Question

The value of $\int_{- 2n}^{2n + \frac{1}{2}}{}$ (sin p x) $\left\{ \frac{x}{2} \right\}$dx is (where {x} denotes the fractional part of x)

• A. $\frac{- 2n\pi + 1}{\pi^{2}}$
• B. $\frac{n}{\pi}$
• C. $\frac{(n + 1)}{\pi}$
• D. $\frac{2n\pi - 1}{\pi^{2}}$

Answer 1

Answer: (A)

$\frac{- 2n\pi + 1}{\pi^{2}}$

Answer 2

I = $\int_{2n}^{2n + \frac{1}{2}}{}$ (sin x) $\left\{ \frac{x}{2} \right\}$dx

= 2n$\int_{0}^{2}{}$ (sin px)$\frac{x}{2}$dx +$\int_{0}^{1/2}{}$ (sin px)$\frac{x}{2}$dx

=$\frac{- 2n\pi + 1}{\pi^{2}}$