2$
To me, That means I need to sub $t$ with $t^2$ by what the integral changes: $ \\\int\langle X(\gamma(g(t))),\gamma'(g(t))\rangle\,\mathrm{dt}=$
$$\begin{align}\begin{array}{cc} \\ \displaystyle{\int_0^\pi}\biggl\langle\left(\begin{array}{cc} {\mathrm{e}}^{\sin\left(t^2\right)}+\sin\left(t^2\right)\,{\mathrm{e}}^{\cos\left(t^2\right)}\\ {\mathrm{e}}^{\cos\left(t^2\right)}+\cos\left(t^2\right)\,{\mathrm{e}}^{\sin\left(t^2\right)}\\ 1 \end{array}\right),\left(\begin{array}{cc} -\sin\left(t^2\right)\\ \cos\left(t^2\right)\\ 1 \end{array}\right)\biggr\rangle \\ \\ =\int_0^\pi\cos\left(t^2\right)\,\left({\mathrm{e}}^{\cos\left(t^2\right)}+\cos\left(t^2\right)\,{\mathrm{e}}^{\sin\left(t^2\right)}\right)-\sin\left(t^2\right)\,\left({\mathrm{e}}^{\sin\left(t^2\right)}+\sin\left(t^2\right)\,{\mathrm{e}}^{\cos\left(t^2\right)}\right)+1\end{array}\end{align}$$
But this integral seems definitely not computable. So I'm stuck right there...

Please note that $X = (e^y + y e^x, e^x + x e^y, 1)$ is a continuously differentiable vector field and its curl is zero -
$\nabla \times X = (0, 0, 0)$.
So the vector field is conservative and its potential function is,
$f = x e^y + y e^x + z$
In other words $X = \nabla f$
By Fundamental Theorem of Line Integrals,
$\displaystyle \int_C \nabla f \cdot dr = f(\vec r(b)) - f(\vec r(a))$
where $a$ is the starting point and $b$ is the end point.
Now for the first curve, starting point is $(1, 0, 0)$ and endpoint is $(-1, 0, \pi)$. So line integral is,
$ = (-1 \cdot e^0 + 0 \cdot e^{-1} + \pi) - (1 \cdot e^0 + 0 \cdot e^1 + 0)$
$= \pi - 2$.
You can similarly find for the second.
