Question ID: #958
Let $[.]$ denote the greatest integer function. Then
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) dx$ is equal to:
- (1) $15\pi+4$
- (2) $11\pi+2$
- (3) $13\pi+1$
- (4) $12\pi+5$
Solution:
We split the integral based on the integer values of $[x]$, $[\sin x]$, and $[\cos x]$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Approximate limits: $-\frac{\pi}{2} \approx -1.57$ to $\frac{\pi}{2} \approx 1.57$.
Region 1: $x \in (-\frac{\pi}{2}, -1)$
$[x] = -2$
$\sin x \in (-1, \sin(-1)) \Rightarrow [\sin x] = -1$
$\cos x \in (0, \cos(-1)) \Rightarrow [\cos x] = 0$
Integrand: $\frac{12(3-2)}{3-1+0} = \frac{12}{2} = 6$
Area $A_1 = \int_{-\frac{\pi}{2}}^{-1} 6 dx = 6[-1 – (-\frac{\pi}{2})] = 3\pi – 6$
Region 2: $x \in (-1, 0)$
$[x] = -1$
$\sin x \in (\sin(-1), 0) \Rightarrow [\sin x] = -1$
$\cos x \in (\cos(-1), 1) \Rightarrow [\cos x] = 0$
Integrand: $\frac{12(3-1)}{3-1+0} = \frac{24}{2} = 12$
Area $A_2 = \int_{-1}^{0} 12 dx = 12[0 – (-1)] = 12$
Region 3: $x \in (0, 1)$
$[x] = 0$
$\sin x \in (0, \sin(1)) \Rightarrow [\sin x] = 0$
$\cos x \in (\cos(1), 1) \Rightarrow [\cos x] = 0$
Integrand: $\frac{12(3+0)}{3+0+0} = \frac{36}{3} = 12$
Area $A_3 = \int_{0}^{1} 12 dx = 12[1 – 0] = 12$
Region 4: $x \in (1, \frac{\pi}{2})$
$[x] = 1$
$\sin x \in (\sin(1), 1) \Rightarrow [\sin x] = 0$
$\cos x \in (0, \cos(1)) \Rightarrow [\cos x] = 0$
Integrand: $\frac{12(3+1)}{3+0+0} = \frac{48}{3} = 16$
Area $A_4 = \int_{1}^{\frac{\pi}{2}} 16 dx = 16[\frac{\pi}{2} – 1] = 8\pi – 16$
Total Integral $I = A_1 + A_2 + A_3 + A_4$
$$I = (3\pi – 6) + 12 + 12 + (8\pi – 16)$$
$$I = 11\pi + 2$$
Ans. (2)
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