The value of

Question:

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{d x}{[x]+[\sin x]+4}$, where $[\mathrm{t}]$ denotes the

greatest integer less than or equal to $t$, is:

  1. (1) $\frac{1}{12}(7 \pi+5)$

  2. (2) $\frac{1}{12}(7 \pi-5)$

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

  4. (4) $\frac{3}{10}(4 \pi-3)$


Correct Option: , 3

Solution:

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

$=\int_{\frac{-\pi}{2}}^{-1} \frac{d x}{-2-1+4}+\int_{-1}^{0} \frac{d x}{-1-1+4}+\int_{0}^{1} \frac{d x}{0+0+4}+\int_{1}^{\frac{\pi}{2}} \frac{d x}{1+0+4}$

$=\left(-1+\frac{\pi}{2}\right)+\frac{1}{2}(0+1)+\frac{1}{4}(1-0)+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$

$=\frac{3 \pi}{5}-\frac{9}{20}=\frac{3}{20}(4 \pi-3)$

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