Let's solve the integral step by step:

$\int_0^{2\pi} \left[ 2\cos t + t \cos t + \sin^2 t \right] \, dt$

Step 1: Break the integral into parts

$\int_0^{2\pi} \left[ 2\cos t + t \cos t + \sin^2 t \right] \, dt = \int_0^{2\pi} 2\cos t \, dt + \int_0^{2\pi} t \cos t \, dt + \int_0^{2\pi} \sin^2 t \, dt$

Step 2: Solve each term

(a) First term: $\int_0^{2\pi} 2\cos t \, dt$

The integral of $2\cos t$ is: $\int 2\cos t \, dt = 2\sin t$ Evaluate over $[0, 2\pi]$: $2\sin t \Big|_0^{2\pi} = 2\sin(2\pi) - 2\sin(0) = 0 - 0 = 0$

(b) Second term: $\int_0^{2\pi} t\cos t \, dt$

This requires integration by parts: Let $u = t$ and $dv = \cos t \, dt$, so $du = dt$ and $v = \sin t$.

Using the integration by parts formula: $\int u \, dv = uv - \int v \, du$ We get: $\int t \cos t \, dt = t \sin t - \int \sin t \, dt$ The integral of $\sin t$ is $-\cos t$, so: $\int t \cos t \, dt = t \sin t + \cos t$ Evaluate over $[0, 2\pi]$: $\left[ t \sin t + \cos t \right]_0^{2\pi} = \left( 2\pi \sin(2\pi) + \cos(2\pi) \right) - \left( 0 \sin(0) + \cos(0) \right)$ Since $\sin(2\pi) = 0$, $\cos(2\pi) = 1$, and $\cos(0) = 1$: $\left[ t \sin t + \cos t \right]_0^{2\pi} = (2\pi \cdot 0 + 1) - (0 + 1) = 1 - 1 = 0$

(c) Third term: $\int_0^{2\pi} \sin^2 t \, dt$

Use the trigonometric identity $\sin^2 t = \frac{1 - \cos(2t)}{2}$: $\int_0^{2\pi} \sin^2 t \, dt = \int_0^{2\pi} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \int_0^{2\pi} 1 \, dt - \frac{1}{2} \int_0^{2\pi} \cos(2t) \, dt$

• First term: $\frac{1}{2} \int_0^{2\pi} 1 \, dt = \frac{1}{2} \cdot \left[t\right]_0^{2\pi} = \frac{1}{2} (2\pi - 0) = \pi$
• Second term: $\frac{1}{2} \int_0^{2\pi} \cos(2t) \, dt$ The integral of $\cos(2t)$ is $\frac{\sin(2t)}{2}$, so: [ \frac{1}{2} \int_0^{2\pi} \cos(2t) , dt = \frac{1}{2} \cdot \left[\frac{\sin(2t)}{2}\right]_0^{2\pi} = \frac{1}{4} \left[\sin(4\pi) - \sin(0)\right] = \frac{1}{4} (0 - 0) = 0 ] Thus: $\int_0^{2\pi} \sin^2 t \, dt = \pi$

Step 3: Combine results

$\int_0^{2\pi} \left[ 2\cos t + t \cos t + \sin^2 t \right] \, dt = 0 + 0 + \pi = \pi$

Final Answer:

$\boxed{\pi}$

Would you like a detailed explanation of any step?

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Here are some related questions to deepen understanding:

1. How do trigonometric identities simplify integrals involving $\sin^2 t$ or $\cos^2 t$?
2. Why does $\int_0^{2\pi} \cos t \, dt$ evaluate to 0 over a full period?
3. What is the role of symmetry in simplifying periodic function integrals?
4. How does integration by parts work for terms like $t \cos t$?
5. What are other common methods to evaluate $\sin^2 t$ integrals?

Tip: Always check for periodicity and symmetry when dealing with trigonometric integrals over $[0, 2\pi]$ or similar intervals.