Question

If $ S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} \, dx $, and $ n $ is an integer, then $ S_{n+1} - S_n = $:

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

Hint:

For integrals of trigonometric functions involving sine terms, recognizing the symmetry often helps in identifying the result.

Answer 1

The Correct Option is C

Step 1: Express \( S_n \).
We are given the expression for \( S_n \) as follows: \[ S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} \, dx \] This integral is known to be related to the sum of the sine functions in the Fourier series expansion.
Step 2: Observe the structure of the integral.
Note that the integral contains a sine term in the numerator and a sine function in the denominator, which suggests the use of standard results from trigonometric integrals. Specifically, this type of integral can be expressed as a sum of individual terms for each value of \( n \).
Step 3: Find the difference \( S_{n+1} - S_n \).
Using known properties of such integrals, we deduce that the difference \( S_{n+1} - S_n \) results in: \[ S_{n+1} - S_n = 0 \] This result is true because of the symmetry of the sine functions in the integrand.
Thus, the correct answer is: \[ \boxed{0} \]