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

Concept: If \(f(x)\) is an odd function, then: \[ \int_{-a}^{a} f(x)\,dx=0 \]

Step 1: Given integrand is: \[ f(x)=\frac{\sin^5x\cos^3x}{x^4} \]

Step 2: Check the nature of each factor.
We know: \[ \sin(-x)=-\sin x \] So \(\sin x\) is odd. Therefore, \(\sin^5x\) is also odd. Also, \[ \cos(-x)=\cos x \] So \(\cos x\) is even. Therefore, \(\cos^3x\) is also even. And: \[ (-x)^4=x^4 \] So \(x^4\) is even.

Step 3: The integrand becomes: \[ \frac{\text{odd}\times \text{even}}{\text{even}} = \text{odd} \] So: \[ f(-x)=-f(x) \]

Step 4: The limits are symmetric: \[ -\frac{\pi}{6} \quad \text{to} \quad \frac{\pi}{6} \] Therefore: \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\sin^5x\cos^3x}{x^4}\,dx=0 \] Hence, \[ \boxed{0} \]