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If g(x)=int(0)^(x)cos^(4) t dt , then ...

If `g(x)=int_(0)^(x)cos^(4)` t dt , then ` (x+pi)` equals

A

`g(x) +g (pi)`

B

`g(x)-g (pi)`

C

`g(x)g(pi)`

D

`(g(x))/(g(pi))`

Text Solution

Verified by Experts

The correct Answer is:
A
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)`
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