4. Find the area of the region in the first quadrant $x>0, y>0$ bounded above the graph of $y=\arcsin (x)$ and below the graph of the $y=\arccos (x)$.
Solution: We can integrate over $y$ rather than $x$. In particular, the solution is $\int_{0} \sin y \mathrm{~d} y+$ $\int_{\pi / 4}^{\pi / 2} \cos y \mathrm{~d} y=\left(1-\frac{1}{\sqrt{2}}\right) 2=2-\sqrt{2}$.
5. What is the minimum vertical distance between the graphs of $2+\sin (x)$ and $\cos (x)$ ?
Solution: The derivative of $2+\sin (x)-\cos (x)$ is $\cos x+\sin x$, which in the interval $0 \leq x<2 \pi$ is zero at $x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$. At $\frac{7 \pi}{4}$, when $\sin (x)$ is negative and $\cos (x)$ is positive, the distance reaches its minimal value of $2-\sqrt{2}$.