Question

What is the value of $\underset{0}{\overset{1}{\int }}\cos \left(\pi x\right)\cos \left(\left[2x\right]\pi \right)dx$ ?
(Here $[t]$ denotes the integral part of the real number $t$ )

• A. $1$
• B. $-1$
• C. $\frac{2}{\pi }$
• D. $\frac{-2}{\pi }$

Answer 1

Answer: (C)

Let
$I=\underset{0}{\overset{1}{\int }}\cos \left(\pi x\right)\cos \left[2x\right]\pi dx$
$\Rightarrow I=\underset{0}{\overset{1/2}{\int }}\cos \left(\pi x\right)\cos 0dx+\underset{1/2}{\overset{1}{\int }}\cos \pi x\cos \pi dx$
$\Rightarrow I=\underset{0}{\overset{1/2}{\int }}\cos \pi xdx-\underset{1/2}{\overset{1}{\int }}\cos \pi xdx$
$\Rightarrow I={\left[\frac{\sin \pi x}{\pi }\right]}_0^{1/2}-{\left[\frac{\sin \pi x}{\pi }\right]}_{1/2}^1$
$\Rightarrow I=\frac{1}{\pi }\left[\sin \frac{\pi }{2}-\sin 0\right]-\frac{1}{\pi }\left[\sin \pi -\sin \frac{\pi }{2}\right]$
$\Rightarrow I=\frac{1}{\pi }\left[1-0\right]-\frac{1}{\pi }\left[0-1\right]=\frac{2}{\pi }$