Question

The value of $\int\limits^{\pi/2}_{-\pi/2} \frac{dx}{\left[x\right]+\left[\sin x\right]+4} $ where [t] denotes the greatest integer less than or equal to t, is :

• A. $\frac{1}{12} (7 \pi + 5)$
• B. $\frac{3}{10} (4\pi - 3)$
• C. $\frac{1}{12} (7\pi - 5)$
• D. $\frac{3}{20} (4\pi - 3)$

Answer 1

The Correct Option is D

$I = \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}} \frac{dx}{\left[x\right]+\left[\sin x\right]+4} $
$ = \int^{-1}_{\frac{-\pi}{2}} \frac{dx}{-2-1+4} + \int^{0}_{-1} \frac{dx}{-1-1+4} +\int^{1}_{0} \frac{dx}{0+0+4} $
$ + \int^{\frac{\pi}{2}}_{1} \frac{dx}{1+0+4} $
$ \int^{-1}_{\frac{-\pi}{2}} \frac{dx}{1} +\int^{0}_{-1} \frac{dx}{2} +\int^{1}_{0} \frac{dx}{4} + \int^{\frac{\pi}{2}}_{1} \frac{dx}{5} $
$ \left(1- + \frac{\pi}{2}\right) + \frac{1}{2} \left(0+1\right) + \frac{1}{4} + \frac{1}{5} \left(\frac{\pi}{2} -1\right) $
$ -1 + \frac{1}{2} +\frac{1}{4}- \frac{1}{5}+ \frac{\pi}{2}+\frac{\pi}{10} $
$ \frac{-20+10+5-4}{20} + \frac{6\pi}{10} $
$ \frac{-9}{20} +\frac{3\pi}{5} $