<p>The value of the integral $\int_{0}^{\pi} e^{\cos 2x} \cos^3(2n+1)x dx$ for any integer $n$ has the value</p>
Step-by-Step Solution
Key Concept: The integrand can be decomposed using the fact that $\cos^3((2n+1)x)$ is an odd function about $x = \pi/2$. When combined with the even function $e^{\cos 2x}$, the product's behavior under substitution reveals symmetry properties that force the integral to vanish.
<p><strong>Step 1:</strong> Write the integral as $I = \int_{0}^{\pi} e^{\cos 2x} \cos^3((2n+1)x) \, dx$.</p><p><strong>Step 2:</strong> Use the substitution $u = \pi - x$, so $du = -dx$. When $x = 0$, $u = \pi$; when $x = \pi$, $u = 0$.</p><p><strong>Step 3:</strong> The integral becomes: $I = \int_{\pi}^{0} e^{\cos 2(\pi-u)} \cos^3((2n+1)(\pi-u)) \, (-du)$.</p><p><strong>Step 4:</strong> Simplify using trigonometric identities: $\cos(2\pi - 2u) = \cos 2u$ and $\cos((2n+1)\pi - (2n+1)u) = -\cos((2n+1)u)$ (since $2n+1$ is odd).</p><p><strong>Step 5:</strong> This gives: $I = \int_{0}^{\pi} e^{\cos 2u} \cdot (-\cos^3((2n+1)u)) \, du = -\int_{0}^{\pi} e^{\cos 2u} \cos^3((2n+1)u) \, du = -I$.</p><p><strong>Step 6:</strong> From $I = -I$, we get $2I = 0$, therefore $I = 0$.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C