If $\int^1_0 \frac{\sin x}{1+x}dx =I $ then.... Problem : 
If $\displaystyle \int^1_0 \dfrac{\sin x}{1+x}dx =I $ then $\displaystyle \int^{4\pi}_{4\pi-2} \dfrac{\sin\frac{x}{2}}{4\pi +2-x}dx  $ =? 
Options are : 
(a) $I$ (b) $-I$ (c) $2I$  (d) $-2I$.
Please suggest how to proceed in such problem and how to make to integrals comparable to get the relation between them, will be of great help thanks. 
 A: Given $$\displaystyle \int_{0}^{1}\frac{\sin x }{1+x} = I\;,$$  and  $$\displaystyle \int_{4\pi-2}^{4\pi}\frac{\sin \frac{x}{2}}{4\pi+2-x}dx$$
Now Let $$\displaystyle J = \int_{4\pi-2}^{4\pi}\frac{\sin \frac{x}{2}}{4\pi+2-x}dx$$
Using the formula $$\displaystyle \bullet\; \int_{a}^{b}f(x)dx = (b-a)\int_{0}^{1}f\left[(b-a)x+a\right]dx$$
So we get $$\displaystyle J = 2\int_{4\pi-2}^{4\pi}\frac{\sin \left(\frac{2 x+4\pi-2}{2}\right)}{4\pi+2-\left(2x+4\pi-2\right)}dx = 2\int_{0}^{1}\frac{\sin \left(2\pi+x-1\right)}{2(2-x)}dx$$
So we get $$\displaystyle J = \int_{0}^{1}\frac{\sin (x-1)}{1+(1-x)}dx = -\int_{0}^{1}\frac{\sin (1-x)}{1+(1-x)}dx$$
Now Put $(1-x) = t$ and $dx = -dt$ and changing limit , we get 
$$\displaystyle J = -\times -\int_{1}^{0}\frac{\sin t}{1+t} dt = -\int_{0}^{1}\frac{\sin t}{1+t}dt = -\int_{0}^{1}\frac{\sin x}{1+x}dx = -I$$
Above we have used the formulae
$$\displaystyle \bullet\ \int_{a}^{b} f(x)dx = -\int_{b}^{a}f(x)dx$$ and $$\displaystyle \bullet\; \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$
A: Hint:
$$
\int_{x} \frac{\sin \frac{x}{2}}{4\pi + 2 -x} = \int_{x = 4\pi - 2u} \frac{-\sin u}{2(1+u)}(-2) = \int_{u}\frac{\sin u}{1+u}.
$$
A: $$\int_{4\pi-2}^{4\pi}\frac{\sin(x/2)}{4\pi+2-x}dx\underset{y=x-4\pi}{=}\int_{-2}^0\frac{\sin(\frac{y}{2})}{2-y}dy$$
Now, do the substitution $u=-\frac{y}{2}$ to conclude.
