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