Question

\[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\sin^5x\cos^3x}{x^4}\,dx= \]

• A. \(\frac{\pi}{2}\)
• B. \(\frac{\pi}{4}\)
• C. \(0\)
• D. \(1\)

Hint:

Before solving symmetric definite integrals, always check whether the integrand is odd or even.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This definite integral is over a symmetric interval \( [-a, a] \).
For such integrals, before performing any calculations, we should check if the integrand is an even or odd function.
Step 2: Key Formula or Approach:
Property: If \( f(x) \) is an odd function (\( f(-x) = -f(x) \)), then \( \int_{-a}^a f(x) dx = 0 \).
Property: \( \sin(-x) = -\sin x \) (odd) and \( \cos(-x) = \cos x \) (even).
Step 3: Detailed Explanation:

Let \( f(x) = \frac{\sin^5 x \cos^3 x}{x^4} \).

We evaluate \( f(-x) \) to determine the nature of the function:
\[ f(-x) = \frac{(\sin(-x))^5 (\cos(-x))^3}{(-x)^4} \]

Apply trigonometric and power properties:
\( \sin(-x) = -\sin x \implies (\sin(-x))^5 = (-\sin x)^5 = -\sin^5 x \).
\( \cos(-x) = \cos x \implies (\cos(-x))^3 = \cos^3 x \).
\( (-x)^4 = x^4 \).

Substitute back:
\[ f(-x) = \frac{(-\sin^5 x)(\cos^3 x)}{x^4} = - \frac{\sin^5 x \cos^3 x}{x^4} \]

Since \( f(-x) = -f(x) \), the function \( f(x) \) is an **odd function**.

According to the property of definite integrals for odd functions over symmetric limits:
\[ \int_{-\pi/6}^{\pi/6} f(x) dx = 0 \]

Step 4: Final Answer:
The integral is 0.