1. Given integrals

We need to compare $I_1=\int_0^1 e^{-x}\cos^2 x\,dx,$ $I_2=\int_0^1 e^{-x^2}\cos^2 x\,dx,$ $I_3=\int_0^1 e^{-x^3}\,dx.$

We will compare them pairwise.

2. Compare $I_2$ and $I_1$

For $x\in[0,1]$, $x^2\le x \implies -x^2\ge -x \implies e^{-x^2}\ge e^{-x}.$ Also, $\cos^2 x\ge 0.$ Hence, $e^{-x^2}\cos^2 x\ge e^{-x}\cos^2 x \quad \text{for all }x\in[0,1].$ Therefore, $I_2\ge I_1.$ In fact, the inequality is strict for $x\in(0,1)$, so $I_2>I_1.$

3. Compare $I_3$ and $I_2$

For $x\in[0,1]$, $x^3\le x^2 \implies -x^3\ge -x^2 \implies e^{-x^3}\ge e^{-x^2}.$ Also, $\cos^2 x\le 1.$ Thus, $e^{-x^3}\ge e^{-x^2}\ge e^{-x^2}\cos^2 x.$ So for every $x\in[0,1]$, $e^{-x^3} \ge e^{-x^2}\cos^2 x.$ Integrating over $[0,1]$ gives $I_3\ge I_2.$ Again this is strict on $(0,1)$ because $\cos^2 x<1$ for $x\in(0,1]$, hence $I_3>I_2.$

4. Combine the inequalities

From Steps 2 and 3, $I_3>I_2>I_1.$

So the correct option is: $\boxed{\text{C: } I_3>I_2>I_1}.$

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.