Evaluate `int_(0)^(npi+t)(|cosx|+|sinx|)dx,` where `n epsilonN` and `t epsilon[0,pi//2]`.

Explanation

To evaluate the integral

\[ I = \int_{0}^{n\pi + t} (|\cos x| + |\sin x|) \, dx \]

where \( n \in \mathbb{N} \) and \( t \in [0, \frac{\pi}{2}] \), we can follow these steps:

### Step 1: Break the Integral into Two Parts
We can split the integral into two parts:

\[ I = \int_{0}^{n\pi} (|\cos x| + |\sin x|) \, dx + \int_{n\pi}^{n\pi + t} (|\cos x| + |\sin x|) \, dx \]

### Step 2: Evaluate the First Integral
The functions \( |\cos x| \) and \( |\sin x| \) are periodic with period \( \pi \). Therefore, we can evaluate the first integral from \( 0 \) to \( n\pi \):

\[ \int_{0}^{n\pi} (|\cos x| + |\sin x|) \, dx = n \int_{0}^{\pi} (|\cos x| + |\sin x|) \, dx \]

### Step 3: Evaluate the Integral from \( 0 \) to \( \pi \)
Now we need to calculate \( \int_{0}^{\pi} (|\cos x| + |\sin x|) \, dx \). Since both \( |\cos x| \) and \( |\sin x| \) are non-negative in the interval \( [0, \pi] \):

\[ \int_{0}^{\pi} (|\cos x| + |\sin x|) \, dx = \int_{0}^{\pi} \cos x \, dx + \int_{0}^{\pi} \sin x \, dx \]

Calculating these integrals:

1. \( \int_{0}^{\pi} \cos x \, dx = [\sin x]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \)
2. \( \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2 \)

Thus,

\[ \int_{0}^{\pi} (|\cos x| + |\sin x|) \, dx = 0 + 2 = 2 \]

### Step 4: Substitute Back
Now substituting back into our expression for \( I \):

\[ I = n \cdot 2 + \int_{n\pi}^{n\pi + t} (|\cos x| + |\sin x|) \, dx = 2n + \int_{n\pi}^{n\pi + t} (|\cos x| + |\sin x|) \, dx \]

### Step 5: Evaluate the Second Integral
Next, we need to evaluate \( \int_{n\pi}^{n\pi + t} (|\cos x| + |\sin x|) \, dx \). In the interval \( [n\pi, n\pi + t] \), where \( t \in [0, \frac{\pi}{2}] \):

- \( \cos x \) will be \( (-1)^n \cos(t) \)
- \( \sin x \) will be \( \sin(t) \)

Thus, we have:

\[ \int_{n\pi}^{n\pi + t} (|\cos x| + |\sin x|) \, dx = \int_{0}^{t} (|\cos(n\pi + x)| + |\sin(n\pi + x)|) \, dx \]

Since \( |\cos(n\pi + x)| = |(-1)^n \cos x| = (-1)^n \cos x \) and \( |\sin(n\pi + x)| = \sin x \):

\[ \int_{0}^{t} (|\cos(n\pi + x)| + |\sin(n\pi + x)|) \, dx = \int_{0}^{t} ((-1)^n \cos x + \sin x) \, dx \]

### Step 6: Calculate the Final Integral
Calculating this integral:

\[ \int_{0}^{t} (-1)^n \cos x \, dx + \int_{0}^{t} \sin x \, dx \]

1. \( \int_{0}^{t} \cos x \, dx = [\sin x]_{0}^{t} = \sin t - 0 = \sin t \)
2. \( \int_{0}^{t} \sin x \, dx = [-\cos x]_{0}^{t} = -\cos t + 1 = 1 - \cos t \)

Thus, we have:

\[ \int_{0}^{t} (|\cos(n\pi + x)| + |\sin(n\pi + x)|) \, dx = (-1)^n \sin t + (1 - \cos t) \]

### Step 7: Combine Everything
Finally, combining everything:

\[ I = 2n + (-1)^n \sin t + (1 - \cos t) \]

### Final Answer
Thus, the value of the integral is:

\[ I = 2n + (-1)^n \sin t + 1 - \cos t \]