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Question

The value of $$\displaystyle \int^{\pi/2}_{-\pi/2} \dfrac{dx}{[x]+[\sin x]+4}$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is:


A
112(7π+5)
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B
310(4π3)
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C
112(7π5)
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D
320(4π3)
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Solution

The correct option is A $$\dfrac{3}{20}(4\pi -3)$$
$$\displaystyle I  = \int_{-\pi/2}^{\pi/2}\dfrac{dx}{[x]+[\sin x]+4}$$
$$=\displaystyle \int_{-\pi/2}^{-1} \dfrac{dx}{-2-1+4} +\int^0_{-1} \dfrac{dx}{-1-1+4}+\int^1_0 \dfrac{dx}{0+0+4} + \int^{\pi/2}_{1} \dfrac{dx}{1+0+4}$$
$$=\displaystyle \int_{-\pi/2}^{-1} \dfrac{dx}{1} + \int_{-1}^0 \dfrac{dx}{2} + \int_0^1 \dfrac{dx}{4} + \int^{\pi/2}_{1} \dfrac{dx}{5}$$
$$=\left(-1+\dfrac{\pi}{2}\right) + \dfrac{1}{2} (0+1) + \dfrac{1}{4} + \dfrac{1}{5} \left(\dfrac{\pi}{2} - 1\right)$$
$$=-1 + \dfrac{1}{2}+\dfrac{1}{4}- \dfrac{1}{5}+\dfrac{\pi}{2} + \dfrac{\pi}{10}$$
$$=\dfrac{-9}{20} + \dfrac{3\pi}{5}=\dfrac {3}{20}(4\pi-3)$$

Mathematics

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