Question

The value of ${\int }_{-\pi /2}^{\pi /2}\frac{dx}{\left[x\right]+\left[\sin x\right]+4}$, where $\left[t\right]$ 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)

$\frac{3}{20}(4\pi -3)$

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