Indefinite Integration 2 Question 1
1. The value of $\int _{-\pi / 2}^{\pi / 2} \frac{d x}{[x]+[\sin x]+4}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is
(a) $\frac{1}{12}(7 \pi-5)$
(b) $\frac{1}{12}(7 \pi+5)$
(c) $\frac{3}{10}(4 \pi-3)$
(d) $\frac{3}{20}(4 \pi-3)$
(2019 Main, 10 Jan II)
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Answer:
Correct Answer: 1. (d)
Solution:
- Let $I=\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{[x]+[\sin x]+4}$
$$ =\int _{\frac{-\pi}{2}}^{-1} \frac{d x}{[x]+[\sin x]+4}+\int _{-1}^{0} \frac{d x}{[x]+[\sin x]+4} $$
$+\int _0^{1} \frac{d x}{[x]+[\sin x]+4}+\int _1^{\frac{\pi}{2}} \frac{d x}{[x]+[\sin x]+4}$
$$ \because \quad[x]=\begin{array}{ll} -2, & -\pi / 2<x<-1 \\ -1, & -1 \leq x<0 \\ 0, & 0 \leq x<1 \\ 1, & 1 \leq x<\pi / 2 \end{array} $$
and $[\sin x]=\begin{array}{cc}-1, & -\pi / 2<x<-1 \ -1, & -1<x<0 \ 0, & 0<x<1 \ 0, & 1<x<\pi / 2\end{array}$
$[\because$ For $x<0,-1 \leq \sin x<0$ and for $x>0,0<\sin x \leq 1]$
So, $\quad I=\int _{\frac{-\pi}{2}}^{-1} \frac{d x}{-2-1+4}+\int _{-1}^{0} \frac{d x}{-1-1+4}+\int _0^{1} \frac{d x}{0+0+4}$
$$ \begin{aligned} & =\int _{\frac{\pi}{2}}^{-1} \frac{d x}{1}+\int _{-1}^{0} \frac{d x}{2}+\int _0^{1} \frac{d x}{4}+\int _1^{\frac{\pi}{2}} \frac{d x}{5} \\ & =-1+\frac{\pi}{2}+\frac{1}{2}(0+1)+\frac{1}{4}(1-0)+\frac{1}{5} \frac{\pi}{2}-1 \\ & =-1+\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{\pi}{2}+\frac{\pi}{10} \\ & =\frac{-20+10+5-4}{20}+\frac{5 \pi+\pi}{10} \\ & =-\frac{9}{20}+\frac{3 \pi}{5}=\frac{3}{20}(4 \pi-3) \end{aligned} $$
