Question

$\int_{-\pi/6}^{\pi/6} \frac{\sin^{5}x \cos^{3}x}{x^{4}} dx =$

• A. $\pi/2$
• B. $\pi/4$
• C. 0
• D. 1

Hint:

Before starting a complex definite integral with symmetric limits, always check for odd/even properties—it can save you minutes of work!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the symmetry of the integrand. For an integral with symmetric limits $[-a, a]$, if the function $f(x)$ is "odd" ($f(-x) = -f(x)$), the integral is zero.
Step 2: Key Formula or Approach:
1. Property: $\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function.
2. $\sin(-x) = -\sin x$ (Odd), $\cos(-x) = \cos x$ (Even), $(-x)^4 = x^4$ (Even).
Step 3: Detailed Explanation:
Let $f(x) = \frac{\sin^5 x \cos^3 x}{x^4}$. Test for symmetry: \[ f(-x) = \frac{[\sin(-x)]^5 [\cos(-x)]^3}{(-x)^4} = \frac{(-\sin x)^5 (\cos x)^3}{x^4} \] \[ f(-x) = \frac{-\sin^5 x \cos^3 x}{x^4} = -f(x) \] Since $f(x)$ is an odd function and the limits are symmetric ($-\pi/6$ to $\pi/6$), the areas above and below the x-axis cancel each other out exactly.
Step 4: Final Answer:
The value of the integral is 0.