Answer
$\pi $
Work Step by Step
Consider $f(x)=\int_{-\pi/2}^{\pi/2} (1-\sin x) dx$
Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
This implies that $\int_{-\pi/2}^{0} (1-\sin x) dx+\int_{0}^{\pi/2} (1-\sin x) dx=[x-\cos x]_{-\pi/2}^{0}+[x-\cos x]_{0}^{\pi/2}$
Hence $f(x)=(0+\dfrac{\pi}{2})-(\cos 0+\cos (\dfrac{\pi}{2})]+(\dfrac{\pi}{2}-0)-(cos (\dfrac{\pi}{2})-\cos 0)]=\pi$
