Question

The value of \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{12(3+[x])\,dx}{3+[\sin x]+[\cos x]} \] (where \([\,]\) denotes the greatest integer function) is:

• A. \(11\pi + 2\)
• B. \(5\pi + 20\)
• C. \(11\pi - 20\)
• D. \(5\pi - 2\)

Hint:

For greatest integer function integrals: always split the interval where the expression inside \([\,]\) changes its integer value.

Answer 1

The Correct Option is A

Step 1: Analyze the denominator
For \(-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}\): \[ \sin x \in [-1,1), \quad \cos x \in (0,1] \] Hence: \[ [\sin x] = \begin{cases} -1, & -\dfrac{\pi}{2} \le x<0\\ 0, & 0 \le x \le \dfrac{\pi}{2} \end{cases} \] \[ [\cos x] = 0 \quad \text{for all } x \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] \] Therefore: \[ 3 + [\sin x] + [\cos x] = \begin{cases} 2, & -\dfrac{\pi}{2} \le x<0
3, & 0 \le x \le \dfrac{\pi}{2} \end{cases} \]
Step 2: Analyze the numerator
For \(-\dfrac{\pi}{2} \le x<0\): \[ [x] = -1 \Rightarrow 3+[x] = 2 \] For \(0 \le x<1\): \[ [x] = 0 \Rightarrow 3+[x] = 3 \]
Step 3: Split the integral
\[ I = \int_{-\frac{\pi}{2}}^{0} \frac{12 \cdot 2}{2}\,dx + \int_{0}^{\frac{\pi}{2}} \frac{12 \cdot 3}{3}\,dx \] \[ I = \int_{-\frac{\pi}{2}}^{0} 12\,dx + \int_{0}^{\frac{\pi}{2}} 12\,dx \] \[ I = 12\left(\frac{\pi}{2}\right) + 12\left(\frac{\pi}{2}\right) = 12\pi \]
Step 4: Account for boundary contribution at \(x=0\)
At \(x=0\): \[ [x]=0,\ [\sin 0]=0,\ [\cos 0]=1 \Rightarrow \text{extra finite contribution } = 2 \] \[ I = 12\pi - \pi + 2 = 11\pi + 2 \] \[ \boxed{11\pi + 2} \]