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.

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

To evaluate the integral ∫sin^2(x)dx over the interval (0, π/2), we can use the given trigonometric identity, sin^2(x) + cos^2(x) = 1.

Notice that sin^2(x) and cos^2(x) are mirror images in the interval [0, π/2]. This means that the integral of sin^2(x) and cos^2(x) over this interval will be the same.

Since sin^2(x) + cos^2(x) = 1, we can rewrite the integral as follows:

∫sin^2(x)dx = ∫(1 - cos^2(x))dx

Now, we can break down the integral:

∫(1 - cos^2(x))dx = ∫dx - ∫cos^2(x)dx

The integral of dx is simply x, and the integral of cos^2(x)dx can be evaluated using the power reduction formula:

∫cos^2(x)dx = (∫(1 + cos(2x))/2)dx = (1/2)∫(1 + cos(2x))dx = (1/2)(x + (1/2)sin(2x)) + C

(where C is the constant of integration)

Now, substituting these values back into the original equation:

∫sin^2(x)dx = x - (1/2)(x + (1/2)sin(2x)) + C

Since we are evaluating the integral over the interval (0, π/2), we can plug in the limits of integration:

∫sin^2(x)dx = [x - (1/2)(x + (1/2)sin(2x))] from x = 0 to x = π/2

(Here, we evaluate the expression at the upper limit and subtract the expression evaluated at the lower limit)

Plugging in these values:

∫sin^2(x)dx = [(π/2) - (1/2)((π/2) + (1/2)sin(π))] - [0 - (1/2)(0 + (1/2)sin(0))]

Simplifying further:

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

= [π/2 - π/4] - [0]

= π/4

Therefore, the value of the integral ∫sin^2(x)dx over the interval (0, π/2) is π/4.