Animated Solution for Mathematics - Definite Integration: Let $[\cdot ]$ denote the greatest integer function. Then ${\int }_{-\pi /2}^{\pi /2}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right)dx$ is equal to :

Taming the Integral

Welcome, fellow explorer of the mathematical universe. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of brackets and trigonometric functions.

We are looking at the integral:

It is natural to feel intimidated by the Greatest Integer Function, but I want you to take a deep breath. In calculus, when we see a function that jumps, we do not fight it; we embrace it by breaking it down.

Mapping the Terrain

First, let us look at our domain. We are integrating from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$. Since $\pi \approx 3.14$, our range is approximately $[-1.57,1.57]$.

The Greatest Integer Function $[x]$ is a step function that changes its value at every integer. Within our range, the integers are $-1,0,1$. These are our critical points.

They divide our journey into four distinct sub-intervals: $[-\frac{\pi }{2},-1)$, $[-1,0)$, $[0,1)$, and $[1,\frac{\pi }{2}]$. By splitting the integral at these points, we transform a complex, variable expression into a series of simple, constant values.

Taming the Denominator

Now, let us look at the denominator: $3+[\sin x]+[\cos x]$. This is where the magic happens.

For $x<0$, we are in the fourth quadrant. Here, $\sin x$ is between $-1$ and $0$, so $[\sin x]=-1$. Meanwhile, $\cos x$ is between $0$ and $1$, so $[\cos x]=0$. The denominator becomes $3-1+0=2$.

For $x>0$, we are in the first quadrant. Both $\sin x$ and $\cos x$ are between $0$ and $1$, so both their greatest integer values are $0$. The denominator becomes $3+0+0=3$. Just like that, the denominator is no longer a variable; it is a constant that changes only once at $x=0$.

The Step-by-Step Construction

With the denominator tamed, we evaluate the integrand for each interval:

In the interval $[-\frac{\pi }{2},-1)$, we have $[x]=-2$. The integrand is $\frac{12(3-2)}{2}=6$. In the interval $[-1,0)$, we have $[x]=-1$. The integrand is $\frac{12(3-1)}{2}=12$. In the interval $[0,1)$, we have $[x]=0$. The integrand is $\frac{12(3+0)}{3}=12$. In the interval $[1,\frac{\pi }{2}]$, we have $[x]=1$. The integrand is $\frac{12(3+1)}{3}=16$.

We have effectively turned a complex curve into a series of horizontal line segments.

The Final Summation

Now, we simply calculate the area under these segments:

Evaluating these integrals, we get:

Simplifying this expression:

Grouping the terms, $3\pi +8\pi =11\pi$ and $-6+12+12-16=2$.

The final result is $11\pi +2$. By breaking the problem into logical steps, we turned a mountain into a series of small, conquerable hills.