If `g(x)=int_0^xcos^4tdt ,` then `g(x+pi)` equals (a)`g(x)+g(pi)` (b) `g(x)-g(pi)` (c)`g(x)g(pi)` (d) `(g(x))/(g(pi))`

Options

A. `g(x) +g (pi)`
B. `g(x)-g (pi)`
C. `g(x)g(pi)`
D. `(g(x))/(g(pi))`

Explanation

Given , f (x) `=int_(0)^(x)cos^(4)t dt`
`rArrg(x+pi)=int_(0)^(pi+x)cos^(4)t dt`
` = int_(0)^(pi)cos^(4)t dt+int_(pi)^(pi+x)cos^(4)t dt = I_(1)+I_(2)` where , `I_(1)=int_(0)^(pi)cos^(4)t dt=g(pi)`
and `I_(2)=int_(pi)^(pi+x)cos^(4)t dt`
Put `t=pi+y`
`rArrdt = dy`
`I_(2)=int_(0)^(x)cos^(4)(y+pi)dy`
`=int_(0)^(x)(-cosy)^(4)dy=int_(0)^(x)cos^(4)ydy=g(x)`
`:. g(x+pi)=g(pi)+g(x)`