Not able to solve it

Question

Find area between curve

y=4cos(6x), y=4sin(12x), x=0, x=pi/12

Answer 1

So you're taking the integral from 0 to pi/12 of the the upper function minus the lower function. Since you have the first part of the integral with the cosine graph on top, and the second part with the sine graph on top, you're likely expected to split the integral into those two parts and then integrate. Now these graphs are equal when cos(6x) = sin(12x). Using double angle formulas, sin(12x) = 2sin(6x)cos(6x), so sin(6x) = 1/2. This would occur when 6x = pi/6, so x=pi/36 is where the two graphs meet.

So your first integral is from 0 to pi/36 of 4cos(6x) - 4sin(12x), giving you an area of 1/6. Your second integral is from pi/36 to pi/12 of 4sin(12x) - 4cos(6x), also giving you an area of 1/6. So the total area between these two graphs is 1/3.