Final answer:
To evaluate the integral ∫(0 to π/2) 9 cos²(x) dx, we can use a trigonometric identity that relates the cosine squared function to the average value of the sine squared function over a complete cycle.
Step-by-step explanation:
To evaluate the integral ∫(0 to π/2) 9 cos²(x) dx, we can use a trigonometric identity that relates the cosine squared function to the average value of the sine squared function over a complete cycle. The identity states that the average value of cos²(x) is equal to the average value of sin²(x) over a complete cycle. So, we can rewrite the integral as:
∫(0 to π/2) 9 sin²(x) dx
Now, we can use a standard integral formula to evaluate this integral: ∫ sin²(x) dx = (1/2) * (x - sin(x) * cos(x)). Plugging this formula into the integral:
∫(0 to π/2) 9 sin²(x) dx = 9 * (1/2) * (π/2 - sin(π/2) * cos(π/2) - 0 + sin(0) * cos(0))
Since sin(π/2) = 1 and cos(π/2) = 0, and sin(0) = 0 and cos(0) = 1, the integral simplifies to:
∫(0 to π/2) 9 sin²(x) dx = (9/2) * (π/2)
Hence, the value of the integral ∫(0 to π/2) 9 cos²(x) dx is (9/2) * (π/2).