Question

The value of $\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}$ where $\left[ + \right]$ denote greatest integer Less than or equal to $+$ , is :
(a) $\dfrac{1}{12}\left( 7\pi +5 \right)$
(b) $\dfrac{3}{10}\left( 4\pi -3 \right)$
(c) $\dfrac{1}{12}\left( 7\pi -5 \right)$
(d) $\dfrac{3}{20}\left( 4\pi -3 \right)$

Answer 1

We are given an integral which contains a greatest integer function, so to simplify our function we will split our limit of the integral into parts then when we have the integral in the perfect form we will take the constant terms outside and then integrate it with respect to and then simplify and get the required answer.

Complete step-by-step answer:
We know that $\left[ + \right]$ is given as the greatest integer function. To find the integral of our function we will check how the function will behave on a different interval.
We know that on $\left[ -\dfrac{\pi }{2},-1 \right]$ we have $\left[ x \right]=-2\text{ }and\text{ }\left[ \sin x \right]=-1............\left( 1 \right)$
While on $\left( -1,0 \right)$
$\left[ x \right]=-1\text{ }and\text{ }\left[ \sin x \right]=-1...........\left( 2 \right)$
On $\left( 0,1 \right)$ we will get as follows,
$\left[ x \right]=0\text{ }and\text{ }\left[ \sin x \right]=0...............\left( 3 \right)$
And lastly on $\left( 1,\dfrac{\pi }{2} \right]$
$\left[ \sin x \right]=0\text{ }and\text{ }\left[ x \right]=1............\left( 4 \right)$
So now, we will split our interval of integral as follows,
$\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}=\int\limits_{-\dfrac{\pi }{2}}^{-1}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}+\int\limits_{-1}^{0}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}+\int\limits_{0}^{1}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}+\int\limits_{1}^{\dfrac{\pi }{2}}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}$
Now using the equations, $\left( 1 \right),\left( 2 \right),\left( 3 \right)\text{ }and\text{ }\left( 4 \right)$ , we will get,
$\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}=\int\limits_{-\dfrac{\pi }{2}}^{-1}{\dfrac{dx}{-2-1+4}}+\int\limits_{-1}^{0}{\dfrac{dx}{-1-1+4}}+\int\limits_{0}^{1}{\dfrac{dx}{0+0+4}}+\int\limits_{1}^{\dfrac{\pi }{2}}{\dfrac{dx}{1+0+4}}$
After simplification , we will get
$\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\dfrac{dx}{\left[ x \right]+\left[ \sin x \right]+4}}=\int\limits_{-\dfrac{\pi }{2}}^{-1}{\dfrac{dx}{1}}+\int\limits_{-1}^{0}{\dfrac{dx}{2}}+\int\limits_{0}^{1}{\dfrac{dx}{4}}+\int\limits_{1}^{\dfrac{\pi }{2}}{\dfrac{dx}{5}}$
Now we will take the constant out of above integral. So, we will get,
$\Rightarrow \int\limits_{-\dfrac{\pi }{2}}^{-1}{dx}+\dfrac{1}{2}\int\limits_{-1}^{0}{dx}+\dfrac{1}{4}\int\limits_{0}^{1}{dx}+\dfrac{1}{5}\int\limits_{1}^{\dfrac{\pi }{2}}{dx}$
As $\int\limits_{a}^{b}{dx}=\left( x \right)_{a}^{b}$ so we can write,
$=\left( x \right)_{-\dfrac{\pi }{2}}^{-1}+\dfrac{1}{2}\left( x \right)_{-1}^{0}+\dfrac{1}{4}\left( x \right)_{0}^{1}+\dfrac{1}{5}\left( x \right)_{1}^{\dfrac{\pi }{2}}$
Now, after putting the limit we will get,

& =\left( -1+\dfrac{\pi }{2} \right)+\dfrac{1}{2}\left( 0+1 \right)+\dfrac{1}{4}\left( 1-0 \right)+\dfrac{1}{5}\left( \dfrac{\pi }{2}-1 \right) \\\ & \\\ \end{aligned}$$ Now on simplifying the above expression, we will get, $=\dfrac{-9}{20}+\dfrac{3\pi }{5}$ $=\dfrac{3}{20}\left( 4\pi -3 \right)$ So the correct option is option (d) $\dfrac{3}{20}\left( 4\pi -3 \right)$ **So, the correct answer is “Option (d)”.** **Note:** On $\left( -\dfrac{\pi }{2},-1 \right)$ , $\left[ n \right]=-2$ as we have that $-\dfrac{\pi }{2}=-1.57$ So for $nt\left( -1.57,-1 \right)$ the $\left[ n \right]$ = greatest integer less than or equal to $x$ is $-2$ So $\left[ n \right]=-2$ here. Similarly on $\left( -1,0 \right)$ , we get the value of $\left[ n \right)=-1$ . While simplifying we add only like terms that are, $$\pi $$ multiple are added with other $$\pi $$ multiple are added with other $$\pi $$ multiple and constant with each other.