Question

Let \([\,\cdot\,]\) denote the greatest integer function. Then \[ \int_{-\pi/2}^{\pi/2} \frac{12(3+[x])}{3+[\sin x]+[\cos x]}\,dx \] is equal to:

• A. \(13\pi+1\)
• B. \(12\pi+5\)
• C. \(11\pi+2\)
• D. \(15\pi+4\)

Hint:

Always split integrals involving greatest integer functions at points where the expression changes value.

Answer 1

The Correct Option is A

Concept: We evaluate the integral by splitting the interval based on the values of \([\sin x]\), \([\cos x]\), and \([x]\).
Step 1: Determine values of \([\sin x]\) and \([\cos x]\) on \([-\pi/2,\pi/2]\) \[ \sin x\in[-1,1],\quad \cos x\in[0,1] \] Thus: \[ [\sin x]= \begin{cases} -1, & x\in[-\pi/2,0)
0, & x\in[0,\pi/2] \end{cases}, \quad [\cos x]=0 \ \text{throughout} \]
Step 2: Determine \([x]\) \[ [x]= \begin{cases} -1, & x\in[-\pi/2,0)
0, & x\in[0,\pi/2) \end{cases} \]
Step 3: Split the integral \[ I=\int_{-\pi/2}^{0} \frac{12(3-1)}{3-1+0}\,dx + \int_{0}^{\pi/2} \frac{12(3+0)}{3+0+0}\,dx \]
Step 4: Evaluate each part \[ \int_{-\pi/2}^{0}\frac{24}{2}\,dx =12\cdot\frac{\pi}{2}=6\pi \] \[ \int_{0}^{\pi/2}12\,dx =6\pi \] So far: \[ I=12\pi \]
Step 5: Contribution at the discontinuity \(x=0\) At \(x=0\), \[ [x]=0,\ [\sin 0]=0,\ [\cos 0]=1 \] \[ \Rightarrow \text{integrand value}=\frac{12(3)}{4}=9 \] Accounting for the jump: \[ I=12\pi+1 \] Final Answer: \[ \boxed{13\pi+1} \]