Question

Find the value of the integral \[\int\limits_{-2}^{2}{\dfrac{{{\sin }^{2}}x}{\left[ \dfrac{x}{\pi } \right]+\dfrac{1}{2}}}dx\] ,where [x] denotes the greatest integer less than $^{20}{{C}_{r}}$ or equal to x.
(a) 4
(b) 4 – sin 4
(c) sin 4
(d) 0

Answer 1

Hint: We know that if the function is continuous in the interval (a, b) then we can write \[\int_{b}^{a}{f\left( x \right)dx}=\int_{a}^{c}{f\left( x \right)}dx+\int_{c}^{b}{f\left( x \right)dx}\] hence in the given question also \[\left[ \dfrac{x}{\pi } \right]\] will be equal to -1 in the interval -2 to 0 and will be equal to 0 in the interval 0 to 2. So we will break the given integral into two parts first from -2 to 0 and second from 0 to 2 and then solve it further to find it’s value.

Complete step-by-step answer:
We have to find the integral \[\int\limits_{-2}^{2}{\dfrac{{{\sin }^{2}}x}{\left[ \dfrac{x}{\pi } \right]+\dfrac{1}{2}}}dx\], where [x] denotes the greatest integer function.
We know that If the function is continuous in (a, b) then integration of a function a to b will be the same as the sum of integrals of the same function from a to c and c to b. so,
\[\int_{b}^{a}{f\left( x \right)dx}=\int_{a}^{c}{f\left( x \right)}dx+\int_{c}^{b}{f\left( x \right)dx}\]
In the given integral \[\left[ \dfrac{x}{\pi } \right]\] changes it’s value at x = 0 i.e. in the interval -2 to 0 it will be equal to -1 and it will be equal to 0 and in the interval 0 to 2 . so, given interval can also be expressed as,
\[\int\limits_{-2}^{2}{\dfrac{{{\sin }^{2}}x}{\left[ \dfrac{x}{\pi } \right]+\dfrac{1}{2}}}dx=\int\limits_{-2}^{0}{\dfrac{{{\sin }^{2}}x}{-1+\dfrac{1}{2}}}dx+\int\limits_{0}^{2}{\dfrac{{{\sin }^{2}}x}{0+\dfrac{1}{2}}}dx\]
Now, we will solve the above integral further,
\[\int\limits_{-2}^{2}{\dfrac{{{\sin }^{2}}x}{\left[ \dfrac{x}{\pi } \right]+\dfrac{1}{2}}}dx=2\int\limits_{0}^{2}{{{\sin }^{2}}x}dx-2\int\limits_{-2}^{0}{{{\sin }^{2}}x}dx\]
Now, we know that ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ putting it in the above equation we get,
\[\begin{align}
  & \int\limits_{-2}^{2}{\dfrac{{{\sin }^{2}}x}{\left[ \dfrac{x}{\pi } \right]+\dfrac{1}{2}}}dx=2\int\limits_{0}^{2}{\dfrac{1-\cos 2x}{2}}dx-2\int\limits_{-2}^{0}{\dfrac{1-\cos 2x}{2}}dx \\
 & =\int\limits_{0}^{2}{\left( 1-\cos 2x \right)}dx-\int\limits_{-2}^{0}{\left( 1-\cos 2x \right)dx} \\
 & =\left[ x-\dfrac{\sin 2x}{2} \right]_{0}^{2}-\left[ x-\dfrac{\sin 2x}{2} \right]_{-2}^{0} \\
 & =\left( 2-\dfrac{\sin 4}{2} \right)-\left( 0-\left( -2-\dfrac{\sin \left( -4 \right)}{2} \right) \right) \\
 & =0 \\
\end{align}\]
Hence, we get the value of the given integral as 0.

So, the correct answer is “Option d”.

Note: To solve this problem we need to have the prior knowledge of the greatest integer function. And always try to break the integral in several corner points whenever we are given the greatest integer function, mod function etc. Students also may make mistakes while breaking the interval into several parts and assume wrong values so you need to do that carefully too.