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Step-by-Step Solution
Step 1: Understand the Integral and the Greatest Integer Function
We wish to evaluate
$ \displaystyle I = \int_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} \frac{dx}{\lfloor x \rfloor + \lfloor \sin x \rfloor + 4}. $
Here, $ \lfloor t \rfloor $ denotes the greatest integer less than or equal to $t$. We will split the integral according to integer intervals of $x$ and note the behavior of $ \lfloor \sin x \rfloor $ on those intervals.
Step 2: Determine Sub-Intervals in $x$ and Corresponding Floor Values
We split the interval $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $ into sub-intervals where $ \lfloor x \rfloor $ remains constant:
1. $ x \in \left[-\tfrac{\pi}{2}, -1\right) $
2. $ x \in [-1, 0) $
3. $ x \in [0, 1) $
4. $ x \in [1, \tfrac{\pi}{2}] $
Next, for each sub-interval, we also determine $ \lfloor \sin x \rfloor $.
Step 3: Evaluate Each Sub-Integral
3.1: For $x \in \left[-\tfrac{\pi}{2}, -1\right)$
• $ \lfloor x \rfloor = -2 $ because $-1.57\ldots \leq x < -1.$
• $ \sin x $ lies in $[-1, \sin(-1)],$ so $ \lfloor \sin x \rfloor = -1. $
Therefore,
$ \lfloor x \rfloor + \lfloor \sin x \rfloor + 4 = (-2) + (-1) + 4 = 1. $
Hence, the integral for this part is:
$$
\int_{-\tfrac{\pi}{2}}^{-1} \frac{dx}{1}
= \left. x \right|_{-\tfrac{\pi}{2}}^{-1}
= \Bigl(-1\Bigr) - \Bigl(-\tfrac{\pi}{2}\Bigr)
= -1 + \tfrac{\pi}{2}.
$$
3.2: For $x \in [-1, 0)$
• $ \lfloor x \rfloor = -1 $ for $-1 \leq x < 0.$
• $ \sin x \in [\sin(-1), 0),$ thus $ \lfloor \sin x \rfloor = -1. $
So,
$ \lfloor x \rfloor + \lfloor \sin x \rfloor + 4 = (-1) + (-1) + 4 = 2. $
The integral becomes:
$$
\int_{-1}^{0} \frac{dx}{2}
= \left. \frac{x}{2} \right|_{-1}^{0}
= \frac{0}{2} - \frac{(-1)}{2}
= \frac{1}{2}.
$$
3.3: For $x \in [0, 1)$
• $ \lfloor x \rfloor = 0 $ for $0 \leq x < 1.$
• $ \sin x \in [0, \sin(1)],$ so $ \lfloor \sin x \rfloor = 0. $
Hence,
$ \lfloor x \rfloor + \lfloor \sin x \rfloor + 4 = 0 + 0 + 4 = 4. $
The corresponding integral is:
$$
\int_{0}^{1} \frac{dx}{4}
= \left. \frac{x}{4} \right|_{0}^{1}
= \frac{1}{4}.
$$
3.4: For $x \in [1, \tfrac{\pi}{2}]
• $ \lfloor x \rfloor = 1 $ for $1 \leq x \leq \tfrac{\pi}{2} \approx 1.57.$
• $ \sin x \in [\sin(1), 1],$ and for almost all points in that interval (except exactly $x = \pi/2$), $ \sin x < 1,$ so $ \lfloor \sin x \rfloor = 0. $
Thus,
$ \lfloor x \rfloor + \lfloor \sin x \rfloor + 4 = 1 + 0 + 4 = 5. $
The integral is:
$$
\int_{1}^{\tfrac{\pi}{2}} \frac{dx}{5}
= \left. \frac{x}{5} \right|_{1}^{\tfrac{\pi}{2}}
= \frac{\tfrac{\pi}{2}}{5} - \frac{1}{5}
= \frac{\pi}{10} - \frac{1}{5}.
$$
Step 4: Sum All Contributions
Summing the four parts together:
$$
I = \Bigl(\tfrac{\pi}{2} - 1\Bigr) \;+\; \tfrac{1}{2} \;+\; \tfrac{1}{4} \;+\; \Bigl(\tfrac{\pi}{10} - \tfrac{1}{5}\Bigr).
$$
Combine like terms:
1) Numeric terms:
$$
-1 + \tfrac{1}{2} + \tfrac{1}{4} - \tfrac{1}{5}
= -1 + 0.5 + 0.25 - 0.2
= -0.45
= -\frac{9}{20}.
$$
2) $ \pi $ terms:
$$
\tfrac{\pi}{2} + \tfrac{\pi}{10}
= \tfrac{5\pi}{10} + \tfrac{\pi}{10}
= \tfrac{6\pi}{10}
= \tfrac{3\pi}{5}.
$$
Therefore,
$$
I = -\frac{9}{20} + \frac{3\pi}{5}.
$$
Factoring out $ \tfrac{3}{20} $ gives us:
$$
I = \frac{3\pi}{5} - \frac{9}{20}
= \frac{3}{20}(4\pi - 3).
$$
This matches the given correct answer.
Final Answer
$ \displaystyle \frac{3}{20}\bigl(4\pi - 3\bigr). $
