If `g(x)=int_(0)^(x)cos^(4)t dt,` then `g(x+pi)` equals to (a)`(g(x))/(g(pi))` (b)`g(x)+g(pi)` (c)`g(x)-g(pi)` (d)`g(x).g(pi)`

Options

A. `(g(x))/(g(pi))`
B. `g(x)+g(pi)`
C. `g(x)-g(pi)`
D. `g(x).g(pi)`

Explanation

To solve the problem, we need to find \( g(x + \pi) \) where \( g(x) = \int_{0}^{x} \cos^4 t \, dt \).

### Step-by-step Solution:

1. **Define the Function**:
We have:
\[ g(x) = \int_{0}^{x} \cos^4 t \, dt \]

2. **Substitute \( x + \pi \) into the Function**:
We need to find:
\[ g(x + \pi) = \int_{0}^{x + \pi} \cos^4 t \, dt \]

3. **Split the Integral**:
We can split the integral into two parts:
\[ g(x + \pi) = \int_{0}^{x} \cos^4 t \, dt + \int_{x}^{x + \pi} \cos^4 t \, dt \]
The first part is simply \( g(x) \):
\[ g(x + \pi) = g(x) + \int_{x}^{x + \pi} \cos^4 t \, dt \]

4. **Evaluate the Second Integral**:
The integral \( \int_{x}^{x + \pi} \cos^4 t \, dt \) can be simplified using the periodicity of the cosine function. Since \( \cos^4 t \) is periodic with period \( \pi \), we have:
\[ \int_{x}^{x + \pi} \cos^4 t \, dt = \int_{0}^{\pi} \cos^4 t \, dt \]

5. **Define \( g(\pi) \)**:
We can express this integral as:
\[ g(\pi) = \int_{0}^{\pi} \cos^4 t \, dt \]

6. **Combine the Results**:
Therefore, we can write:
\[ g(x + \pi) = g(x) + g(\pi) \]

7. **Conclusion**:
Thus, the final result is:
\[ g(x + \pi) = g(x) + g(\pi) \]

### Answer:
The correct option is (b) \( g(x) + g(\pi) \).