The value of `int_(-pi)^( pi)sum_(r=0)^(999)cos rx(1+sum_(r=1)^(999)sin rx)`dx, is

Options

A. `2pi`
B. `999pi`
C. `0`
D. `pi`

Answer

A

Explanation

`I=int_(-pi)^(pi) sum_(r=0)^(999) cos rx (1+ sum_(r=1)^(999) sin rx)dx`
`:. I=int_(-pi)^(pi) sum_(r=0)^(999) cosrx(1-sum_(r=1)^(999) sinrx)dx` (Replacing `x` by `-x`)
Adding we get
`2I=2int_(-pi)^(pi)sum_(r=0)^(999) cos rx dx`
`implies I=int_(-pi)^(pi) 1 dx+int_(-pi)^(pi)sum_(r=1)^(999)cos rx dx`
`=2pi+2 int_(0)^(pi) sum_(r=1)^(999)cos rx dx`
`=2pi + [sum_(r=1)^(999) sin rx ]_(0)^(pi)=2pi+0=2pi`