A calculus problem by Akshat Sharda

$\begin{aligned} I & = \int_{4\pi-2}^{4\pi} \frac {\sin \frac x2}{4\pi+2-x} \ dx & \small \color{#3D99F6} \text{Let } 2u = 4\pi + 2 - x \implies 2\ du = - dx \\ & = \int_2^1 \frac {-2\sin (2\pi+1-u)}{2u} \ du \\ & = \int_1^2 \frac {\sin (1-u)}u \ du & \small \color{#3D99F6} \text{Let }-t =1-u \implies dt = du \\ & = \int_0^1 \frac {\sin (-t)}{t+1} \ dt & \small \color{#3D99F6} \text{Replace }t \text{ with }x \\ & = - \int_0^1 \frac {\sin x}{x+1} \ dx \end{aligned}$

Now consider

$\begin{aligned} \alpha & = \int_0^1 \frac {\cos x}{(x+1)^2} \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \frac {\cos x}{x+1} \bigg|_0^1 + \int_0^1 \frac {-\sin x}{x+1} \ dx \\ & = - \frac {\cos 1}2 + 1 + I \end{aligned}$

$\begin{aligned} \implies I & = \boxed{\dfrac {\cos 1}2 - 1 + \alpha} \end{aligned}$