Find the value of the integral 1 2cosx 2 cosx2 dx from 0 to 2

To find the value of the integral ∫ (1 + 2cos(x) + 2cos(x²)) dx from 0 to 2, we can break it down into simpler parts.

First, we split the integral:

∫ (1 + 2cos(x) + 2cos(x²)) dx = ∫ 1 dx + 2∫ cos(x) dx + 2∫ cos(x²) dx

Now, we evaluate each part:

1. ∫ 1 dx from 0 to 2:

   This is straightforward, and it equals:

   [x] (from 0 to 2) = 2 – 0 = 2.

2. 2∫ cos(x) dx from 0 to 2:

   The integral of cos(x) is sin(x), so we calculate:

   2[sin(x)] (from 0 to 2) = 2(sin(2) – sin(0)) = 2sin(2).

3. 2∫ cos(x²) dx from 0 to 2:

   This integral does not have a simple antiderivative in terms of elementary functions. However, it can be computed numerically. Let’s denote this part as:

   Let I = 2∫ cos(x²) dx from 0 to 2.

Now, we need to add all three parts together:

Result = 2 + 2sin(2) + I.

To get a numerical approximation for I (the integral of 2cos(x²)), numerical integration methods, like Simpson’s rule or using a calculator, can be employed.

In conclusion, the total value of the integral depends on the numerical computation of the last term, plus the constants calculated.