If `int_0^1(sint)/(1+t)dt=alpha,` then the value of the integral `int_(4pi-2)^(4pi)(sin(t/2))/(4pi+2-t)dt` is (1)`2alpha` (2) `-2alpha` (3) `alpha` (4) `-alpha`

Explanation

`I=1/2int_(4pi-2)^(4pi)("sin"t/2)/(1+(2pi-t/2))dt`
Put `2pi-t/2=z`
`:.-1/2dt=dz` or `dt=-2dz`
When `t=4pi-2,z=2pi-2pi+1=1`
when `t=4pi,z=2pi-2pi=0`
`:.I=1/2int_(1)^(0)(sin(2pi-z)(-2dz))/(1+z)`
`=int_(0)^(1)(-sin z dz)/(z+1)`
`=-int_(0)^(1)(sint)/(1+t)dt`
`=-alpha`