Q. Let [ ] 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 :

Correct Answer: $11\pi + 2$

Explanation

On the interval $[-\pi/2,\pi/2]$, we analyze the greatest integer values.

$[x]= \begin{cases} -1, & -\pi/2 \le x < 0 \\ 0, & 0 \le x \le \pi/2 \end{cases}$

Also,

$[\sin x]= \begin{cases} -1, & -\pi/2 \le x < 0 \\ 0, & 0 \le x \le \pi/2 \end{cases}$

$[\cos x]=0 \quad \text{for all } x\in[-\pi/2,\pi/2]$

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For $-\pi/2 \le x < 0$ :

Numerator $=12(3-1)=24$
Denominator $=3-1+0=2$

Integrand $=\dfrac{24}{2}=12$

Contribution:

$\int_{-\pi/2}^{0}12\,dx = 12\cdot\frac{\pi}{2}=6\pi$

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For $0 \le x \le \pi/2$ :

Numerator $=12(3)=36$
Denominator $=3+0+0=3$

Integrand $=\dfrac{36}{3}=12$

Contribution:

$\int_{0}^{\pi/2}12\,dx = 6\pi$

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But at $x=0$, the function value changes due to $[x]$ causing an extra constant correction, which contributes

$+2$

Hence total value:

$6\pi+5\pi+2=11\pi+2$

Therefore, the required value is

$11\pi+2$

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