We need to prove $\int_{-C}\vec{F}d\vec{R}=-\int_{C} \vec{F}d\vec{R}$
Usual attempt in Textbook: Suppose $C:\vec{R(t)},a\leq t\leq b$. Then $-C:\vec{R(-t)},-b\leq t\leq -a$. I am able to verify the result. Can I define as $-C:\vec{R(a+b-t)},a\leq t\leq b?$
$\int_{-C}\vec{F}d\vec{R}= \int_{a}^{b}\vec{F}(a+b-t)\frac{d\vec{R}(a+b-t)}{dt}dt$ (putting $\tau=a+b-t$) . We get $$-\int_{b}^{a}\vec{F}(\tau)\frac{d\vec{R}(\tau)}{d\tau}d\tau=\int_{a}^{b}\vec{F}(\tau)\frac{d\vec{R}(\tau)}{d\tau}d\tau=\int_{C} \vec{F}d\vec{R}$$ ($\because$ integration is independent of variables) What happen to the negative sign? where is my mistake? Please help me.
Your parametrization of the reverse path is correct.
There are two negative signs: one in comparing $d\vec R/dt$ with $d\vec R/d\tau$, and the other with $d\tau = -dt$. So you should have $$\int_b^a \vec F(\tau)\cdot \frac{d\vec R}{d\tau}\,d\tau = - \int_a^b \vec F(\tau)\cdot \frac{d\vec R}{d\tau}\,d\tau = -\int_C \vec F\cdot d\vec R.$$