Question

What is the value of $\int\limits_{0}^{1}\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=\int\limits_{0}^{1}\cos \left(\pi x\right)\cos\left[2x\right]\pi\,dx $
$\Rightarrow I=\int\limits_{0}^{1 /2}\cos\left(\pi x\right)\cos\,0\,dx+\int\limits_{1 /2}^{1}\cos\,\pi x\,\cos\,\pi dx$
$\Rightarrow I=\int\limits_{0}^{1 /2}\cos\,\pi\,xdx -\int\limits_{1 /2}^{1}\cos\,\pi\,xdx$
$\Rightarrow I=\left[\frac{\sin\,\pi x}{\pi}\right]_{0}^{1 /2}-\left[\frac{\sin\,\pi x}{\pi}\right]_{1 /2}^{1}$
$\Rightarrow I=\frac{1}{\pi}\left[\sin \frac{\pi}{2}-\sin\,0\right]-\frac{1}{\pi}\left[\sin\,\pi-\sin \frac{\pi}{2}\right]$
$\Rightarrow I=\frac{1}{\pi}\left[1-0\right]-\frac{1}{\pi}\left[0-1\right]=\frac{2}{\pi}$