The value of `int_0^x[cost]dt ,x in [(4n+1)pi/2,(4n+3)pi/2]a n dn in N ,` is equal to where [.] represents greatest integer function. `pi/2(2n-1)-2x` `pi/2(2n-1)+x` `pi/2(2n+1)-x` (d) `pi/2(2n+1)+x`

Options

A. `(pi)/2(2n-1)-2x`
B. `(pi)/2(2n-1)+x`
C. `(pi)/2(2n+1)-x`
D. `(pi)/2(2n+1)+x`

Answer

C

Explanation

`I=int_(0)^(x)[cost]dt`
`=int_(0)^(2npi)[cost]dt+int_(2npi)^(x)[cost]dt`
`=n int_(0)^(2pi) [cost]dt+int_(2npi)^(2npi+pi//2) [cost]dt+int_(2npi+(pi)/2)^(x)[cost]dt`
`=-npi+0+(x-(2npi+pi//2))(-1)`
`=-npi+2npi+pi//2-x`
`=(2n+1)pi//2 -x`