Let the area of the region enclosed by the curve $y=\min \{\sin x, \cos x\}$ and the $x$ axis between $x=-\pi$ to $x=\pi$ be $A$. Then $A^2$ is equal to ___________

Solution

$\begin{aligned} & y=\min \{\sin x, \cos x\} \\ & x-\text { axis } \quad x-\pi \quad x=\pi\end{aligned}$

$\begin{aligned} & \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} \\ & \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} \\ & =(\cos x+\sin x)_{-3 \pi / 4}^{-\pi} \\ & =(-1+0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) \\ & =-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ & \int_{\pi / 4}^{\pi / 2} \cos x d x=(\sin x)_{\pi / 4}^{\pi / 2}=1-\frac{1}{\sqrt{2}} \\ & A=4 \\ & A^2=16\end{aligned}$

Hence, the correct answer is 16.