Definite Integral

Look at the second integral, it can rewritten as:

$\displaystyle \int_{-2}^1\sin((x+6)^2-3)\, dx$

Use the substitution, $x+6=-t$ , the lower limit change to $-4$ and upper limit to $-5$ . Also, $dx=-dt$ . Hence, the second integral is equivalent to

$\displaystyle -\int_{-4}^{-5} \sin(t^2-3)\,dt$

This is same as the first integral with just the opposite sign. Hence, $A=0$ .

$\displaystyle \Rightarrow \boxed{A+1=1}$