The given integral is:

$\int_{-1}^{1} x^{17} \cos^4(x) \, dx$

Step 1: Symmetry Check

• The term $x^{17}$ is an odd function because it is raised to an odd power ($x^{17} = -(-x)^{17}$).
• The term $\cos^4(x)$ is an even function because $\cos(x)$ is an even function and raising it to any power preserves its evenness.

The product of an odd function ($x^{17}$) and an even function ($\cos^4(x)$) is odd. An integral of an odd function over a symmetric interval $[-a, a]$ equals zero:

\int_{-a}^a f(x) \, dx = 0 \quad \text{if f(x) is odd}.

Final Result:

$\int_{-1}^{1} x^{17} \cos^4(x) \, dx = 0$

Would you like me to expand on any particular aspect or explore a related problem?

Here are 5 related questions for deeper learning:

1. Why does the integral of an odd function over a symmetric interval evaluate to zero?
2. How does the symmetry property apply to even functions in integrals?
3. What would the result be if the power of $x$ were even (e.g., $x^{16}$)?
4. How can we evaluate $\int_{-1}^{1} \cos^4(x) \, dx$ instead?
5. Can you express $\cos^4(x)$ in terms of trigonometric identities to simplify its integral?

Tip: Always analyze the symmetry of a function before integrating over symmetric intervals—it can save time and simplify computations.