Key Concepts and Formulas

• Greatest Integer Function (GIF): For any real number $x$, $[x]$ denotes the greatest integer less than or equal to $x$. Key properties include:
  • $[x + n] = [x] + n$ for any integer $n$.
  • $[x] + [-x] = 0$ if $x$ is an integer.
  • $[x] + [-x] = -1$ if $x$ is not an integer.
• Symmetry Property of Definite Integrals: For an integral of the form $\int_{-a}^a f(x) dx$, it can be rewritten as $\int_0^a (f(x) + f(-x)) dx$. This is particularly useful when $f(x)$ involves even or odd functions, or functions with symmetric properties.
• Integral over a Single Point: The definite integral of a function over a single point is zero. This means that isolated discontinuities or specific point values do not affect the value of the integral.

Step-by-Step Solution

Let the given integral be $I$. $I = \int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx}$

Step 1: Simplify the integrand using GIF properties. The integrand is of the form $[n - y]$, where $n = [x]$ is an integer and $y = \sin x$. Using the property $[n - y] = n + [-y]$, we can rewrite the integrand as: $[[x] - \sin x] = [x] + [ - \sin x]$ Why this step? This property allows us to separate the terms within the outermost GIF, making the expression easier to integrate. The integral now becomes: $I = \int_{ - \pi /2}^{\pi /2} {([x] + [ - \sin x])dx}$

Step 2: Apply the property for symmetric limits. The limits of integration are from $-\pi/2$ to $\pi/2$, which are symmetric about $0$. We use the property $\int_{-a}^a f(x) dx = \int_0^a (f(x) + f(-x)) dx$. Here, $f(x) = [x] + [-\sin x]$. First, let's find $f(-x)$: $f(-x) = [-x] + [-\sin(-x)]$ Since $\sin(-x) = -\sin x$, we get: $f(-x) = [-x] + [\sin x]$ Now, we sum $f(x)$ and $f(-x)$: $f(x) + f(-x) = ([x] + [ - \sin x]) + ([-x] + [\sin x])$ Rearranging the terms: $f(x) + f(-x) = ([x] + [-x]) + ([\sin x] + [ - \sin x])$ Why this step? Applying the symmetry property transforms the integral into one with a lower limit of 0, and the new integrand $f(x) + f(-x)$ often simplifies significantly due to the properties of $[x] + [-x]$ and $[\sin x] + [-\sin x]$. So, the integral becomes: $I = \int_0^{\pi /2} {([x] + [-x] + [\sin x] + [ - \sin x])dx}$

Step 3: Evaluate the terms involving GIF within the new interval $[0, \pi/2]$. The interval of integration is now $[0, \pi/2]$. Note that $\pi/2 \approx 1.57$.

• Term 1: $[x] + [-x]$ For $x \in [0, \pi/2]$:

  • If $x=0$, then $[0] + [-0] = 0 + 0 = 0$.
  • If $x \in (0, \pi/2)$, then $x$ is not an integer. In this case, $[x] + [-x] = -1$. Since the integral is not affected by the value at a single point ($x=0$), for the purpose of integration over $(0, \pi/2]$, we consider $[x] + [-x] = -1$.

• Term 2: $[\sin x] + [ - \sin x]$ For $x \in [0, \pi/2]$, the value of $\sin x$ ranges from $\sin 0 = 0$ to $\sin(\pi/2) = 1$.

  • If $x=0$, then $\sin x = 0$. In this case, $[\sin 0] + [-\sin 0] = [0] + [0] = 0 + 0 = 0$.
  • If $x \in (0, \pi/2]$, then $\sin x$ ranges from $(0, 1]$. For any value $v$ such that $0 < v \leq 1$, $v$ is not an integer.
    • If $\sin x \in (0, 1)$, then $\sin x$ is not an integer, so $[\sin x] + [-\sin x] = -1$.
    • If $\sin x = 1$ (which occurs at $x = \pi/2$), then $[\sin(\pi/2)] + [-\sin(\pi/2)] = [1] + [-1] = 1 + (-1) = 0$. Why this step? We are breaking down the integral into simpler parts by evaluating the GIF for each component within the specified interval. The behavior of $\sin x$ in $[0, \pi/2]$ is crucial here. Let's analyze the integrand term by term in the interval $[0, \pi/2]$: $[x] + [-x] + [\sin x] + [ - \sin x]$
  • At $x=0$: $0 + 0 + 0 + 0 = 0$.
  • For $x \in (0, \pi/2)$: $x$ is not an integer, so $[x] + [-x] = -1$. Also, for $x \in (0, \pi/2)$, $\sin x \in (0, 1)$. Since $\sin x$ is not an integer in this range, $[\sin x] + [-\sin x] = -1$. So, for $x \in (0, \pi/2)$, the integrand is $(-1) + (-1) = -2$.
  • At $x=\pi/2$: $\sin x = 1$. $[x] + [-x]$ at $x=\pi/2$: $[\pi/2] + [-\pi/2] = 1 + (-2) = -1$. $[\sin x] + [-\sin x]$ at $x=\pi/2$: $[1] + [-1] = 1 + (-1) = 0$. So, at $x=\pi/2$, the integrand is $(-1) + 0 = -1$.

