Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.

Answer 1

Let V = ∫sin^2x dx
Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V

Now, since sin^2x+cos^2x = 1,

2V = ∫[0,π/2] 1 dx = π/2
V = π/4

check:

sin^2x = (1-cos2x)/2
∫sin^2x dx = ∫(1-cos2x)/2 dx
= 1/2 (x - 1/2 sin2x) [0,π/2]
= 1/2[(π/2)-(0)]
= π/4