Question

The value of \(\displaystyle \int_{0}^{2\pi} \frac{\sin 2x}{x(2\pi - x)}\,dx\) is equal to:

• A. \(\pi\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{8}\)
• E. 0

Hint:

For integrals on symmetric intervals or using the property \(\int_0^{2a} f(x) dx\), check if the function satisfies \(f(2a-x) = -f(x)\). If it does, the integral over that range is always zero.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We can use a fundamental property of definite integrals: \(\int_a^b f(x) dx = \int_a^b f(a + b - x) dx\).
This property allows us to transform the variable and often simplifies the integrand significantly.

Step 2: Key Formula or Approach:
Let \(I = \int_0^{2\pi} \frac{\sin 2x}{x(2\pi - x)} dx\).
Apply the property \(x \to 2\pi - x\).

Step 3: Detailed Explanation:
Applying the transformation \(x = 2\pi - u \implies dx = -du\):
When \(x=0, u=2\pi\). When \(x=2\pi, u=0\).
\[ I = \int_{2\pi}^0 \frac{\sin(2(2\pi - u))}{(2\pi - u)(2\pi - (2\pi - u))} (-du) \]
\[ I = \int_0^{2\pi} \frac{\sin(4\pi - 2u)}{(2\pi - u)(u)} du \]
Recall the trigonometric identity \(\sin(4\pi - \theta) = -\sin \theta\):
\[ I = \int_0^{2\pi} \frac{-\sin 2u}{u(2\pi - u)} du \]
Notice that the resulting expression is exactly \(-I\):
\[ I = -I \]
\[ 2I = 0 \implies I = 0 \]

Step 4: Final Answer:
The value of the integral is 0.