The first quadrant is 0<x<90°. For y=sin(x), the curve starts at the origin (0,0) and rises to (90,1); for y=cos(x) it starts at (1,0) and finishes at (90,0). The curves intersect at (45,sqrt(2)/2) because sin(x)=cos(x) and tan(x)=1, making x=45°, where sin(x)=cos(x)=sqrt(2)/2.
The area we need can be split into two and because it's symmetrical we only need to find the area of half of it. Note that the area under cos(x) is greater than the area under sin(x) in the range 0<x<45°.
The area under y=sin(x) is given by the integral S[0,45](sin(x)dx) and under y=cos(x) S[0,45](cos(x)dx), where S denotes integral and the [low,high] the limits of integration. Evaluating the integrals we have -cos(x)[0,45]=1-sqrt(2)/2=(2-sqrt(2))/2 and sin(x)[0,45]=sqrt(2)/2. The difference of these (the area under cos(x) minus that under sin(x)) gives us the area we need sqrt(2)/2-(2-sqrt(2))/2=sqrt(2)-1. To find the total area, we double this: 2sqrt(2)-2. (We could have evaluated S[45,90] for both and added the results.) The area is 2sqrt(2)-2=0.8284 approx.