<p>Evaluate \(\displaystyle\int_{-\pi}^{\pi}\frac{x^3 + x\cos x + \tan^5 x}{2+\cos^2 x}\,dx\)</p>
Step-by-Step Solution
Key Concept: Check numerator: x^3 is odd, x cos x is odd (odd \times even=odd), tan^5x is odd; denominator 2+cos^2x is even. Odd/even = odd \to integral over symmetric interval = 0.
<div class='solution'>
<p>Let $f(x) = \dfrac{x^3+x\cos x+\tan^5 x}{2+\cos^2 x}$.</p>
<p><strong>Check parity:</strong></p>
<ul>
<li>$x^3$: odd; $x\cos x$: odd × even = odd; $\tan^5 x$: odd (tan is odd)</li>
<li>Numerator = odd + odd + odd = <strong>odd</strong></li>
<li>Denominator: $2+\cos^2(-x) = 2+\cos^2 x$ = <strong>even</strong></li>
</ul>
<p>∴ $f(-x) = \dfrac{\text{odd}}{\text{even}} = -f(x)$ → $f$ is <strong>odd</strong>.</p>
<p>$$\int_{-\pi}^{\pi}f(x)\,dx = 0\qquad\text{(odd function, symmetric limits)}$$</p>
<p>$$\boxed{I = 0}$$</p>
</div>
Correct Answer: B