Question
The value of $$\int_{ - 2}^2 {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}dx,} $$ where $$\left[ x \right] = $$ the greatest integer greater than or equal to $$x,$$ is :
A.
1
B.
0
C.
$$4 - \sin \,4$$
D.
none of these
Answer :
0
Solution :
$$\eqalign{
& I = \int_{ - 2}^0 {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}dx} + \int_0^2 {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}dx} \cr
& \,\,\,\,\, = \int_{ - 2}^0 {\frac{{{{\sin }^2}x}}{{ - 1 + \frac{1}{2}}}dx} + \int_0^2 {\frac{{{{\sin }^2}x}}{{0 + \frac{1}{2}}}dx} \cr
& \,\,\,\,\, = 2\int_0^{ - 2} {{{\sin }^2}x\,dx} + 2\int_0^2 {{{\sin }^2}x\,dx} \cr
& \,\,\,\,\, = 2\int_0^2 {{{\sin }^2}\left( { - x} \right)\,d\left( { - x} \right)} + 2\int_0^2 {{{\sin }^2}x\,dx} \cr
& \,\,\,\,\, = 0 \cr} $$