I would like to evaluate the follwoing integral for $kr \gg 1$ and in the limit $\Lambda \rightarrow \infty$: $$ \int_{-2k-\Lambda}^{2k+\Lambda} \frac{\sin[(q-2k)r]\cos(2kr) + \cos[(q-2k)r]\sin(2kr)}{q-2k}dq.$$
The result should be $\pi cos(2kr)$. However, I don't know to to get to this result. What I have tried so far is a substitution $x = (q-2k)r$ which did not help much.
Hint:
Use the property of sines that $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
In your case with $$A=qr-2kr$$ and $$B=2kr$$