Trigonometric area problems frequently appear in JEE. Key results: Area under one arch of y = sin(x) (from 0 to pi) = 2. Area under y = cos(x) from 0 to pi/2 = 1. Area between y = sin(x) and y = cos(x) from 0 to pi/4: integral from 0 to pi/4 of (cos(x) - sin(x)) dx = [sin(x) + cos(x)] from 0 to pi/4 = sqrt(2) - 1. For area between y = sin(2x) and y = sin(x) from 0 to pi/3: first find intersection where sin(2x) = sin(x), i.e., 2sin(x)cos(x) = sin(x), giving sin(x)(2cos(x) - 1) = 0, so x = 0 or cos(x) = 1/2, i.e., x = pi/3. Then integrate (sin(2x) - sin(x)) from 0 to pi/3.