Question

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

• A. \( \pi/2 \)
• B. \( \pi/4 \)
• C. \( \pi/8 \)
• D. \( 0 \)

Hint:

Always check for symmetry of the function and the interval of integration. If the function is odd and the interval is symmetric about zero, the integral is zero.

Answer 1

The Correct Option is D

Let $f(x) = \frac{(\sin x)^5 (\cos x)^3}{x^4}$. We want to evaluate the integral $\int_{-\pi/6}^{\pi/6} f(x) dx$. Let's check if $f(x)$ is an even or odd function. $$f(-x) = \frac{(\sin (-x))^5 (\cos (-x))^3}{(-x)^4} = \frac{(-\sin x)^5 (\cos x)^3}{x^4} = \frac{-\sin^5 x \cos^3 x}{x^4} = -f(x)$$ Since $f(-x) = -f(x)$, the function $f(x)$ is an odd function. The integral of an odd function over a symmetric interval $[-a, a]$ is always zero. $$\int_{-a}^{a} f(x) dx = 0 \quad \text{if } f(x) \text{ is an odd function}$$ In this case, the interval is $[-\pi/6, \pi/6]$, which is symmetric about $0$, and the function $f(x) = \frac{(\sin x)^5 (\cos x)^3}{x^4}$ is an odd function. Therefore, $$\int_{-\pi/6}^{\pi/6} \frac{(\sin x)^5 (\cos x)^3}{x^4} dx = 0$$