Answer
$$\int_{0}^{\pi/2}(2-sin\theta)^{2}d\theta=\frac{9}{4}\pi-4$$
Work Step by Step
$$\int_{0}^{\pi/2}(2-sin\theta)^{2}d\theta=\int_{0}^{\pi/2}(4-4sin\theta+sin^{2}\theta)d\theta$$
$$=\int_{0}^{\pi/2}(4-4sin\theta+\frac{1-cos2\theta}{2})d\theta$$
$$=\left |\frac{9}{2}\theta+4cos\theta-\frac{sin2\theta}{4} \right |_{0}^{\pi/2}$$
$$=\frac{9}{4}\pi+4*0-\frac{sin\pi}{4}-4*1+\frac{sin0}{4}$$
$$=\frac{9}{4}\pi-4$$
