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: Concept
Check if the integrand is an even or odd function over a symmetric interval $[-a, a]$.

Step 2: Meaning
A function $f(x)$ is odd if $f(-x) = -f(x)$. The integral of an odd function from $-a$ to $a$ is always zero.

Step 3: Analysis
Let $f(x) = \frac{\sin^5 x \cos^3 x}{x^4}$. $f(-x) = \frac{\sin^5(-x) \cos^3(-x)}{(-x)^4} = \frac{(-\sin x)^5 (\cos x)^3}{x^4} = -\frac{\sin^5 x \cos^3 x}{x^4} = -f(x)$.

Step 4: Conclusion
Since the function is odd and the interval is symmetric, the integral is 0.
Final Answer: (C)