If `f(x) = int_(0)^(x)(2cos^(2)3t+3sin^(2)3t)dt`, `f(x+pi)` is equal to :

Options

A. `f(x) + 2f(x)`
B. `f(x) + 2f(pi/2)`
C. `f(x)+4f(pi/4)`
D. `2f(x)`

Answer

To solve the problem, we need to compute \( f(x + \pi) \) where

\[ f(x) = \int_{0}^{x} (2 \cos^2(3t) + 3 \sin^2(3t)) \, dt. \]

### Step 1: Rewrite the integral
We can rewrite the integrand using the identities for \( \cos^2 \) and \( \sin^2 \):

\[ \cos^2(3t) = \frac{1 + \cos(6t)}{2}, \quad \sin^2(3t) = \frac{1 - \cos(6t)}{2}. \]

Substituting these into the integral, we have:

\[ f(x) = \int_{0}^{x} \left( 2 \cdot \frac{1 + \cos(6t)}{2} + 3 \cdot \frac{1 - \cos(6t)}{2} \right) dt. \]

### Step 2: Simplify the integrand
Now simplify the expression:

\[ f(x) = \int_{0}^{x} \left( 1 + \cos(6t) + \frac{3}{2} - \frac{3}{2} \cos(6t) \right) dt. \]

Combining the terms gives:

\[ f(x) = \int_{0}^{x} \left( \frac{5}{2} - \frac{1}{2} \cos(6t) \right) dt. \]

### Step 3: Integrate term by term
Now we can integrate each term:

\[ f(x) = \int_{0}^{x} \frac{5}{2} dt - \int_{0}^{x} \frac{1}{2} \cos(6t) dt. \]

Calculating the first integral:

\[ \int_{0}^{x} \frac{5}{2} dt = \frac{5}{2} x. \]

For the second integral, we have:

\[ \int_{0}^{x} \frac{1}{2} \cos(6t) dt = \frac{1}{2} \cdot \frac{1}{6} \sin(6t) \bigg|_{0}^{x} = \frac{1}{12} \sin(6x). \]

Thus,

\[ f(x) = \frac{5}{2} x - \frac{1}{12} \sin(6x). \]

### Step 4: Compute \( f(x + \pi) \)
Now we need to find \( f(x + \pi) \):

\[ f(x + \pi) = \frac{5}{2} (x + \pi) - \frac{1}{12} \sin(6(x + \pi)). \]

Using the property of sine, \( \sin(6(x + \pi)) = -\sin(6x) \), we have:

\[ f(x + \pi) = \frac{5}{2} (x + \pi) + \frac{1}{12} \sin(6x). \]

### Step 5: Simplify the expression
This simplifies to:

\[ f(x + \pi) = \frac{5}{2} x + \frac{5}{2} \pi + \frac{1}{12} \sin(6x). \]

### Conclusion
Thus, we have:

\[ f(x + \pi) = f(x) + \frac{5}{2} \pi. \]

### Final Answer
The final expression for \( f(x + \pi) \) is:

\[ f(x + \pi) = f(x) + \frac{5}{2} \pi. \]