If $\int\limits_{0}^{1}\frac{\sin t}{1+t}dt=α$, then the value of the integral $\int\limits_{4π-2}^{4π}\frac{\sin\frac{t}{2}}{4π+2-t}dt$ in terms of $α$ is given by

$-α$

Let

$I=\int\limits_{4π-2}^{4π}\frac{\sin\frac{t}{2}}{4π+2-t}dt=\frac{1}{2}\int\limits_{4π-2}^{4π}\frac{\sin\frac{t}{2}}{(2π-\frac{t}{2})-1}dt$

$⇒I=\int\limits_{0}^{1}\frac{\sin (2 π-u)}{1+u}du$, where $2π-\frac{t}{2}=u$

$⇒I=-\int\limits_{0}^{1}\frac{\sin u}{1+u}du=-α$