  Why this step? We have identified that the value of the integrand changes based on whether $x$ is an integer or not, and whether $\sin x$ is an integer or not. We need to integrate the function over the interval $[0, \pi/2]$.

Step 4: Evaluate the integral. The integral can be split based on the behavior of the integrand. However, a simpler approach is to consider the average value over the interval. The integrand is: $([x] + [-x]) + ([\sin x] + [-\sin x])$

In the interval $[0, \pi/2]$:

• $x$ takes integer values only at $x=0$ and $x=1$.
• $\sin x$ takes integer values only at $\sin x = 0$ (i.e., $x=0$) and $\sin x = 1$ (i.e., $x=\pi/2$).

Let's analyze the integrand in subintervals:

• For $x \in (0, 1)$: $[x] + [-x] = -1$. $\sin x \in (0, \sin 1)$. Since $\sin 1 \approx 0.84$, $\sin x$ is not an integer in $(0, 1)$. So, $[\sin x] + [-\sin x] = -1$. The integrand is $-1 + (-1) = -2$.
• For $x = 1$: $[1] + [-1] = 0$. $\sin 1$ is not an integer. $[\sin 1] + [-\sin 1] = -1$. The integrand is $0 + (-1) = -1$.
• For $x \in (1, \pi/2)$: $[x] = 1$. $[-x]$ is between $[-1.57, -1)$. So $[-x] = -2$. Thus, $[x] + [-x] = 1 + (-2) = -1$. $\sin x \in (\sin 1, 1)$. So $\sin x$ is not an integer. Thus, $[\sin x] + [-\sin x] = -1$. The integrand is $-1 + (-1) = -2$.
• For $x = \pi/2$: $[x] = 1$. $[-x] = -2$. $[x] + [-x] = -1$. $\sin x = 1$. $[\sin x] + [-\sin x] = [1] + [-1] = 0$. The integrand is $-1 + 0 = -1$.

Since the integral is not affected by single points, we can integrate the value $-2$ over $(0, \pi/2)$ and then consider the endpoints. The integral $I$ is: $I = \int_0^{\pi /2} {(-2)dx}$ Why this step? Over the interval $(0, \pi/2)$, the value of the integrand $[x] + [-x] + [\sin x] + [-\sin x]$ is consistently $-2$. The points where the integrand might change value ($x=0, x=1, x=\pi/2$) are isolated and do not contribute to the integral's value.

$I = [-2x]_0^{\pi/2}$ $I = -2(\pi/2) - (-2)(0)$ $I = -\pi - 0$ $I = -\pi$

Common Mistakes & Tips

• Incorrect GIF Properties: Be very careful with the properties of $[x] + [-x]$. It's $0$ for integers and $-1$ for non-integers.
• Ignoring the Interval: Always consider the specific interval of integration when evaluating GIF expressions. The value of $[x]$ and $[\sin x]$ depends heavily on the range of $x$.
• Symmetry Property Application: Ensure the function $f(x)$ is correctly defined and that $f(x) + f(-x)$ is correctly simplified before integrating from $0$ to $a$.

Summary

The problem was solved by first simplifying the integrand using the property of the greatest integer function $[n-y] = n + [-y]$. Then, the symmetry property of definite integrals was applied to transform the integral from $[-\pi/2, \pi/2]$ to $[0, \pi/2]$ by considering $f(x) + f(-x)$. Within the interval $[0, \pi/2]$, the terms $[x] + [-x]$ and $[\sin x] + [-\sin x]$ were analyzed. It was found that for $x \in (0, \pi/2)$, $[x] + [-x] = -1$ and $[\sin x] + [-\sin x] = -1$, making the integrand $-2$. Integrating this constant value over the interval $[0, \pi/2]$ yielded the final result.

The final answer is $\boxed{-\pi}$.