To find the integral of sin(x) cos(x) from 0 to π/2, we can use a simple trigonometric identity. We know that:
sin(2x) = 2 sin(x) cos(x)
This leads us to rewrite sin(x) cos(x) as:
sin(x) cos(x) = (1/2) sin(2x)
Now, we can set up the integral:
∫ sin(x) cos(x) dx = ∫ (1/2) sin(2x) dx
Now we can integrate:
∫ (1/2) sin(2x) dx = -(1/4) cos(2x) + C
Next, we will evaluate this definite integral from 0 to π/2:
∫_{0}^{π/2} sin(x) cos(x) dx = [-(1/4) cos(2x)]_{0}^{π/2}
Calculating the upper limit:
-(1/4) cos(2(π/2)) = -(1/4) cos(π) = -(1/4)(-1) = 1/4
Now calculating the lower limit:
-(1/4) cos(2(0)) = -(1/4) cos(0) = -(1/4)(1) = -1/4
Now, putting it all together:
∫_{0}^{π/2} sin(x) cos(x) dx = (1/4) - (-1/4) = 1/4 + 1/4 = 1/2
Thus, the value of the integral of sin(x) cos(x) from 0 to π/2 is:
1/2