Answer
$\sqrt 2-1$
Work Step by Step
Step 1. See figure. Find the intersections between the two trig functions. Letting $sin(x)=cos(x)$, we have $x=\frac{\pi}{4}$ for the enclosed region of concern.
Step 2. The enclosed area between the two trig functions and the y-axis can be found as
$A=\int_0^{\pi/4} cos(x)-sin(x)dx =sin(x)|_0^{\pi/4}+cos(x)|_0^{\pi/4}=sin(\pi/4)-sin(0)+cos(\pi/4)-cos(0)=\sqrt 2-1$
