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.
To evaluate the integral ∫ sin^2(x) dx on the interval [0, π/2] using the identity sin^2(x) + cos^2(x) = 1 and the fact that sin^2(x) and cos^2(x) are mirror images in this interval, here's how you can proceed:
1. Start by rewriting sin^2(x) using the identity: sin^2(x) = 1 - cos^2(x). This identity follows from sin^2(x) + cos^2(x) = 1, by subtracting cos^2(x) from both sides.
2. Substitute the rewritten expression into the integral: ∫ (1 - cos^2(x)) dx.
3. Distribute the integral into two separate integrals: ∫ dx - ∫ cos^2(x) dx.
4. Integrate each term separately:
- For the first integral, ∫ dx, you are integrating with respect to x, so the result is simply x.
- For the second integral, ∫ cos^2(x) dx, you can make use of the fact that cos^2(x) is the mirror image of sin^2(x) on the interval [0, π/2]. Therefore, you can replace cos^2(x) with sin^2(x), resulting in the integral ∫ sin^2(x) dx.
5. Now you have x - ∫ sin^2(x) dx.
6. Observe that the remaining integral (the one you started with) is the same as the original integral you were trying to evaluate.
7. Let's call the original integral I: I = ∫ sin^2(x) dx.
8. Now, substitute I back into the equation: x - I = ∫ sin^2(x) dx.
9. Rearrange the equation to isolate I: I = x - I.
10. Solve for I by adding I to both sides of the equation: 2I = x.
11. Divide both sides of the equation by 2: I = x/2.
12. Finally, evaluate the integral by substituting the limits of integration: I = π/2 - 0.
Therefore, the value of the integral ∫ sin^2(x) dx on the interval [0, π/2] is π/2.