Question

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

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

Answer 1

The correct option is A $\frac{3}{20}(4\pi -3)$

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