find the area of the region enclosed by the curves y=cos(px÷2) and y=1-x^2where x is variable in interval [-1,1]\nHINT:If f(x) is even then f(x)=f(-x).

since, from curve you can see the both the function are even valued , you can integrate function between 0 to 1 and take the twice the value of integral, To calculate area integrate
1-x^2-cos (px÷2)between 0 to 1 and take twice of this integral , [x-x^3/3-(2/p)*sin(px÷2)], between 0 and 1, it comes = 1-1/3-(2/p)=.03, and twice this is .06.