First integrate `g(x) = int_(1-2x)^(1+2x) tsint dt` by parts

Use

`int u (dv)/(dt) dt = uv - int v (du)/(dt) dt`

where `u = t` `implies (du)/(dt) = 1`

and `(dv)/(dx) = sint` `implies v = -cost`

`therefore int_(1-2x)^(1+2x) tsint dt = -tcost|_(1-2x)^(1+2x) - int_(1-2x)^(1+2x) -cost dt `

`= (1-2x)cos(1-2x) - (1+2x)cos(1+2x) - (-sint)|_(1-2x)^(1+2x)`

`= sin(1+2x) - sin(1-2x) + (1-2x)(cos(1-2x) - cos(1+2x))`

Then `g'(x) = 2cos(1+2x) + 2cos(1-2x) -2cos(1-2x) + 2cos(1+2x)`

`+ (1+2x)(2sin(1-2x) +2sin(1+2x))`

`= 4cos(1+2x) + 2(1+2x)(sin(1-2x) + sin(1+2x))` **answer**

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