Let f(x)= {1/3, x ≤ π / 2; (b (1- sin x))/((π-2 x)^2), x>π / 2}. If f is continuous at x=π / 2, then the value of ∫_0^3 b -6 |x^2+2 x-3| d x is:

Let $\displaystyle f(x)=\left\{\begin{array}{cc}\frac{1}{3} & , x \leq \pi / 2 \\ \frac{\mathrm{~b}(1-\sin x)}{(\pi-2 x)^2} & , x>\pi / 2\end{array}\right.$. If $\displaystyle f$ is continuous at $\displaystyle x=\pi / 2$, then the value of $\displaystyle \int_0^{3 \mathrm{~b}-6}\left|x^2+2 x-3\right| \mathrm{d} x$ is :