Question

The value of $\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dx}{\left[x\right]+\left[\sin x\right]+4}$ 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

Answer: (D)

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{\left[x\right]+\left[\sin x\right]+4}$
$={\int }_{\frac{-\pi }{2}}^{-1}\frac{dx}{-2-1+4}+{\int }_{-1}^0\frac{dx}{-1-1+4}+{\int }_0^1\frac{dx}{0+0+4}$
$+{\int }_1^{\frac{\pi }{2}}\frac{dx}{1+0+4}$
${\int }_{\frac{-\pi }{2}}^{-1}\frac{dx}{1}+{\int }_{-1}^0\frac{dx}{2}+{\int }_0^1\frac{dx}{4}+{\int }_1^{\frac{\pi }{2}}\frac{dx}{5}$
$\left(1-+\frac{\pi }{2}\right)+\frac{1}{2}\left(0+1\right)+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi }{2}-1\right)$
$-1+\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{\pi }{2}+\frac{\pi }{10}$
$\frac{-20+10+5-4}{20}+\frac{6\pi }{10}$
$\frac{-9}{20}+\frac{3\pi }{5}$