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Show that int0^(npi+v)|sinx|dx=2n+1-cosv...

Show that `int_0^(npi+v)|sinx|dx=2n+1-cosv ,` where `n` is a positive integer and `,lt=v

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`int_(0)^(npi+v)|sinx|dx = int_(0)^(pi) |sinx| dx = int_(pi)^(2pi)|sinx|dx + . . . +int_((n-1)_(pi))^(npi)|sinx| dx + int_(npi)^(npi+v) |sinx|dx`
`= sum_(r=1)^(n)int_((r-1)_(pi))^(rpi)|sinx|dx+ int_(npi)^(npi+v)|sinx|dx`
Now to solve , `int_((r-1)_(pi))^(rpi)|sinx|dx ` , we have
` x= (r-1)pi+t`
`rArr sinx = sin [ (r-1)pi+t] = (-1)^(r-1)sin t` and when `x=(r-1)pi, t =0` and when
`x=rpi, t = pi`
`int_((r-1)_(pi))^(rpi)|sinx|dx=int_(0)^(pi) |(-1)^(r-1)sint dt`
`= int_(0)^(pi) |sint|dt = int_(0)^(pi) sin t dt`
` = [ - cos t] _(0)^(pi) =- cos pi + cos 0=2`
Again , `int_(npi)^(npi+v)|sinx|dx, ` putting ` x = n pi +t`
Then , ` int _(npi)^(npi+v)|sinx|dx= int _(0)^(v)|(-1)^(n) sin t | dt = int _(0)^(v) sin t dt`
`[-cost ] _(0)^(v)=- cos v+ cos 0=1- cos v`
`:. int _(0)^(npi+v)|sinx| dx = sum_(r=1)^(n)int_((r-1)_(pi))^(rpi)|sinx| dx = int _(npi)^(npi+v)|sinx| dx`
`=sum _(r=1)^(n)2+int_(npi)^(npi+v)|sinx| dx`
`= 2n +1- cos v`
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