Let `f` be the function defined on `[-pi,pi]` given by `f(0)=9` and `f(x)=sin((9x)/2)/sin(x/2)` for `x!=0`. The value of `2/pi int_-pi^pif(x)dx` is (asked as Match the following question)

Options

A. 0
B. 2
C. 4
D. 6

Answer

C

Explanation

Let `I=(2)/(pi)overset(pi)underset(-pi)int f(x)dx`. Then,
`I=(2)/(pi)overset(pi)underset(-pi)int ("sin"((9x)/(2)))/("sin"((x)/(2)))dx`
`rArrI=(4)/(pi)overset(pi)underset(0)int ("sin"(9x)/(2))/("sin"(x)/(2))dx`
`rArr I=(8)/(pi)overset(pi//2)underset(0)int (sin9theta)/(sintheta)d theta`
`rArr I=((sin9 theta-sin7theta)+(sin7theta-sin5theta)+(sin5theta-sin3theta)+(sin3theta-sintheta)+sintheta)/(sin theta)d theta`
`rArr I=(8)/(pi)overset(pi//2)underset(pi)(cos8 theta+cos 6theta+cos 4theta+cos 2 theta+1)d theta`
`rArrI=(8)/(pi)xx(pi)/(2)=4`