The value of $\int_{- π / 2}^{π / 2} \frac{d x}{[x] + [\sin x] + 4}$ , where [t] denotes the greatest integer less than or equal to t, is

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$\frac{3}{20} (4 π - 3)$

$\frac{1}{12} (7 π - 5)$

$\frac{1}{12} (7 π + 5)$

$\frac{3}{10} (4 π - 3)$

answer is A.

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Detailed Solution

Let $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{d x}{[x] + [\sin x] + 4}$

$\begin{array}{l} = \int_{- \frac{π}{2}}^{- 1} \frac{d x}{- 2 - 1 + 4} + \int_{- 1}^{0} \frac{d x}{- 1 - 1 + 4} + \int_{0}^{1} \frac{d x}{0 + 0 + 4} + \int_{1}^{\frac{π}{2}} \frac{d x}{1 + 0 + 4} \\ = \int_{- \frac{π}{2}}^{- 1} d x + \int_{- 1}^{0} \frac{d x}{2} + \int_{0}^{1} \frac{d x}{4} + \int_{1}^{\frac{π}{2}} \frac{d x}{5} \\ = (- 1 + \frac{π}{2}) + \frac{1}{2} (0 + 1) + \frac{1}{4} (1) + \frac{1}{5} (\frac{π}{2} - 1) \\ = - 1 + \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{π}{2} + \frac{π}{10} = \frac{3}{20} (4 π - 3) \end{array}$