Prove that $\int_{-1}^{1} x^{17} \cos^{4} x \, dx = 0$.

Let $I = \int_{-1}^{1} x^{17} \cos^{4} x \, dx$.
Define the function $f(x) = x^{17} \cos^{4} x$.
Now,check for parity by evaluating $f(-x)$:
$f(-x) = (-x)^{17} \cos^{4}(-x)$.
Since $(-x)^{17} = -x^{17}$ and $\cos(-x) = \cos x$,we have:
$f(-x) = -x^{17} \cos^{4} x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} x^{17} \cos^{4} x \, dx = 0$.
Hence,the result is proved.